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# f5ch06 - 142 CHAPTER 6 FORCE AND MOTION II CHAPTER 6 Answer...

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142 CHAPTER 6 FORCE AND MOTION – II C HAPTER 6 Answer to Checkpoint Questions . ( a ) zero (because there is no attempt at sliding); ( b ) N; ( c ) no; ( d ) yes . ( a ) ( N); ( b ) decreases; ( c ) decreases . greater . ( a ) a downward, N upward; ( b ) a and N upward . ( a ) R ; ( b R . ( a ) same; ( b ) increases; ( c ) increases Answer to Questions . they slide at the same angle for all orders . ( a ) same; ( b ) increases; ( c ) increases; ( d ) no . ( a ) upward; ( b ) horizontal, toward you; ( c ) no change; ( d ) increases; ( e ) increases . ( a ) increases; ( b ) increases; ( c ) decreases; ( d ) decreases; ( e ) decreases . the frictional force f s is initially directed up the ramp, decreases in magnitude to zero, and then is directed down the ramp, increasing in magnitude until the mag- nitude reaches f s; max ; thereafter, the magnitude of the frictional force is f k , which is a constant smaller value . f s is initially directed up the ramp and increases in magnitude until the magnitude reaches f s; max ; thereafter the magnitude of the frictional force is f k (a constant smaller value) . ( a ) decreases; ( b ) decreases; ( c ) increases; ( d ) increases . ( a ) decreases; ( b ) decreases; ( c ) decreases; ( d ) decreases . ( a ) zero; ( b ) in nite . , , . , , then , , and tie . ( a ) decreases; ( b ) decreases

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m g f N F m g f N CHAPTER 6 FORCE AND MOTION – II 143 . ( a ) less; ( b ) greater . it increases until f s; max is reached, then it suddenly equals f k , a constant smaller value, while the coin slides outward in a spiral across the turntable Solutions to Exercises & Problems E ( a ) The free-body diagram for the bureau is shown to the right. F is the applied force, f is the force of friction, N is the normal force of the oor, and m g is the force of gravity. Take the x axis to be horizontal and the y axis to be vertical. As- sume the bureau does not move and write Newton's second law. The x component is F f = and the y component is N mg = . The force of friction is then equal in magnitude to the applied force: f = F . The normal force is equal in magnitude to the force of gravity: N = mg . As F increases, f increases until f = s N . Then the bureau starts to move. The minimum force that must be applied to start the bureau moving is F = s N = s mg = ( : )( kg)( : m/s ) = N. ( b ) The equation for F is the same but the mass is now kg kg = kg. Thus F = s mg = ( : )( kg)( : m/s ) = N : E The free-body diagram for the player is shown to the right. N is the normal force of the ground on him, m g is the force of gravity, and f is the force of friction. The force of friction is related to the normal force by f = k N . Use Newton's second law to nd the normal force. The vertical component of his acceleration is , so the vertical component of Newton's second law is N mg = and the normal force equals his weight in magnitude: N = mg . Thus k = f N = f mg = N ( kg)( : m/s ) = : :
N F f s m g θ x y 144 CHAPTER 6 FORCE AND MOTION – II E Solve the smallest angle min in question from Eq. - in Sample Problem - : min = tan ( s ) = tan ( : ) = : E ( a ) The free-body diagram is shown to the right. The equation of motion in the y di- rection is N + F sin = mg : Thus the magnitudes of the normal force corresponding to the various values of are N = mg F sin = ( : kg)( : m/s ) ( N)(sin N (for = ) ; N = mg F sin = ( : kg)( : m/s ) ( : N (for : ) ; and N = mg F sin = ( : kg)( : m/s ) ( : N (for : ) : The horizontal component F x

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## This note was uploaded on 01/22/2012 for the course PHYS 2101 taught by Professor Grouptest during the Fall '07 term at LSU.

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f5ch06 - 142 CHAPTER 6 FORCE AND MOTION II CHAPTER 6 Answer...

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