Chapter 4 FORCES AND NEWTON S LAWS OF MOTION PREVIEW QUICK REFERENCE. Important Terms

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1 Chapter 4 FORCES AND NEWTON S LAWS OF MOTION PREVIEW Dynaics is the study o the causes o otion, in particular, orces. A orce is a push or a pull. We arrange our knowledge o orces into three laws orulated by Isaac Newton: the law o inertia, the law o orce and acceleration (F net = a), and the law o action and reaction. Friction is the orce applied by two suraces parallel to each other, and the noral orce is the orce applied by two suraces perpendicular to each other. Newton s law o universal gravitation states that all asses attract each other with a gravitational orce which is proportional to the product o the asses and inversely proportional to the square o the distance between the. The gravitational orce holds satellites in orbit around a planet or star. The content contained in all sections o chapter 4 o the textbook is included on the AP Physics B exa. QUICK REFERENCE Iportant Ters coeicient o riction the ratio o the rictional orce acting on an object to the noral orce exerted by the surace in which the object is in contact; can be static or kinetic dynaics the study o the causes o otion (orces) equilibriu the condition in which there is no unbalanced orce acting on a syste, that is, the vector su o the orces acting on the syste is zero. orce any inluence that tends to accelerate an object; a push or a pull ree body diagra a vector diagra that represents all o the orces acting on an object riction the orce that acts to resist the relative otion between two rough suraces which are in contact with each other gravitational ield space around a ass in which another ass will experience a orce gravitational orce the orce o attraction between two objects due to their asses 47

2 inertia the property o an object which causes it to reain in its state o rest or otion at a constant velocity; ass is a easure o inertia inertial reerence rae a reerence rae which is at rest or oving with a constant velocity; Newton s laws are valid within any inertial reerence rae kinetic riction the rictional orce acting between two suraces which are in contact and oving relative to each other law o universal gravitation the gravitational orce between two asses is proportional to the product o the asses and inversely proportional to the square o the distance between the. ass a easure o the aount o substance in an object and thus its inertia; the ratio o the net orce acting on an accelerating object to its acceleration net orce the vector su o the orces acting on an object newton the SI unit or orce equal to the orce needed to accelerate one kilogra o ass by one eter per second squared non-inertial reerence rae a reerence rae which is accelerating; Newton s laws are not valid within a non-inertial reerence rae. noral orce the reaction orce o a surace acting on an object static riction the resistive orce that opposes the start o otion between two suraces in contact weight the gravitational orce acting on a ass Equations and Sybols F F W g F s ax k G net F F k G r a s N N ( kinetic ) ( static) where F = orce = ass a = acceleration W = weight g = acceleration due to gravity s ax = axiu static rictional orce k = kinetic rictional orce F N = noral orce F G = gravitational orce r = distance between the centers o two asses 48

3 Ten Hoework Probles Chapter 4 Probles, 4, 6, 44, 5, 58, 67, 68, 70, 80 DISCUSSION OF SELECTED SECTIONS Newton s Laws o Motion The irst law o otion states that an object in a state o constant velocity (including zero velocity) will continue in that state unless acted upon by an unbalanced orce. The property o the book which causes it to ollow Newton s irst law o otion is its inertia. Inertia is the sluggishness o an object to changing its state o otion or state o rest. We easure inertia by easuring the ass o an object, or the aount o aterial it contains. Thus, the SI unit or inertia is the kilogra. We oten reer to Newton s irst law as the law o inertia. The law o inertia tells us what happens to an object when there are no unbalanced orces acting on it. Newton s second law tells us what happens to an object which does have an unbalanced orce acting on it: it accelerates in the direction o the unbalanced orce. Another nae or an unbalanced orce is a net orce, eaning a orce which is not canceled by any other orce acting on the object. Soeties the net orce acting on an object is called an external orce. Newton s second law can be stated like this: A net orce acting on a ass causes that ass to accelerate in the direction o the net orce. The acceleration is proportional to the orce (i you double the orce, you double the aount o acceleration), and inversely proportional to the ass o the object being accelerated (twice as big a ass will only be accelerated hal as uch by the sae orce). In equation or, we write Newton s second law as F net = a where F net and a are vectors pointing in the sae direction. We see ro this equation that the newton is deined as a kg. s The weight o an object is deined as the aount o gravitational orce acting on its ass. Since weight is a orce, we can calculate it using Newton s second law: F net = a becoes Weight = g, where the speciic acceleration associated with weight is, not surprisingly, the acceleration due to gravity. Like any orce, the SI unit or weight is the newton. 49

