Discover the answer to this question in this chapter. Site historique maritime de la Pointe-au-Père

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1 In 2008, the submarine Onondaga was hauled on land beside Pointe-au-Père s wharf, not far from Rimouski. To achieve this, the 1400-ton submarine was towed up a 4 slope with the help of pulles. In all, the steel cable pulled 18 times on the pulle attached to the submarine. The submarine was on small carts rolling on rails, thereb limiting the friction between the ground and the submarine so that it will be neglected here. With what force did the have to pull on the cable in order to move the submarine at a constant speed? Site historique maritime de la Pointe-au-Père Discover the answer to this question in this chapter.

2 Universal Attraction In ever model of the universe made in the last 2500 ears, there are planets revolving around a central object. In the geocentric sstem, the most popular before the discover of Newton s laws, planets revolve around the Earth whereas the revolve around the Sun in the Copernican sstem (published in 1543). Whatever the model used, it was necessar to explain how the planets could revolve around the central object. In the models associating force and velocit, there had to be a force acting on the planets in the direction of their motion. It was even imagined at some point that angels exerted this force! Force according to the theories associating force and velocit (false) Starting from his idea that an object that is not moving awa for Earth nor approaching Earth would continue its motion at a constant speed around the Earth, Galileo came to the conclusion that no force was needed to explain planetar motions. In the absence of force, the planets would simpl continue their circular motion at constant speed. Galileo s ideas were corrected later when it was proposed that the motion would be in a straight line at constant speed if there is no force. Thus, on a circular orbit, there should be a force acting of the planets. For a circular motion, there must be an acceleration towards the centre of the circle. Since the net force is alwas in the same direction as the acceleration according to Newton s second law, there must a force directed towards the centre of the orbit acting on the planets Version 4 - Forces, part 1 2

3 Force according to theories associating force and acceleration Clearl, the force ou re looking for depends on the theor of force ou believe in. You are searching for a different force depending on what ou associate the force with. When force and velocit are associated, a tangential force is sought, while a centripetal force is sought if force and acceleration are associated. As this centripetal force is pointing directl towards the central object, it is ver tempting to sa that the central object attracts the planet. Hence, some scientists before Newton concluded that the Sun must attract the planets. Some of these scientists even got further b saing that ever object must exert an attractive force on ever other object in the universe (not because the had made the connection between force and acceleration, but merel because the were tring to explain the origin of gravitation). How Did Newton Discover the Law of Gravitation? Newton, however, was the first to find the magnitude of this centripetal force. It makes sense that the others did not succeed because the were working with bad definitions of force based on an association between force and velocit. B properl defining force, Newton had a much better chance of getting it right. Two formulas set the stage for Newton s discover of the law of gravitation: the centripetal acceleration formula law (published in 1673) and Kepler s third law (published in 1619). Kepler had obtained this law b comparing the time taken b a planet to orbit the Sun (the period T) and the distance between the planet and the Sun (the radius of the trajector r). He then got that r³ T². With F = ma, Newton was then able to find that the force on a planet in orbit around the Sun is 2018 Version 4 - Forces, part 1 3

4 F F F F g g g g = m a 2 2 m2 4π r = 2 T 2 m2 4π r 3 r m2 2 r (The centripetal acceleration was used) (Kepler's law was used) The force is, therefore, proportional to the mass of the object on which it acts (m ). However, according to Newton s third law, if the planet having a mass m exerts a force on the planet having a mass m, then the planet having a mass m exerts the same force on the planet having a mass m. As the force is proportional to the mass of the planet it acts on, and as the force is acting on the two masses, then the force must be proportional to both masses. aspx F g m m r The Law of Gravitation Thus, all objects attract ever other object. Finall, the law of attraction can be written using a constant of proportionalit to obtain Law of Gravitation (General Formula) 1) Magnitude of the force m m 1 2 Fg = G r 2 2) Direction of the force Attraction of the two masses one towards the other. 3) Application point of the force See the chapter on gravitation. For planets: from the centre of the planet. (Note that the distance r between the planets is the distance between the centres of the planets.) Measurements made b Cavendish in 1789 and improved thereafter lead to the determination of the value of the constant G. The value is 2018 Version 4 - Forces, part 1 4

5 Gravitational Constant G = x N m²/kg² Small historical note: Robert Hooke was not ver happ to see that Newton gave him no credit for the discover of the law of gravitation. It was indeed him who set Newton on the track of this discover in 1679 when he asked him how the planets would behave if the were under the influence of a force directed towards the Sun and decreasing with 1/r². So it seems that Hooke knew that the force decreases with the square of the distance before Newton! It is true that Hooke had arrived at this conclusion, but the proof was completel wrong. It was based on a force proportional to v² and not to the acceleration. Rather than giving credit to Hooke, Newton (who hated Hooke and considered that he had not reall used the information provided b Hooke) invented a falling apple stor to tell how he had his first thought on gravitation. Another small historical note: Do not think that Newton s law of gravitation was an immediate and universal success. Newton did not explain how a planet could exert a force on another object while there is no contact between the two objects. In France, where Descartes s sstem occupied the central place, it was almost impossible to accept such a force acting at a distance without contact. For a Cartesian, the onl acceptable forces were the result of contact between bodies. How could the Sun attract remote planets? Mabe there was a hidden mechanism to explain this attraction. Newton did not propose an (and never managed to find one). It took fort ears for Newton s law of gravitation to be accepted in France. The successes achieved in England with Newton s law of gravitation showed that it worked fine, even if nobod could explain how objects attract each other. Common Mistake: Thinking that when Two Objects Attract Each Other with a Gravitational Force, the Less Massive Object Exerts a Smaller Force on the Other Object. Intuitivel, people suspect that the Earth exerts a force on the Moon and that the Moon also exerts a force on Earth. However, man would sa that the Earth exerts a much greater force on the Moon than the Moon exerts on the Earth because it is more massive. Admire the answers of some people in this video. Of course, all this is false; the force must have the same magnitude according to Newton s third law Version 4 - Forces, part 1 5

6 The force of Gravit near the Surface of the Earth (and Other Planets) A formula simpler than the formula given above can be used for objects near the surface of the Earth (or an planet). The force between the Earth (mass M e ) and an object (mass m) near the surface of the Earth is, according to the law of gravitation, F g M m = G R e 2 e For the distance, the radius of the Earth (6380 km) is used. Even if the object is 1 km above the surface of the Earth, the value of distance does not change much. A ver good approximation is thus obtained for all objects near the surface of the Earth. Obviousl, the approximation is not as good if the object is 1000 km from the surface of the Earth. With the values of the Earth s mass and radius, the force is F g M m = G R e 2 e M e = m G R 2 e = m N = m 9.8 kg kg Nm² kg ² 2 6 ( m) 9.8 N/kg is the same as 9.8 m/s². This is the value of g used in the previous chapters. The gravitational force acting on a bod is called the weight, which should not be confused with mass. The mass is in kilograms and represents the amount of material in a bod while the weight is in newton and represents the gravitational force acting on the object Version 4 - Forces, part 1 6

