4.1 Forces. Chapter 4 The Laws of Motion
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1 4.1 Forces Chapter 4 he Laws of Motion
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3 4.2 Newton s First Law it s not the nature of an object to stop, once set in motion, but rather to continue in its original state of motion. An object moves with a velocity that is constant in magnitude and direction, unless acted on by a nonzero net force. he net force on an object is defined as the vector sum of all external forces exerted on the object. IP 4.1 Force Causes Changes in Motion Motion can occur even in the absence of forces. Force causes changes in motion.
4 surface is called the weight, w, of the object, given by 4.3 Newton s Second Law : g is the acceleration of gravity. : newton (N) he acceleration a of an object is directly proportional to the net force acting on it and inversely proportional to its 44337_04_p /13/04 2:31 PM Page 88 mass. F : ma : IP 4.2 Is Not a Force : m a m : a Newton s law of universal gravitation states that every particle in the Universe attracts every other particle with a force that is directly F proportional to the product of the masses of the particles g gravitational force, F and inversely proportional to g is the square of the distance between them. F ACIVE FIGURE 4.6 g he gravitational force between two F g m 2 particles is attractive. ring Equations 4.6 and 4.7, we see that m 1 r w mg 88 Chapter 4 he Laws of Motion Law of universal gravitation Equation 4.1 does not say that the product is a force. All forces how weak it is can be carried out exerted with a small on an balloon. object are Rubbing summed the as balloon in vectors to generate the net force on your hair gives the balloon a tiny electric charge. hrough electric forces, the the left side of the equation. his net balloon then adheres to a wall, resisting the gravitational pull of the entire force is then equated to the product Earth! of the mass and resulting acceleration In addition to contributing to the understanding of motion, Newton studied of the object. Do not include an gravity extensively. Newton s law of universal m : gravitation [4.7] states that every particle a force in your analysis. in the Universe attracts every other particle with a force that is directly proportional to the product of the masses of the particles and inversely proportional to the square of the distance between them. If the particles have masses m 1 and m 2 and are separated by a distance r, as in Active Figure 4.6, the magnitude of the F g m 2 F g G m 1m 2 r in the denominator of Equation 2 r4.7, the weight decreases with increasing m weight of an object on a mountaintop 1 is less than the weight of the same Log into to PhysicsNow at and go to Active Figure 4.6 to change the masses of the and the separation between g the particles G M E to see the effect on the gravitational force. F g G m 1m 2 r 2 [4.5] where G N m 2 /kg 2 is the universal gravitation constant. We examine the gravitational force in more detail in Chapter 7. Weight he magnitude of the gravitational force acting on an object of mass m near Earth s surface is called the weight, w, of the object, given by where g is the acceleration of gravity. SI unit: newton (N) w mg [4.6] quation 4.5, an alternate definition of the weight of an object with mass written as is the mass of Earth and r is the distance from the object to Earth s cenobject is at rest on Earth s surface, then r is equal to Earth s radius R E. ea level. w G M Em r 2 r 2 [4.8] mass, weight is not an inherent property of an object because it can take values, depending on the value of g in a given location. If an object has [4.6] From Equation 4.5, an alternate definition of the weight of an object with mass m can be written as
5 4.4 Newton s hird Law he action force is equal in magnitude to the reaction force and opposite in direction. In all cases, the action and reaction forces act on different objects If object 1 and object 2 interact, the force F12 exerted by object 1 on object 2 : is equal in magnitude but opposite in direction to the force F21 exerted by object 2 on object 1. :
6 Figure 4.9 Newton s second law applied to a rope gives ma. However, if m 0, then. hus, the tension in a massless rope is the same at all points in the rope.