4 Newton s third law is soeties called the law o action and reaction. It states that or every action orce, there is an equal and opposite reaction orce. For exaple, let s say your calculator weighs N. I you set it on a level table, the calculator exerts N o orce on the table. By Newton s third law, the table ust exert N back up on the calculator. I the table could not return the N o orce on the calculator, the calculator would sink into the table. We call the orce the table exerts on the calculator the noral orce. Noral is another word or perpendicular, because the noral orce always acts perpendicularly to the surace which is applying the orce, in this case, the table. The orce the calculator exerts on the table, and the orce the table exerts on the calculator are called an actionreaction pair Newton s Second Law o Motion and the Vector Nature o Newton s Second Law o Motion Since orce is a vector quantity, we ay break orces into their x and y coponents. The horizontal coponent o a orce can cause a horizontal acceleration, and the vertical coponent o a orce can cause a vertical acceleration. These horizontal and vertical coponents are independent o each other. Exaple A orklit lits a 0-kg box with an upward vertical acceleration o.0 /s, while pushing it orward with a horizontal acceleration o.5 /s. (a) Draw a ree-body diagra or the box on the diagra below. (b) What is the agnitude o the horizontal orce F x acting on the box? (c) What is the agnitude o the upward noral orce F N the plator exerts on the box? (d) I the box starts ro rest at ground level (x = 0, y = 0, and v = 0) at tie t = 0, write an expression or its vertical position y as a unction o horizontal distance x. (e) On the axes below, sketch a y vs x graph o the path which the box ollows. Label all signiicant points on the axes o the graph. 50

5 y() Solution: F N a y =.0 /s (a) F x a x =.5 /s W (b) The horizontal orce F x exerted by the wall causes the horizontal acceleration a x =.5 /s. Thus, the agnitude o the horizontal orce is F x = a x = (00 kg)(.5 /s ) = 00 N (c) In order to accelerate the box upward at.0 /s, the noral orce F N ust irst overcoe the downward weight o the box. Writing Newton s second law in the vertical direction gives F net y = a y (F N W) = a y (F N g) = a y F N = a y + g = (00 kg)(.0 /s ) + (00 kg)(9.8 /s ) = 60 N 5

6 (d) Since the box starts ro rest on the ground, we can write x axt and y a y t Substituting or a x and a y, we get x.5 / s t and y.0 / s t Solving both sides or t and setting the equations equal to each other yields y.0.5 x 4 x (e) The graph o y vs x would be linear beginning at the origin o the graph and having a positive slope o 4 : y() 4 x() 4.7 The Gravitational Force Newton s law o universal gravitation states that all asses attract each other with a gravitational orce which is proportional to the product o the asses and inversely proportional to the square o the distance between the. The gravitational orce holds satellites in orbit around a planet or star. The equation describing the gravitational orce is G F G r where F G is the gravitational orce, and are the asses in kilogras, and r is the distance between their centers. The constant G siply links the units or gravitational orce to the other quantities, and in the etric syste happens to be equal to 6.67 x 0 - N /kg. Like several other laws in physics, Newton s law o universal gravitation is an inverse square law, where the orce decreases with the square o the distance ro the centers o the asses. 5

7 Exaple An artiicial satellite o ass = 400 kg orbits the earth at a distance r = 6.45 x 0 6 above the center o the earth. The ass o the earth is = 5.98 x 0 4 kg. Find (a) the weight o the satellite and (b) the acceleration due to gravity at this orbital radius. Solution (a) The weight o the satellite is equal to the gravitational orce that the earth exerts on the satellite: 4 G (6.67x0 kg)(400 kg)(5.98x0 kg) F G 85 N 6 r (6.45x0 ) (b) The acceleration due to gravity is g W F G 85 N 400 kg 9.59 s Note that even high above the surace o the earth, the acceleration due to gravity is not zero, but only slightly less than at the surace o the earth The Noral Force, Static and Kinetic Frictional Forces The noral orce F N is the perpendicular orce that a surace exerts on an object. I a box sits on a level table, the noral orce is siply equal to the weight o the box: F N W I the box were on an inclined plane, the noral orce would be equal to the coponent o the weight o the box which is equal and opposite to the noral orce: F N y gsinθ x θ gcosθ g θ 5