7 Weight (w) or Gravitational Force (F g ) (Formula Valid onl near the Surface of the Earth) 1) Magnitude of the force w = mg = m 9.8 N kg 2) Direction of the force Downwards (towards the centre of the Earth) 3) Application point of the force From the centre of mass of the object. (See the chapter on the centre of mass. For now, take a point roughl at the centre of the object.) It is no accident to find g in this equation. Let s look at what happens to a free-falling object. Newton s second law gives F = ma mg = ma a = g Galileo s conclusion is found again: all bodies fall with the same acceleration, regardless of their mass (since the masses cancelled each other). This is how Newton knew that the force had to be proportional to the mass when he made his second law. It was alread quite clear that the force of gravit increases with the mass of the object. For all free-falling object to have the same acceleration, the masses had to cancel, and so there must be an m in Newton s second law. Therefore, here s wh all free-falling objects have the same acceleration: a greater force on a heavier object is exactl offset b the fact that the heavier object is harder to accelerate. This gives the same acceleration to all objects. The formula mg is onl an approximation. In realit, the gravitational force decreases ver slowl with altitude since it decreases with the square of the distance from the centre of the edublognss.wordpress.com/201 3/04/16/gravitation/ Earth according to the general formula. To give ou an idea of the precision of the approximation, g is 9.5 N/kg at an altitude of 100 km. Newton was the first to discover that the force of gravit decreases with altitude Version 4 - Forces, part 1 7

8 The same calculation can be done to find the gravitational force on the surface of an planet. For example, the weight on the Moon is given b = M Moon Fg G R 2 Moon = m = m 1.6 N kg m M Moon = m G R 2 Moon kg Nm² kg ² 2 6 ( m) This shows that the weight near the surface of the Moon is smaller than the weight near the surface of the Earth because g is onl 1.6 N/kg at the surface of the Moon. On the Moon, the mass of the bod is the same as on Earth, but the weight is approximatel 6 times smaller. You can see what happens with a smaller gravit on the Moon in this video. Therefore, it can be seen that g can take different values at different places in the universe. What Is the Normal Force? When two objects touch each other, there is a force between the two objects. Generall, the forces can be represented as shown in the figure. There are two forces according to Newton s third law. If the top object exerts a force on the bottom object, then the bottom object exerts a force of the same magnitude and opposite direction on the top object. Now, consider onl the force exerted on the top object. This force can be resolved into two components: the component perpendicular to the contact surface and the component parallel to the contact surface Version 4 - Forces, part 1 8

9 The parallel component will be studied in the next chapter (it is the friction force). For now, the focus will be on the perpendicular component. Since normal is snonmous with perpendicular, this component of the force is called the normal force. It is denoted F. (Sometimes N is used, but it ma be confused with the smbol for the newton). This component represents a repulsion force between objects in contact. To understand wh objects repulse each other, imagine ou are standing motionless on a mattress. Your presence then crushes the springs of the mattress. If ou exert a force on the springs of the mattress directed downwards to compress them, then the springs of the mattress exert a force on ou directed upwards according to Newton s third law. It is this force that cancels the force of gravit acting on ou. This force must exactl cancel the force of gravit. If the force exerted b the springs is smaller than the gravitational force, there is a net force on ou directed downwards. You then accelerate downwards and further compress the springs, thereb increasing the force the exert on ou. If the force exerted b the springs is greater than the gravitational force, there is a net force on ou directed upwards. You then accelerate upwards and decompress the springs thereb decreasing the force the exert on ou. Both effects thus lead to a state where the force exerted b the springs on ou exactl cancels our weight. This ma be surprising, but this is exactl what happens if ou push on a surface with our finger. When ou push on the surface, there is a slight deformation, even if the object is ver rigid. Since objects have elasticit, the act like a spring: b compressing the object, the forces between the atoms exert a force opposed to the deformation. In fact, the forces between the atoms can be represented b springs, as in this figure. When ou push on the surface, the atoms move closer together, which compress all the springs between the atoms. If our finger makes a downwards force on the atoms to 2018 Version 4 - Forces, part 1 9

10 compress the springs, then, b Newton s third law, the atoms exert an upward force on our finger. This latter force is the normal force. As the contact between objects can onl compress the bodies in contact, the normal force can onl be a repulsion between two objects in contact. It can never be an attraction. So, if ou are standing still on the ground. Your feet exert a force on the ground, which compresses it a bit (much less than a mattress, but it compresses). If our feet exert a force on the ground downwards, then the ground exerts an equall large upwards force on our feet according to Newton s third law. This force, the normal force, just cancel the force of gravit acting on ou so that ou can sta still. This video gives ou the same explanation. Here s a summar of the characteristics of the normal force. Normal Force (N or F ) 1) Magnitude of the force To be determined with Newton s laws 2) Direction of the force Repulsion between objects, perpendicular to the contact surface 3) Application point of the force Contact surface between objects. The force is alwas perpendicular to the surface of contact between the objects, as for these three balls. clickercentral.net/librar/item_0044 The force on the ball is alwas a repulsive force (the ground pushes on the ball), and this force is alwas perpendicular to the contact surface (the ground in this case) Version 4 - Forces, part 1 10

11 Common mistake: Thinking that the Weight and the Normal Force Are Related b Newton s Third Law This would mean that the normal force would alwas have the same magnitude as the weight and be directed in the opposite direction. It can sometimes be true (as it is the case for an object stationar on a horizontal ground), but it is false most of the time. The example of an object on an inclined plane without friction illustrates this. The forces on the object are the weight (downwards) and the normal force (perpendicular to the contact surface). It can be seen that these two forces are not in opposite direction to each other. Therefore, the cannot be associated b Newton s third law. cnx.org/content/m14087/latest/ If the normal force is not associated with the weight b Newton s third law, what force is associated with the weight? Just appl the trick mentioned earlier to find it: the force associated is obtained b inverting the two objects in the following sentence: to obtain The Earth exerts a gravitational force on the box. The box exerts a gravitational force on the Earth. The force associated is, therefore, the gravitational attraction exerted b the box on the Earth. Calculation of the Normal Force There s no formula to calculate the magnitude of the normal force directl. Newton s second law must be used to find it. Here are some examples. Example A 10 kg box is resting on the ground. What is the magnitude of the normal force exerted b the ground on the box? Two forces are acting on the box. 1) The weight, directed downwards, whose magnitude is 10 kg x 9.8 N/kg = 98 N Version 4 - Forces, part 1 11