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8 4.5 Applications of Newton s Laws Problem-Solving Strategy Newton s Second Law Problems involving Newton s second law can be very complex. he following protocol breaks the solution process down into smaller, intermediate goals: 1. Read the problem carefully at least once. 2. Draw a picture of the system, identify the object of primary interest, and indicate forces with arrows. 3. Label each force in the picture in a way that will bring to mind what physical quantity the label stands for (e.g., for tension). 4. Draw a free-body diagram of the object of interest, based on the labeled picture. If additional objects are involved, draw separate free-body diagrams for them. Choose convenient coordinates for each object. 5. Apply Newton s second law. he x- and y-components of Newton s second law should be taken from the vector equation and written individually. his often results in two equations and two unknowns. 6. Solve for the desired unknown quantity, and substitute the numbers. IP 4.6 A Particle in Equilibrium A zero net force on a particle does not mean that the particle isn t moving. It means that the particle isn t accelerating. If the particle has a nonzero initial velocity and is acted upon by a zero net force, it continues to move with the same velocity.
9 est y x sled. y Strategy When an object is on a slope, it s convenient to use tilted coordinates, as in Figure 4.13b, so that the : normal force n: is in the y-direction and the tension force is in the x-direction. In the absence of friction, the hill exerts no force on the sled in the x-direction. n Because the sled is at rest, the conditions for equilibrium, F x 0 and F y 0, apply, giving two equations for the two unknowns the tension and the normal force. mg sin u x mg cos u y n x Figure 4.12 Fg Figure 4.13 for the sled. (c) 3 (Example 4.5) A traffic light suspended by cables. A free-body 30.0 Fg = m g (Example 4.6) A child holding a sled on a frictionless hill. A free-body diagram mg sin u mg cos u 30.0 Fg = mg Figure 4.13 (Example 4.6) A child holding a sled on a frictionless hill. A free-body diagram for the sled. F y Solution 0 : 3 F g 0 3 Apply F g 1.00 Newton s 10 2 Nsecond law to the sled, with a: 0: : F : : n : F g 0 Force x-component y-component Extract the x-component from this equation to find. he x-component of the normal force is zero, and the : 1 sled s 2 weight is given by mg 77.0 N. : : 3 1 cos sin cos sin N Write the y-component of Newton s second law. he F x y-component 1 cos 37.0 of the 2 cos tension 53.0 is zero, 0 so this equation (1) will give the normal force. F y 1 sin sin N 0 (2) Remarks Unlike its value on a horizontal surface, n is less than the weight of the sled when the sled is on the slope. 2 his 1 cos is 37.0 because cos 53.0 only part of the 1.33 force of gravity (the x-component) is acting to pull the sled down the slope. he y- 1 component of the force of gravity balances the normal force. 1 sin 37.0 Exercise ( )(sin 53.0 ) N 0 Suppose another child of weight w climbs onto the sled. If the tension force is measured to be 60.0 N, find the weight of the N child and the magnitude of the normal force acting on the sled (60.0 N) 79.9 N Answers w 43.0 N, n 104 N Solution Apply Newton s second law to the sled, with a: 0: : F : : n : F g 0 Extract the x-component from this equation to find. he x-component of the normal force is zero, and the sled s weight is given by mg 77.0 N. Write the y-component of Newton s second law. he y-component of the tension is zero, so this equation will give the normal force. F x 0 mg sin (77.0 N)sin N F x 0 mg sin (77.0 N)sin n 66.7 N 38.5 N F y 