8 In this case, the coponent o the weight which is equal and opposite to the noral orce is gcos θ. Friction is a resistive orce between two suraces which are in contact with each other. There are two types o riction: static riction and kinetic riction. Static riction is the resistive orce between two suraces which are not oving relative to each other, but would be oving i there were no riction. A block at rest on an inclined board would be an exaple o static riction acting between the block and the board. I the block began to slide down the board, the riction between the suraces would no longer be static, but would be kinetic, or sliding, riction. Kinetic riction is typically less than static riction or the sae two suraces in contact. The ratio o the rictional orce between the suraces divided by the noral orce acting on the suraces is called the coeicient o riction. The coeicient o riction is represented by the Greek letter (u). Equations or the coeicients o static and kinetic riction are s ax k s and k, where s is the static rictional orce and k is the kinetic FN FN rictional orce. Note that the coeicient o static riction is equal to ratio o the axiu rictional orce and the noral orce. The static rictional orce will only be as high as it has to be to keep a syste in equilibriu. When you draw a ree body diagra o orces acting on an object or syste o objects, you would want to include the rictional orce as opposing the relative otion (or potential or relative otion) o the two suraces in contact. Exaple A block o wood rests on a board. One end o the board is slowly lited until the block just begins to slide down. At the instant the block begins to slide, the angle o the board is θ. What is the relationship between the angle θ and the coeicient o static riction μ s? Solution Let s draw the ree-body diagra or the block on the inclined plane: F N x y gsinθ θ gcosθ g θ 54

9 At the instant the block is just about to ove, the axiu rictional orce directed up the incline is equal and opposite to the +x-coponent o the weight down the incline, and the noral orce is equal and opposite to the y-coponent o the weight. F s ax N Then gsin g cos s F s ax N gsin tan g cos This expression is only valid or the case in which the static rictional orce is axiu. Exaple 4 Ater the block in Exaple just begins to ove, should the board be lowered or raised to keep the block oving with a constant velocity down the incline? Explain your answer. Solution Since the coeicient o kinetic riction is generally less than the coeicient o static riction or the sae two suraces in contact, the block would require less orce directed down the incline (gsin θ) to keep it sliding at a constant speed. Thus, the board should be lowered to a saller θ just ater the block begins to slide to keep the block oving with a constant velocity The Tension Force, Equilibriu Applications o Newton s Laws o Motion The orce in a rope or cable that pulls on an object is called the tension orce. Like any other orce, tension can accelerate or contribute to the acceleration o an object or syste o objects. Exaple 5 An elevator cable supports an epty elevator car o ass 00 kg. Deterine the tension in the cable when the elevator car is (a) at rest and (b) the car has a downward acceleration o.0 /s. 55

10 Solution (a) The ree-body diagra or the car would look like this: F T When the elevator car is at rest, the tension in the cable F T is equal to the weight W o the car: F T = W = g = (00 kg)(0 /s ) = 000 N. (b) When the elevator car is being lowered with an acceleration o.0 /s, the downward weight orce is greater than the upward tension orce. We can use Newton s second law to ind the tension in the cable. Choosing the downward direction as positive, F a ( W F F T T W ) a W a (000 N) (00 kg)(.0 ) 400 s N upward. A syste is said to be static i it has no velocity and no acceleration. According to Newton s irst law, i an object is in static equilibriu, the net orce on the object ust be zero. 56

11 Exaple 6 Three ropes are attached as shown below. The tension orces in the ropes are T, T, and T, and the ass o the hanging ball is =.0 kg. 0º 40º T T T Since the syste is in equilibriu, the net orce on the syste ust be zero. We can ind the tension in each o the three ropes by inding the vector su o the tensions and setting this su equal to zero: T T y T y T T x T x T T T + T + T = 0 Since T is in the negative y-direction, T ust be equal to the weight g o the ass. As we resolve each tension orce into its x- and y-coponents, we see that T = g = (.0 kg)(0 /s ) = 0 N T x = T cos40 and T y = T sin40 T x = T cos0 and T y = T sin0 Since the orces are in equilibriu, the vector su o the orces in the x-direction ust equal zero: ΣF x = 0 57

12 T x = T x T cos40 = T cos0 T T cos0 cos40. T The su o the orces in the y-direction ust also be zero: ΣF y = 0 T y + T y = W T sin40 + T sin0 = W Knowing that T. T, we can solve the equations above or each o the tensions in the ropes:.t T sin 40 T sin 0 g g.sin 40 sin 0.0kg0 s 4.4 N. T.T.(4.4N ) 7. 6N 4. Nonequilibriu Applications o Newton s Laws o Motion I the net orce acting on a syste is not zero, the syste ust accelerate. A coon exaple used to illustrate Newton s second law is a syste o blocks and pulley. Exaple 7 In the diagra below, two blocks o ass = kg and = 6 kg are connected by a string which passes over a pulley o negligible ass and riction. What is the acceleration o the syste? kg 6 kg 58