12 2) A normal force. This is a force of repulsion exerted b the ground and is therefore directed upwards. Since the box is not accelerating, Newton s second law gives F = ma 98N + F = 0 F N N = 98N hsics/chapter6section4.rhtml Example A 10 kg box is resting on the ground. A 40 N force is applied downwards on the box. What is the magnitude of normal force exerted b the ground on the box? Three forces are acting on the box. 1) The weight (98 N), directed downwards 2) A normal force, directed upwards 3) A 40 N force, directed downwards Since the box is not accelerating, Newton s second law gives 98N + F 40N = 0 F N F N = ma = 138N s/chapter6section4.rhtml Example A 10 kg box is resting on the ground. A 40 N force is applied upwards on the box. What is the magnitude of normal force exerted b the ground on the box? Three forces are acting on the box. 1) The weight (98 N), directed downwards 2) A normal force, directed upwards 3) A 40 N force, directed downwards 2/phsics/chapter6section4.rhtml 2018 Version 4 - Forces, part 1 12

13 Since the box is not accelerating, Newton s second law gives 98N + F + 40N = 0 F N F N = ma = 58N In these last two examples, the normal force did not have the same magnitude as the weight. This shows once again that the weight and the normal force cannot be related b Newton s third law. Example A 10 kg box is resting on the ground. A 100 N force is applied upwards on the box. What is the magnitude of the normal force exerted b the ground on the box? Three forces are acting on the box. 1) The weight (98 N), directed downwards 2) A normal force, directed upwards 3) A 100 N force, directed upwards Since the box is not accelerating, Newton s second law gives 98N + F + 100N = 0 F N F N = ma = 2N s/chapter6section4.rhtml This answer is impossible; the normal force cannot have a negative value. If that were the case, it would mean that the normal force would be in the direction opposite to the direction indicated in the figure. However, since the normal force is a repulsive force between the bodies, its onl possible direction was alread indicated correctl in the figure. If it is negative (so in the opposite direction), the normal force then becomes an attraction between bodies, and this can never happen with normal forces. This situation is, therefore, impossible. Of course, a 100 N force can act upwards on a 10 kg box. It is, however, impossible to sa that the box remains at rest. If ou lift a box weighing 98 N with a force of 100 N, there is a resultant force of 2 N upwards with no other force to cancel it. The box then accelerates upwards in this situation. It is no longer in contact with the surface, and the normal force is zero Version 4 - Forces, part 1 13

14 There is a normal force exerted at each contact with other objects Normal forces are contact forces. At ever spot where the object touches another object, there is a normal force exerted. This normal is alwas a repulsive force and is alwas perpendicular to the contact surface. The example of this ball illustrates this. This ball touches two other objects, which means that two normal forces are acting on this ball. As normal forces must alwas be perpendicular to the contact surface, the normal forces are in the following directions. Example Draw all the forces acting on the two boxes in the figure. (You do not have to find the magnitudes of the forces.) Version 4 - Forces, part 1 14

15 Let s start with the top box. As the box has mass, it has a weight directed downwards. As it touches the box below it, there is a normal force directed upwards. This force is upwards since the top box and the bottom box repel each other. The forces are therefore: 1) The weight (w ), directed downwards. 2) The normal force exerted b the bottom box (N ), directed upwards. Let s follow with the bottom box. There is, of course, a weight directed downwards. As the box touches two things (the top box and the table) there are 2 normal forces. Since these are repulsive forces, the table exerts a normal force directed upwards, and the top box exerts a downwards force. The forces are therefore: 1) The weight (w ), directed downwards. 2) The normal force exerted b the ground (N ), directed upwards. 3) The normal force exerted b the top box (N ), directed downwards. Note that the normal force exerted b the top box is also called N. It was unnecessar to use a new smbol such as N since this force has the same magnitude as the normal force made on the top box. This is true because of Newton s third law: if the bottom box exerts the normal force N directed upwards on the top box, then the top box exerts a normal force N directed downwards on the bottom box. Example Draw all the forces acting on the three boxes in the figure. (You don t have to find the magnitudes of the forces.) 2018 Version 4 - Forces, part 1 15

16 Let s start with the top box. There is a gravitational force directed downwards. As this box touches two other objects (the left and right boxes), there are also two normal forces acting on the top box. These forces are directed upwards since the boxes on the left and the right repel the top box. The forces on the top box are then: 1) The weight (w ), directed downwards. 2) The normal force exerted b the box on the left (N ), directed upwards. 3) The normal force exerted b the box on the right (N ), directed upwards. Now let s examine the boxes supporting the top box. Each of these boxes has a weight directed downwards. As these boxes also touch two things (the floor and the top box), two normal forces are acting on each of these boxes. The forces on the box on the left are: 1) The weight (w ), directed downwards. 2) The normal force exerted b the ground (N ), directed upwards. 3) The normal force exerted b the top box (N ), directed downwards. The forces on the box on the right are: 1) The weight (w ), directed downwards. 2) The normal force exerted b the ground (N ), directed upwards. 3) The normal force exerted b the top box (N ), directed downwards Version 4 - Forces, part 1 16

17 The tension force is the force exerted b a rope or a string. It can be stated from the outset that a string can onl pull an object in the direction of the string; it cannot push or appl force in a direction perpendicular to the string because the string simpl bends if one tries to appl such forces. Tension Force (T or F T ) 1) Magnitude of the force Given or to be determined with Newton s laws. 2) Direction of the force The rope pulls in the direction of the rope. 3) Application point of the force Where the rope is fixed to the object. Calculation of the Tension Force There s no formula to calculate the magnitude of the tension force directl. Newton s second law must be used to find it. Here are some examples. Example What is the tension force exerted b the rope that supports this 5 kg box at rest? Version 4 - Forces, part 1 17

18 The forces on the box are: 1) The weight, directed downwards, whose magnitude is 5 x 9.8 N/kg = 49 N. 2) The tension force of the rope directed upwards (since a rope must pull). There is no normal force because the box does not touch anthing (excepting the rope). Using the axes shown in the figure, Newton s second law is The solution of this equation is T = 49 N. F = 49N + T = ma The sum of the x-component of the forces was not done because there is no force having an x-component. Since the box does not accelerate, the equation becomes F = 49N + T = 0 Example What is the tension force exerted b the rope that supports this box if it accelerates downwards at 2 m/s²? The forces on the box are: 1) The weight, directed downwards, whose magnitude is 5 x 9.8 N/kg = 49 N. 2) The tension force of the rope directed upwards. Using the axes shown in the figure, Newton s second law is F = 49N + T = ma 2018 Version 4 - Forces, part 1 18 Using the values given, this equation can solve for the tension force. 49N + T = ma m ( ) 49N + T = 5kg 2 T = 39N s²

19 The Forces Exerted b a String Is the Same at Each End (If the Mass of the Rope Is Neglected) It will now be shown that the force exerted b the string on the block in the figure is the same as the force exerted at the other end of the rope. In other words, we want to know if some force is lost in the string. Consider first the forces on the string. As the string pulls on the block towards the right, the block must pull on the string towards the left according to Newton s third law. The equation of the forces acting on the string is F1 F2 = mstringa If the mass of the string is neglected, b assuming m string 0 (this will alwas be assumed unless stated otherwise), the equation becomes F F = F = F 1 2 The two forces at each end of the string must then have the same magnitude. This pair of force stretches the string and puts it under tension. This is wh this force is called the tension force or the tension of the string and is denoted T or F T. Since the force exerted b the box on the string directed towards the left has a magnitude T, the force exerted b the rope on the box is directed towards the right and also has a magnitude T. This is the same force as the one exerted at the other end of the string. This shows that no force is lost in the string. (This is no longer true if the mass of the string is not neglected. In this case, a part of the force is used to accelerate the string, and some of the force is lost.) 2018 Version 4 - Forces, part 1 19