0 n mg cos n (77.0 N)(cos 30.0 ) 0 Remarks Unlike its value on a horizontal surface, n is less than the weight of the sled when the sled is on the slope. his is because only part of the force of gravity (the x-component) is acting to pull the sled down the slope. he y- component of the force of gravity balances the normal force. Exercise 4.6 Suppose another n child of 66.7 weight Nw climbs onto the sled. If the tension force is measured to be 60.0 N, find the weight of the child and the magnitude of the normal force acting on the sled. Answers F y 0 n mg cos n (77.0 N)(cos 30.0 ) 0 w 43.0 N, n 104 N
10 - e t t scale doesn t measure the true weight, it measures the force that it exerts on the fish, so in each case solve for this force, which is the apparent weight as measured by the scale. a Figure 4.17 (Example 4.9) a r e e - - Solution Find the scale reading as the elevator accelerates upwards. mg mg Apply Newton s second law to the fish, taking upwards as the positive direction: Figure 4.17 (Example 4.9) ma F mg Solve for : ma mg m(a g) tes pwards Find the mass of the fish from its weight of 40.0 N: m w mg g ma F mg Compute the value of, substituting a 2.00 m/s 2 : Figure 4.17 (Example 4.9) Find the scale reading as the elevator accelerates downwards N 9.80 m/s kg m(a g) (4.08 kg)(2.00 m/s m/s 2 ) 48.1 N mg : m w g /s 2 : ma mg m(a g) he analysis is the same, the only change being the acceleration, which is now negative: a 2.00 m/s N 9.80 m/s kg (c) Find the scale reading after the elevator cable breaks. Now a 9.8 m/s 2, the acceleration due to gravity: m(a g) (4.08 kg)(2.00 m/s m/s 2 ) 48.1 N m(a g) (4.08 kg)( 2.00 m/s m/s 2 ) 31.8 N m(a g) (4.08 kg)( 9.80 m/s m/s 2 ) 0 N ates he m(a g) (4.08 kg)( 2.00 m/s m/s 2 ) /s 2.
11 m 1 m 2 a 1 m 1 m 1 g m 2 ACIVE FIGURE 4.18 a 2 m 2 g m 1 a 1 m 1 g (1) m 2 a 2 m 2 g (2) m 2 a 1 m 2 g (m 1 m 2 )a 1 m 2 g m 1 g 2m 1 m 2 m 1 m 2 g a 1 m 2 m 1 m 1 m 2 g
12 4.6 Forces of Friction
13 n n Motion f s F f k F mg mg f f s,max f s = F f k = mkn 0 Static region Kinetic region F (c)
14 y f s n mg cos u u mg sin u u F g x F x mg sin s n 0 (1) F y n mg cos 0 (2) n mg cos F x mg sin s mg cos 0 : tan s tan : tan 1 (0.350) 19.3
15 m 1 n F x f k m 1 a 1 F y n m 1 g 0 m 2 fk m 1 m 1 g y x k m 1 g m 1 a 1 (1) F y m 2 g m 2 a 2 m 2 a 1 (2) m 2 m 2 g m 2 g km 1 g (m 1 m 2 )a 1 a 1 m 2g km 1 g m 1 m 2 a 1 (7.00 kg)(9.80 m/s2 ) (0.300)(4.00 kg)(9.80 m/s 2 ) (4.00 kg 7.00 kg) 5.17 m/s N (m 1 m 2 )a m 2 g kn m 2 g km 1 g a m 2g km 1 g m 1 m 2
16 Draw a free-body diagram for each block. he static friction force causes the top block to move horizontally, and and the maximum such force corresponds to f s s n. his same static friction retards the motion of thebottom block. As long as the top block isn t slipping, the acceleration of both blocks is the same. Write Newton s second law for each block, and eliminate the acceleration a by substitution, solving for the tension. f s n 1 n 2 m M mg n 1 f s x m M x Mg Solution Write the two components of Newton s second law for the top block: Figure 4.24 (Example 4.14) (Exercise 4.14) x-component: ma s n 1 y-component: 0 n 1 mg Solve the y-component for n, substitute the result into the x-component, and then solve for a: n 1 mg : ma s mg : a s g Write the x-component of Newton s second law for the bottom block: Ma s mg (1) Substitute the expression for a s g into Equation (1) and solve for the tension : M s g s mg : (m M) s g Now evaluate to get the answer: (5.00 kg 10.0 kg)(0.350)(9.80 m/s 2 ) 51.5 N Remarks Notice that the y-component for the 10.0-kg block wasn t needed, because there was no friction be-
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