13 Solution Let s draw a ree-body orce diagra or each block. There are two orces acting on each o the asses: weight downward and the tension in the string upward. Our ree-body orce diagras should look like this: T T W W Writing Newton s second law or each o the blocks: F a T g a T g a a g T a and F T g a Notice that the tension T acting on the kg block is greater than block its weight, but the 6 kg block has a greater weight than the tension T. This is, o course, the reason the 6 kg block accelerates downward and the kg block accelerates upward. The tension acting on each block is the sae, and their accelerations are the sae. Setting their tensions equal to each other, we get g a g a Solving or a, we get a 5.0 s 59

14 Exaple 8 A block o ass = kg rests on a horizontal table. A string is tied to the block, passed over a pulley, and another block o ass = 4 kg is hung on the other end o the string, as shown in the igure below. The coeicient o kinetic riction between the kg block and the table is 0.. Find the acceleration o the syste. μ k = 0. kg 4 kg Solution Once again, let s draw a ree-body orce diagra or each o the blocks, and then apply Newton s second law. F N T T k W W 60

15 Block : F a Block : T k T F k Since F a a g, T g a k T g a k N N Setting the two equations or T equal to each other: g a g a k F W a T a g T a T g a Solving or a and substituting the values into the equation, we get g k g a s kg 4kg 4kg 0 0. kg 0 s 6 s Exaple 9 Three blocks o ass,, and are connected by a string passing over pulley attached to a plane inclined at an angle θ as shown below. θ θ The riction between each block on the inclined plane and the surace o the plane has a agnitude. (a) Draw the orces acting on each block. (b) Assuing that is large enough to descend and cause the syste to accelerate, deterine the acceleration o the syste in ters o the given quantities and undaental constants. 6

16 (c) As reaches the top o the inclined plane, it coes to rest up against the pulley. The vertical string supporting is cut, and and begin to slide down the incline. Assuing =, what is the speed o the blocks ater they have slid a distance d down the plane? Solution (a) F N T T T T F N W W W (b) Since the pulley serves only to change the direction o the tension in the string that passes over it, we ay ind the acceleration o the syste by treating the asses as i they lie in a straight line and are accelerated by the weight o the hanging block : a W sinθ W Newton s second law: F a W W sin W W sin a a 6

17 (c) Since =, and riction acts equally on both, they slide down the plane with equal acceleration and speed. Thus, the two blocks slide down the incline as i they were one block o ass + accelerated down the plane by a orce o ( + )gsin θ against a rictional orce o directed up the plane. ( + )gsinθ Newton s second law or the syste is F a g sin g sin a i ad ad Since the initial velocity is zero, The speed can beound by the kineatic v v v a g sin d equation 6

18 CHAPTER 4 REVIEW QUESTIONS For each o the ultiple choice questions below, choose the best answer. Unless otherwise noted, use g = 0 /s and neglect air resistance.. The aount o orce needed to keep a 0. kg hockey puck oving at a constant speed o 7 /s on rictionless ice is (A) zero (B) 0. N (C) 0.7 N (D) 7 N (E) 70 N T 0 45 T T W=0N. A orce o 6 N is needed to overcoe a rictional orce o 5 N to accelerate a kg ass across a loor. What is the acceleration o the ass? (A) 4 /s (B) 5 /s (C) 7 /s (D) 0 /s (E) 60 /s. A orce o 00 N directed at an angle o 45 ro the horizontal pulls a 70 kg sled across a rozen rictionless pond. The acceleration o the sled is ost nearly (sin 45 = cos 45 = 0.7) (A).0 /s (B) 0.7 /s (C) 7 /s (D) 5 /s (E) 50 /s 4. Which o the ollowing is true o the agnitudes o tensions T, T, and T in the ropes in the diagra shown above? (A) T ust be greater than 0 N. (B) The tension T is greater than T. (C) The y-coponent o T and T is equal to 0 N. (D) The su o the agnitudes o T and T is equal to T. (E) The su o the agnitudes o T and T is equal to T. 5. Two blocks o ass and 5 are connected by a light string which passes over a pulley o negligible ass and riction. What is the acceleration o the asses in ters o the acceleration due to gravity, g? (A) 4g (B) 5g (C) 6g (D) 4/5 g (E) / g 64