20 This also means that the force exerted b a string has the same magnitude at each end of the string. Thus, if a string connects two objects, the string pulls on each object with forces of equal magnitude. Example What are the tensions of each of the two strings in the situation shown in the figure? When there are several objects, the problem can be solved b finding the equations of forces for each object separatel. We ll start here with the bottom block. The forces on this block are: 1) A 294 N weight, directed downwards. 2) A tension force, directed upwards. The sum of the -component of the forces is F = 294N + T = ma Since the acceleration is zero, the tension is 1 294N + T = T = 294N 2018 Version 4 - Forces, part 1 20

21 Let s now look at the upper block. The forces are: 1) A 98 N weight, directed downwards. 2) A tension force, directed downwards. 3) A tension force, directed upwards. The sum of the -component of the forces is F = 98N T + T = ma 1 2 Since the acceleration is zero, this becomes 98N T + T = As the string that connects the two blocks should exert the same force of tension at each end of the string, then = 294 N. The tension force is then 98N 294N + T = 0 T 2 = 392N 2 Therefore, the tension of the string 1 is 294 N, and the tension of the string 2 is 392 N. This means that the string 1 pulls with a force of 294 N at each of its ends and that the string 2 pulls with a force of 392 N at each of its ends. Note that the force made at each end of a string remains of the same magnitude even if the string passes through pulles (if the mass of the pulle is neglected). scripts.mit.edu/~sraan/perwiki/index.php?title=module_5_-- _Objects_Constrained_to_Move_with_Different_Accelerations The Force of Tension Cannot Be Negative When the force of tension is calculated using Newton s laws, the answer cannot be negative since the direction of the force (a string alwas pulls) was alread taken into account. A negative answer means that the force is in the opposite direction than it had been assumed. This would mean that the rope pushes, which is impossible Version 4 - Forces, part 1 21

22 Example What is the tension force exerted b the string that supports the box in the figure if it accelerates downwards at 20 m/s²? The forces acting on the box are: 1) The weight, directed downwards with a magnitude of 5 x 9.8 N/kg = 49 N 2) The tension force of the string, directed upwards With the axes shown in the figure, Newton s second equation gives Solving for T, the result is F = 49N + T = ma 49N + T = ma m ( ) 49N + T = 5kg 20 T = 51N s² This value is not possible. This situation is, in fact, impossible since the box cannot have an acceleration greater than 9.8 m/s² downwards if the weight is the onl force directed downwards. Following a problem-solving method in dnamics is important. This method is 2018 Version 4 - Forces, part 1 22

23 1) Find all the forces acting on the object that is studied. For now, the onl known forces are the weight, the normal force and the tension force. a) Weight There is a force of gravit on all objects unless it is explicitl stated to neglect the mass. b) Normal force To find out if there is a normal force, ask ourself if the object touches another object (except a string). If so, there is a repulsive force acting on the object at ever contact with another object. c) Tension force All strings exert a tension force. 2) Resolve these forces into their x and -components. This implies that axes must be chosen. To greatl simplif the solution, one of the axes should be parallel to the acceleration (this can be in the direction of the acceleration, which will give a positive acceleration, or in the opposite direction to the acceleration, which will give a negative acceleration). If the object does not accelerate, the axes can be directed in an direction, provided that the x-axis is alwas perpendicular to the -axis. 3) Appl Newton s second law F = ma F = ma x x 4) Solve these equations to find the unknowns. Examples with Onl One Object Example What are the tension forces exerted b the two strings supporting this mass? Version 4 - Forces, part 1 23

24 The forces on the mass are 1) The weight, directed downwards, whose magnitude is 30 kg x 9.8 N/kg = 294 N. 2) The tension force exerted b string 1 (T 1 ). 3) The tension force exerted b string 2 (T 2 ). The direction of those forces can be seen in the following figure. To sum the forces, a table like this one can be used. Forces x Weight N String 1 - T 1 0 String 2 T 2 cos 45 T 2 sin 45 In this table, all the components of the forces are found according to the rules given earlier. To sum the x and -components of the forces, just add the components in each column. (Here s an important note: b using the angle between the force and the x-axis, ou will have onl cosines for x-component and onl sines for the -component. Moreover, ou can never have different angles for x and for the same force. For example, the angle is 45 for both components of T 2 here.) The sum of the x-component of the force is (sum of the x column in the table) F = 0 T + T cos 45 = ma x 1 2 The sum of the -component of the force is (sum of the column in the table) F = 294N + T sin 45 = ma 2 x 2018 Version 4 - Forces, part 1 24

25 Since the acceleration is zero, these equations become T + T cos 45 = N + T sin 45 = 0 2 The second equation can then be solved for T N + T sin 45 = 0 T 2 2 = N This value can be substituted into the x-component equation, and the resulting equation can be solved for T 1. 1 T + T cos 45 = T N cos 45 = 0 T = 294N 1 The tension forces of the string are then 294 N (string 1) and N (string 2). Example Even though Lucien pulls with a 120 N force on Adele s sled, the sled does not move because a friction force opposing the motion is acting. ww.chegg.com/homework-help/questions-and-answers/phsics-archive-2013-october-15 Calculate the magnitude of the normal force and of the friction force acting on the sled. The forces acting on the sled are: 1) The weight (w), directed downwards, whose magnitude is 40 kg x 9.8 N/kg = 392 N. 2) A normal force (F N ), directed upwards. 3) A tension force made b string 1 (T). 4) A friction force (F f ) opposed to the displacement of the sled Version 4 - Forces, part 1 25

26 The direction of these forces can be seen on this figure. It is quite eas to find the components of the friction and normal forces, but it is a little more complicated for the 120 N force. To resolve this force into components, the angle between the force and the positive x-axis must be found. This angle is 150. The table of forces is then Forces x Weight N Normal force 0 F N Tension force 120 N cos 150 = N 120 N sin 150 = 60 N Friction force F f 0 The sum of the x-component of the force is F = N + F = ma The sum of the -component of the force is x f x F = 392N + F + 60N = ma N Since the acceleration is zero, these equations become N + F = 0 392N + F + 60N = 0 This first equation gives us the magnitude of friction force Version 4 - Forces, part 1 26 N N + F = 0 F = N f f f

27 The second equation gives us the magnitude of the normal force. 392N + F + 60N = 0 F N N = 332N Example What are the tension forces exerted b the strings which support this traffic light? equal-814-i-think-814-sin-q The forces acting on the traffic light are: 1) The weight (w), directed downwards, whose magnitude is 20 kg x 9.8 N/kg = 196 N. 2) The tension force of string 1 (T 1 ). 3) The tension force of string 2 (T 2 ). It is important to find the correct angles for the tension forces. For T 1, the angle between the force and the positive x-axis is Version 4 - Forces, part 1 27

28 For T 2, the angle is 45. The table of force is then Forces x Weight N String 1 T 1 cos 150 T 1 sin 150 String 2 T 2 cos 45 T 2 sin 45 The sum of the x-component of the force is F = 0 + T cos150 + T cos 45 = ma x 1 2 The sum of the -component of the force is F = 196N + T sin150 + T sin 45 = ma 1 2 Since the traffic light is not accelerating, the components of the acceleration are zero. Thus, the two equations are T cos150 + T cos 45 = N + T sin150 + T sin 45 = x With two equations and two unknowns, this sstem of equations can be solved. Take the method of our choice (substitution, Gauss-Jordan, Cramer or an other, remember our linear algebra course ) to solve. Here, the substitution method will be used. Thus, the first equation is solved for T 2 T 2 T1 cos150 = cos 45 and the result is substituted in the second equation. Then, T 1 can be found. T cos150 + T cos 45 = T cos = cos N T1 sin150 sin Version 4 - Forces, part 1 28

29 Using this value, T 2 can be found. 1 ( ) 196N + T sin150 cos150 tan 45 = 0 196N T1 = sin150 cos150 tan 45 T = 143.5N T 1 T cos150 cos = = 175.7N Once the answers are obtained, it is usuall a good idea to substitute these values in the original two equations (sum of the x and -components of the forces) to check if the work These examples ma seem long, but the were done in great detail. Later, we will go a little faster. If ou alread feel comfortable to skip some of the steps, such as the table of force, go ahead. Here s a slight variation of the last example. Suppose that the traffic light is now suspended in this wa i-think-814-sin-q (There s an additional string in this version, linking the traffic light to the node that connects the strings.) In this variant, the tension force of the third string can be found with the sum of the forces on the traffic light. It is then quite easil found that the tension force T 3 must be 196 N Version 4 - Forces, part 1 29

30 Then the forces acting on the node connecting the three strings are considered. The following forces act on the node (figure). These forces are the same forces as the forces acting on the traffic light in the preceding example. The tension forces T 1 and T 2 are therefore the same as the were in the example. All that to sa that if there is a node connecting several strings, the solution of the problem is often found b summing the forces acting on the node. Example A skier of mass m is on a slope inclined at an angle α. What is the acceleration of the skier and what is the normal force exerted on the skier if there is no friction? vhcc2.vhcc.edu/ph1fall9/frames_pages/openstax_problems.htm Even if there are no numerical values in this example, the solution is still done b using the problem-solving method given earlier. The forces are on the skier are: 1) The weight mg, directed downwards. 2) The normal force exerted b the slope. These forces are shown in the figure Version 4 - Forces, part 1 30

31 The x-axis is tilted in the downhill direction because it is suspected that this is the direction of the acceleration of the skier. B using an axis in the direction of the acceleration, the solution is much simpler to obtain. The normal is then directed exactl in the direction of the -axis, and it is eas to find its components. Alas, the situation is more complicated for the weight. On the figure on the left, the directions of the x-axis and the weight are shown. The angle between the x-axis and the force of gravit is -(90 -α) (Remember the sign convention. Here, the positive direction is counter clockwise since it is in the rotation direction going from the positive x-axis towards the positive -axis. When rotating from the positive x-axis towards the force, the rotation is clockwise. As this is in the opposite direction of the positive direction here, the angle is negative). The components of the forces are then Forces x Weight mg cos(-(90 -α)) mg sin(-(90 -α)) Normal force 0 F N The equations are then ( ( α )) ( α ) F = mg cos 90 = ma x ( ) F = mg sin 90 + F = 0 The -component of the acceleration is zero and all the acceleration is the direction of the x-axis (we know this because the axes were chosen in this wa). The acceleration can then be found with the first equation. mg cos 90 ( ( α )) = cos ( 90 α ) g cos( 90 α ) ( ) a = g a = a = g sinα This is the desired result. Some trigonometric identities were used to simplif it. These identities will be looked at more closel after this example. Finall, the normal force can be found. N ma ( ( α )) + FN = mg sin ( 90 α ) = mg sin ( 90 α ) mg sin 90 0 F N F N F N ( ) = = mg cosα 2018 Version 4 - Forces, part 1 31

32 An interesting note: the acceleration of an object on a frictionless slope is independent of the mass. In a toboggan race on a frictionless track, all the racers arrive at the same time at the bottom of the slope, regardless of their mass if the start simultaneousl. This is not reall shocking because the gravitational acceleration is the same for everbod. Reminder of some trigonometric identities The following formulas are sometimes useful to simplif the solution of a problem without numerical data. cos x = cos x cos(90 x) = sin x cos(180 x) = cos x sin x = sin x sin(90 x) = cos x sin(180 x) = sin x cos(90 + x) = sin x cos(180 + x) = cos x sin(90 + x) = cos x sin(180 + x) = sin x Example A mass is hung with a string from the roof of the box of a truck. What is the angle between the pendulum and the vertical when the truck is accelerating at 5 m/s²? The forces on the mass at the end of the string are: 1) The weight (mg), directed downwards. 2) The force of tension, pulling in the direction of the string. According to the figure, the angle of the force of tension is 90 + θ. The table of force is then Forces x Weight 0 - mg Force of tension T cos(90 +θ) T sin(90 +θ) 2018 Version 4 - Forces, part 1 32

33 The equations are then ( θ ) ( θ ) F = T cos 90 + = ma x F = mg + T sin 90 + = 0 There is an acceleration for the x-component since the mass has to follow the motion of the truck. If the truck accelerates at 5 m/s² towards the left, then the mass also accelerates at 5 m/s² towards the left. Using trigonometric identities, the equation becomes Solving for T in the second equation F = T sinθ = ma x F = mg + T cosθ = 0 T = mg cosθ and substituting in the first equation, the angle can be found. mg sinθ = ma cosθ a tanθ = g m ( 5 ) θ = arctan 9.8 θ = (As the acceleration is towards the left and the positive x-axis is towards the right, the acceleration is negative.) s ² m s² Examples with Several Objects When there are several objects in the sstem, two options are available: A) Treat all the objects as a single object. B) Treat all the objects separatel. In the following examples, these two was of considering the sstem will be emploed according to what is asked Version 4 - Forces, part 1 33

34 Example The boxes in the figure are pushed with a force F. The boxes then accelerate at 2 m/s² towards the right (and there is no friction between the ground and the boxes). whs.wsd.wednet.edu/facult/busse/mathhomepage/busseclasses/apphsics/studguides/chapter4/apphsicsch4part6.html a) What is the magnitude of the force F? To calculate this, the two boxes can be considered as a single 4 kg object. The forces on this 4 kg object are: 1) The weight (39.2 N), directed downwards. 2) The normal force, directed upwards. 3) The force F, directed towards the right. The equations are F = F = ma x F = 39.2N + F = 0 The second equation could allow us to find the normal force made b the ground on the boxes, but this information is not needed. The force F can be found with the first equation. F = ma = 4kg 2 = 8N m s² N 2018 Version 4 - Forces, part 1 34