19 6. A -kg block rests on a rictionless table and is connected by a light string to another block o ass kg. The string is passed over a pulley o negligible ass and riction, with the kg ass hanging vertically. What is the acceleration o the asses? (A) 5 /s (B) 6.7 /s (C) 0 /s (D) 0 /s (E) 0 /s 7. Friction (A) can only occur between two suraces which are oving relative to one another. (B) is equal to the noral orce divided by the coeicient o riction. (C) opposes the relative otion between the two suraces in contact. (D) only depends on one o the suraces in contact. (E) is always equal to the applied orce. 9. Which o the ollowing diagras o two planets would represent the largest gravitational orce between the asses? (A) c (B) (C) (D) r r r r (E) r 0 8. A -kg wooden block rests on an inclined plane as shown above. The rictional orce between the block and the plane is ost nearly (sin 0 = 0.5, cos 0 = 0.87, tan 0 = 0.58) (A) N (B) 0 N (C) N (D) 7 N (E) 0 N 0. A satellite is in orbit around the earth. Consider the ollowing quantities: I. distance ro the center o the earth II. ass o the earth III. ass o the satellite The gravitational acceleration g depends on which o the above? (A) I only (B) I and II only (C) III only (D) I and III only (E) I, II, and III 65

20 Free Response Question Directions: Show all work in working the ollowing question. The question is worth 5 points, and the suggested tie or answering the question is about 5 inutes. The parts within a question ay not have equal weight.. (5 points) A block o ass rests on an air table (no riction), and is pulled with a orce probe, producing the Force vs. acceleration graph shown below. (a) Deterine the ass o the block. The block is now placed on a rough horizontal surace having a coeicient o static riction μ s = 0., and a coeicient o kinetic (sliding) riction μ k = 0.. (b) What is the iniu value o the orce F which will cause the block to just begin to ove? L F F θ μ s =0. ; μ k =0. 66

21 (c) Ater the block begins to ove, the sae orce deterined in part (b) continues to act on the block. What is the acceleration o the block? (d) The orce F is now tripled to F, which then pulls the block up an incline o angle θ = 0 and having a coeicient o kinetic riction μ k = 0.. i. Draw the ree-body diagra or the block as it is being pulled up the incline. ii. Deterine the agnitude o the rictional orce k acting on the block as it slides up the incline. iii. Deterine the acceleration o the block as it is pulled up the incline. ANSWERS AND EXPLANATIONS TO CHAPTER 4 REVIEW QUESTIONS Multiple Choice. A The law o inertia states that no net orce is needed to keep an object oving at a constant velocity.. C The net orce is 6 N 5 N = N, and the acceleration is Fnet N a 7/ s kg. A Only the x-coponent o the orce accelerates the sled horizontally: F x =(00 N)cos45 = 70 N. Fx 70 N Then a / s 70 kg 67

22 4. C T y + T y ust equal the weight o the block (0 N), since the syste is in equilibriu. 5. E The net orce acting on the syste is 5g g = 4 g. Then F net 4g 5 a, and a = g. 6. B The weight o the kg block is the net orce accelerating the entire syste. g a, so a = g = 6.7 /s. 7. C Friction acts on each o the suraces in contact that are oving or have the potential or oving relative to each other. 8. B Since the block is in static equilibriu, the rictional orce ust be equal and opposite to the coponent o the weight pointing down the incline: gsin kg 0/ s sin0 0 N 9. D The greater the ass, and the saller the separation distance r, the greater the orce according to Newton s law o universal gravitation. 0. B GM Since the acceleration due to gravity g r E, it does not depend on the ass o the satellite. Free Response Proble Solution (a) points Newton s nd Fnet law states that. This ratio is the slope o the F vs. a graph. So, a F 5 N 0 slope kg a 5 / s 0 (b) points The block will just begin to ove when the orce F overcoes the axiu static rictional orce: F F g 0. kg 0 / s in N s s N s 6 68

23 (c) points Once the block begins to ove we ust use the coeicient o kinetic riction to deterine the rictional orce. k k FN kg 0. kg0 / s N Then the net orce acting on the block is F k = 6 N N = N to the right. The acceleration o the block is Fnet N a / s kg (d) F N F i. points gsinθ k gcosθ g ii. points As the block sides up the incline, the noral orce F N is no longer equal to g, but gcosθ. F g cos 0. kg0 / s cos0. N k k N k 8 iii. points The net orce is now N.8 N kg0 / s sin 0 N F k gsin 6 5 F a net 5 N kg.67 / s 69