35 b) What is the normal force between the two boxes? When asked to find a force between two objects, ou must imperativel separate the sstem and consider the forces acting on each box. Considering the 3 kg box first, the forces are: 1) The weight w 1 = 29.4 N, directed downwards. 2) A normal force made b the ground (N 1 ), directed upwards. 3) A normal force made b the 1 kg box (N 2 ), directed towards the left. 4) The force F, directed towards the right. The equations are F = F N = m a x 2 1 F = 29.4N + N = 0 N 2 can then be calculated with the first equation. F N = m a 2 1 8N N = 3kg 2 N 1 m 2 s² 2 = 2N The 1 kg box could also have been considered to calculate this normal force. The forces on the 1 kg box are: 1) The weight w 2 = 9.8 N, directed downwards. 2) A normal force made b the ground (N 3 ), directed upwards. 3) A normal force made b the 3 kg box (N 2 ), directed to the right. The same smbol was used for the normal force made b the 3 kg box on the 1 kg box and for the normal force made b the 1 kg box on the 3 kg box since these two forces are related b Newton s third law and must have the same magnitude. The force F does not act on this box. If the person pushes on the 3 kg box, he is not touching the 1 kg box. The equations are therefore N 2 is found with the first equation. F = N = m a x 2 2 F = 9.8N + N = 0 N = m a 2 2 = 1kg 2 = 2N 2018 Version 4 - Forces, part 1 35 m s² 3

36 Example Two boxes are pulled with a string as illustrated in the figure. (There is no friction between the ground and the boxes.) whs.wsd.wednet.edu/facult/busse/mathhomepage/busseclasses/apphsics/studguides/chapter4/apphsicsch4part6.html a) What is the acceleration of the boxes? To find the acceleration of the boxes, we will consider that a 40 kg mass is pulled with a 50 N force. The forces on this 40 kg object are: 1) The weight (392 N), directed downwards. 2) A normal force made b the ground, directed upwards. 3) The force of tension of 50 N directed towards the right. The acceleration is found with the sum of the x-component of the force. F = 50N = ma x 50N = 40kg a a x = 1.25 m s² x x b) What is the tension of the string connecting the two boxes? To find the tension force T 2, we have to look at the x-component of the forces acting on one of the boxes. If the 30 kg box is considered, we have T F = T = m a x 2 2 = 30kg 1.25 m 2 s² T 2 = 37.5N x Note that the same result would have been obtained b considering the other box. F = T T = m a x N T = 10kg 1.25 T m 2 s² 2 = 37.5N x 2018 Version 4 - Forces, part 1 36

37 Example A train, initiall at rest, is composed of a 100-ton locomotive and of ton wagons. To move this train, a 1,000,000 N force is exerted b the locomotive. Then, what is the tension in the coupler between the 10 th and 11 th wagon? Let s start b finding the acceleration of the train b considering the whole train as a single object whose mass is 100 tons + 70 tons x 10 = 800 tons. The forces on the train are: 1) The weight, directed downwards. 2) A normal force exerted b the ground, directed upwards. 3) The force of 10 6 N made b the locomotive, directed towards the right. The sum of the x-component of the force is 6 Fx = 10 N = ma With this equation, the acceleration can be found N = ma ( ) 6 10 N = 800 1,000 a = 1.25 m s ² kg a Then, the train is separated into two parts: the first part consisting of the locomotive and the first 10 cars and a second part consisting of the last 60 cars. An of these parts can be used to find the tension of the coupler. We ll consider the rear of the train Version 4 - Forces, part 1 37

38 The forces are: 1) The weight, directed downwards. 2) A normal force exerted b the ground, directed upwards. 3) The tension force (T) of the coupler, directed towards the right. The equation of the sum of the x-components of the forces is The tension is, therefore, F = T = ma x T = ma = 600,000 kg 1.25 = 750,000 N m s² Here s a little trick to use if there are several objects which accelerate in different directions: take a different axis sstem for each object. The solution is much simpler if the x-axis is alwas in the direction of the acceleration. Example What are the tension of the string and the acceleration of the boxes in the sstem shown in the figure on the right? (There is no friction between the ground and the 3 kg box.) cm-rotationalinertia-380-q The sum of the forces exerted on each box is needed. The forces acting on the 3 kg box are: 1) The weight (w 1 = 29.4 N), directed downwards. 2) A normal force made b the ground (F N ), directed upwards. 3) The tension of the string (T), directed towards the right. The equations of forces are F = T = m a x 1 F = 29.4N + F = 0 N The x-axis was set in the direction of the acceleration of the box, i.e. towards the right Version 4 - Forces, part 1 38

39 The forces acting on the 1 kg box are: 1) The weight (w 2 = 9.8 N), directed downwards. 2) The tension of the string (T), directed upwards. The equations are F = 9.8N T = m a x 2 The x-axis is now directed downwards since this is the direction of the acceleration of this box. We thus used two different axes sstems for each box. The following sstem of equations must be solved to find the acceleration and the tension force. T = m a 9.8N T = m a 1 2 B adding these two equations, the acceleration can be found. ( ) ( 9.8 ) 9.8 = ( ) 9.8 = ( ) T + N T = m a + m a N m m a a = 2.45 m s ² 1 2 N kg kg a (This trick, adding the equations, will alwas work in this chapter. When this is done, all the forces that act on two objects of the sstem cancel each other. If the do not cancel, and ou alwas used x-axes directed in the direction of the acceleration of each object, something is wrong with our equations.) With the acceleration, the tension can be found. T = m a 1 = 3kg 2.45 = 7.35N m s² Note that the tension of the string is not just equal to the weight of the suspended mass. The tension is equal to the weight onl if the sstem is not accelerating. When it is accelerating, the tension of the string is no longer equal to the weight of the suspended mass Version 4 - Forces, part 1 39

40 Example What are the tensions of the strings and the acceleration of the boxes in this sstem? (There is no friction between the slope and the 10 kg box.) phsicstasks.eu/uloha.php?uloha=510 Here, we are supposed to use x-axes in the direction of the acceleration of each object. However, it is not obvious to know the direction of the acceleration here. Does the 12 kg box accelerate upwards or downwards? In such a case, ou can guess a direction for the acceleration. If ou get a positive acceleration at the end of the calculation, then the acceleration is actuall in the direction ou have guessed. Conversel, if ou get a negative acceleration, then the sstem accelerates in the direction opposite to the direction ou have guessed. Here it will be assumed that the 12 kg box accelerates upwards. The forces acting on the 12 kg box are shown in the figure. The equation of force is, therefore, F = 117.6N + T = m a x 1 1 The forces acting on the 10 kg box are shown in the figure to the right. Most of the forces are directl in the direction of an axis except for the weight. The angle between this force and the positive x-axis is -50 as shown in the figure to the left Version 4 - Forces, part 1 40

41 The table of force for the 10 kg box is Forces x Weight 98 N cos(-50 ) 98 N sin(-50 ) Normal force 0 F N Tension force 1 - T 1 0 Tension force 2 T 2 0 The equations for the 10 kg box are ( ) ( ) F = 98N cos 50 T + T = m a x F = 98N sin 50 + F = 0 The forces acting on the 5 kg box are shown in the figure. The equation for the sum of the x-component of the forces is F = 49N T = m a x 2 3 The acceleration and the tension forces can now be found. The following sstem of three equations must be solved to obtain the values of theses unknowns N + T = m a N 1 1 ( ) T + T + 98N cos 50 = m a N T = m a 2 3 As it was done previousl, this sstem can be solved b adding the three equations to eliminate the tension forces. ( ) ( ) ( ) 98 cos( 50 ) cos( 50 ) + 49 = ( + + ) N + T + N T + T + N T = m a + m a + m a N N N m m m a 5.607N = 27kg a a = m s ² Now that the acceleration is known, the tension forces can be calculated N + T = m a 1 1 m ( ) 117.6N + T = 12kg s² T = N Version 4 - Forces, part 1 41

42 49N T = m a 2 3 m ( ) 49N T = 5kg s² T 2 = 50.04N The string can also pass several times in the pulle as shown in this figure. In such a case, the string passing through the pulle must be looked at ver carefull to find how often the string pulls on the pulle. Sometimes, the equation of the forces exerted on the pulle must be found to calculate a tension force. The mass of the pulle will alwas be neglected unless stated otherwise. The following examples illustrate these two ideas. Example What is the acceleration of the 10 kg crate in this figure? science.howstuffworks.com/transport/engines-equipment/pulle.htm 2018 Version 4 - Forces, part 1 42

43 The sum of the forces on the pulle must be done to find the magnitude of the tension force acting on the 10 kg crate. The sum of the -component of the forces is F = 100N + 100N T = m a 2 pulle Neglecting the mass of the pulle (m pulle = 0), this becomes 100N + 100N T = 0 T 2 = 200N 2 The figure on the left shows the forces acting on the crate. The force equation then allows us to find the acceleration. 200N 98N = ma 102N = 10kg a a = 10.2 m s ² Example In 2008, the submarine Onondaga was hauled on land beside Pointe-au-Père s wharf, not far from Rimouski. To achieve this, the 1400-ton submarine was towed up a 4 slope with the help of pulles. In all, the steel cable pulled 18 times on the pulle attached to the submarine. The submarine was on small carts rolling on rails, thereb limiting the friction between the ground and the submarine so that it will be neglected here. With what force did the have to pull on the cable in order to move the submarine at a constant speed? Version 4 - Forces, part 1 43

44 The force needed to pull the submarine at constant speed will be found first. The forces acting on the submarine are shown in this figure. The table of forces is then Forces x Weight mg cos(-94 ) mg sin(-94 ) Normal force 0 F Tension force T 1 0 The equation of the x-component of the force is ( ) 1 F = mg cos 94 + T = 0 The tension can now be found with this equation. 1 x 1 1 ( ) + T1 = mg cos( 94 ) kg N ( ) mg cos 94 0 T = T = 1, 400, 000 9,8 cos 94 kg T = 957, 058N Next, the forces acting on the pulle must be looked at. The equation of the x-component of the forces is F = T + 18T = m a x 1 2 pulle Neglecting the mass of the pulle, it becomes T + 18T = 0 T 1 2 T = = 53,170N 2018 Version 4 - Forces, part 1 44

45 Sometimes, it can be difficult to find the direction of the net force (i.e. the sum of forces) acting on an object if the magnitude of each force is not known precisel. For example, suppose the direction of the net force on a skier going downhill has to be found. There are not man forces acting on this skier. There are onl the weight and a normal force (if friction is neglected). Obviousl, the weight is directed downwards, and the normal force has a direction perpendicular to the slope. However, it is hard to find the direction of the resultant force without prior knowledge of the magnitude of these forces. For example, the three following directions can be obtained b changing the magnitude of the normal force. Fortunatel, Newton s second law can help. This law sas that F = ma net In such a vector equation where m is necessaril positive, the direction of the vector F net must be identical to the direction the vector a. It follows that Direction of the Net Force The net force is alwas in the same direction as the acceleration Version 4 - Forces, part 1 45

46 It then becomes much easier to find the direction of the net force. As it is obvious that the skier is accelerating downhill, the direction of the net force is also directed towards the bottom of the hill. Law of Gravitation (General Formula) 1) Magnitude of the force m m 1 2 Fg = G r 2 2) Direction of the force Attraction of the two masses one towards the other. 3) Application point of the force See the chapter on gravitation. For planets: from the centre of the planet. Weight (w) or Gravitational Force (F g ) (Formula Valid Onl Near the Surface of the Earth) 1) Magnitude of the force w = mg = m 9.8 N kg 2) Direction of the force Downwards (centre of the Earth). 3) Application point of the force From the centre of mass of the object. (See the chapter on the centre of mass. For now, take a point roughl at the centre of the object.) 2018 Version 4 - Forces, part 1 46

47 Normal Force (N or F ) 1) Magnitude of the force To be determined with Newton s laws. 2) Direction of the force Repulsion between objects, perpendicular to the contact surface. 3) Application point of the force Contact surface between objects. Tension Force (T or F T ) 1) Magnitude of the force Given or to be determined with Newton s laws. 2) Direction of the force The rope pulls in the direction of the rope. 3) Application point of the force Where the rope is fixed to the object. Direction of the Net Force The net force is alwas in the same direction as the acceleration Version 4 - Forces, part 1 47

48 4.2 The Normal Force 1. William, whose mass is 72 kg, is in an elevator. a) What is the normal force acting on William if the elevator moves up with a constant speed of 5 m/s? b) What is the normal force acting on William when the elevator goes upwards with a speed of 5 m/s if the speed is increasing at a rate of 2 m/s²? c) What is the normal force acting on William when the elevator goes upwards with a speed of 5 m/s if the speed is decreasing at a rate of 3 m/s²? whs.wsd.wednet.edu/facult/busse/mathhomepage/busseclasses/apphsics/studguides/apphsics2012/chapter5_2012/ch apter5_2012.html 2. Two boxes are placed one on top of the other in an elevator. What are the normal forces acting on each block if the elevator goes up at 8 m/s and is slowing down at a rate of 1 m/s²? 4.3 The Tension Force 3. A helicopter lifts a 300 kg item during the construction of a dome. a) What is the tension force exerted b the rope if the helicopter does not accelerate? b) What is the tension force exerted b the rope if the helicopter has a 3 m/s² acceleration directed upwards? c) What is the tension force exerted b the rope if the helicopter has a 2 m/s² acceleration directed downwards? Version 4 - Forces, part 1 48

49 4. Two boxes are suspended from the ceiling of an elevator. a) What are the tensions of the strings if the elevator has an acceleration of 2.4 m/s² directed downwards? b) What is the maximum upwards acceleration that the elevator can have without breaking the strings (if the break when their tension exceeds 200 N)? Applications of Newton s Laws 5. What are normal forces acting on this 400 g ball? 6. What is the normal force made b the ground acting on the 12 kg box in this situation? whs.wsd.wednet.edu/facult/busse/mathhomepage/busseclasses/apphsics/studguides/apphsics2012/chapter5_2012/ch apter5_2012.html 2018 Version 4 - Forces, part 1 49

50 7. What are the normal forces acting on the 12 kg and the 20 kg boxes in this situation? whs.wsd.wednet.edu/facult/busse/mathhomepage/busseclasses/apphsics/studguides/apphsics2012/chapter5_2012/ch apter5_2012.html 8. Gontran (on the left) and Philemon (on the right) move their big snowball b exerting the forces shown in the figure. There is no friction between the snowball and ground. cnx.org/content/m42139/latest/ a) What is the acceleration of the snowball? b) What is the magnitude of normal force exerted b the ground on the snowball? 9. Irina, whose mass is 60 kg, is climbing a cliff. At some point, she finds herself in the position shown in the figure. a) What is the tension of the rope? b) What is the magnitude of the normal force acting on Irina s feet? cnx.org/content/m42139/latest/?collection=col11406/latest 2018 Version 4 - Forces, part 1 50

51 10. Indiana, whose mass is 65 kg, is in the unfortunate position shown in the figure. What is the tension of the rope? An object is suspended with two strings as shown in the figure. What are the tensions of the two strings? An object is suspended as shown in the figure. What are the tensions of the three strings? Version 4 - Forces, part 1 51

52 13. A force F is maintaining a 10 kg block in the equilibrium position shown in the figure. a) What is the magnitude of the force F? b) What is the tension of the string? 14. Yannick is skiing. At the bottom of a slope, he has an uphill speed of 30 m/s. Neglect friction in this problem. a) How far will Yannick travel before stopping? b) How much time does it take for Yannick to stop? vhcc2.vhcc.edu/ph1fall9/frames_pages/openstax_problems.htm 2018 Version 4 - Forces, part 1 52

53 15. Wolfgang is skiing downhill. While he moves, there s a constant friction force acting in a direction opposite to the velocit. According to what is shown in the figure, what is the magnitude of the friction force if Wolfgang s mass is 70 kg? vhcc2.vhcc.edu/ph1fall9/frames_pages/openstax_problems.htm 16. A 30 kg block is held in place b a horizontal string on a 25 slope. a) What is the tension of the rope? b) What is the magnitude of the normal force between the surface and the block? An 80 kg crate is pushed with an 800 N horizontal force on a 40 inclined surface. a) What is the acceleration of the crate? b) What is the magnitude of the normal force acting on the crate? 2018 Version 4 - Forces, part 1 53

54 18. The two boxes shown in the figure are pushed with a 50 N force. There is a 10 N frictional force opposed to the motion of box A and also an 8 N frictional force opposed to the motion of box B. a) What is the acceleration of the boxes? b) What is the normal force between the two boxes? 19. Here are two blocks connected b a rope passing over a pulle. a) What is the acceleration of the blocks? b) What is the tension of the rope? A 300 N force is exerted on a 24 kg block connected to an 18 kg block b a string, as shown in this figure. phsics.stackexchange.com/questions/45534/tensions-and-pulles-with-masses a) What is the magnitude of the acceleration of the blocks? b) What is the tension of the rope? c) What is the magnitude of the normal force acting on each block? 2018 Version 4 - Forces, part 1 54

55 21. Two blocks are connected b a rope as shown in this figure. a) What is the mass m of the block on the slope if the 2 kg block has an acceleration of 2 m/s² directed downwards? b) What is the mass m of the block on the slope if the tension of the rope is 25 N? 22. Three blocks are connected b ropes as shown in this figure. a) What is the magnitude of the acceleration of the blocks? b) What are the tensions of the two strings? 23. Two blocks are connected b a rope as shown in this figure. a) What is the acceleration of the blocks? b) What is the tension of the rope? cnx.org/content/m14124/latest/?collection=col10322/ Version 4 - Forces, part 1 55

56 24. The tractor of this airport luggage carrier exerts an 800 N force. a) What is the acceleration of the airport luggage carrier? b) What are the tensions T 1, T 2 and T 3? 25. When block A is pulled with a 100 N force, block B has an acceleration of 1 m/s² directed downwards. When the block A is pulled with a 200 N force, block B has an acceleration of 2 m/s² directed upwards. What are the masses of the two blocks? ww.chegg.com/homework-help/questions-and-answers/figure kg-blocksare-connected-massless-string-pulleradius255-cm-rotational-ine-q What force F should be exerted to keep this sstem in equilibrium? en.wikipedia.org/wiki/mechanical_advantage_device 2018 Version 4 - Forces, part 1 56

57 27. In the situation shown in the figure, a) what force F should be exerted to keep this sstem in equilibrium? b) what is the acceleration of the block if the force F has a magnitude of 20 N? 28. What should be the mass of the bucket for this sstem to be in equilibrium? phsicstasks.eu/uloha.php?uloha= What is the acceleration of Romeo, whose mass if 40 kg, if he pulls on the rope with a 250 N force? phsicstasks.eu/uloha.php?uloha= Version 4 - Forces, part 1 57

58 30. In the situation shown in the figure, what is the tension T and the angle θ? What force F should be exerted to keep this sstem in equilibrium? ux.brookdalecc.edu/fac/engtech/mqaissaunee/engi101/jpegs/chapter06/ Challenges (Questions more difficult than the exam questions.) 32. With what force should this triangle be pushed so that the small block does not slide down or up the slope? (There is no friction.) Version 4 - Forces, part 1 58

59 33. In the situation shown in the figure, there is no friction between the table and the 5 kg rope sliding on the table. What will be the speed of the rope when the end arrives at the edge of the table? The Normal Force 1. a) N directed upwards b) N directed upwards c) N directed upwards 2. 5 kg box: a 44 N normal force directed upwards, made b the 10 kg box 10 kg box: a 44 N normal force directed downwards, made b the 5 kg box and a 132 N normal force directed upwards made b the ground 4.3 The Tension Force 3. a) 2,940 N b) 3,840 N c) 2,340 N 4. a) T 1 = N T 2 = 74 N b) 2.7 m/s² 4.4 Applications of Newton s Laws 5. vertical wall: 6.79 N towards the left inclined surface: 7.84 N at F N = 78.4 N directed upwards kg box: a 78.4 N normal force directed upwards, made b the 20 kg box 20 kg box: a 78.4 N normal force directed downwards, made b the 12 kg box and a N normal force directed upwards, made b the ground 8. a = 3.89 m/s² F N = N 9. T = N F N = N 10. 3,654 N 11. Rope to the right: 199 N rope to the left: 374 N 2018 Version 4 - Forces, part 1 59