On the box f is the magnitude of the force of

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on the box, f is the magnitude of the force of friction between the slab and the block, m s is the mass of the slab, and m b is the mass of the block. Notice that Newton's third law has been taken into account: the forces of friction on the two objects have the same magnitude but are in op- posite directions; the block pushes down on the slab with a force of magnitude N b . Take the x axis to be to the left and the y axis to be upward. The x component of Newton's second law for the slab is f = m s a s
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164 CHAPTER 6 FORCE AND MOTION – II and the y component is N s N b m s g = ; where a s is the acceleration of the slab. The x component of Newton's second law for the block is F f = m b a b and the y component is N b m b g ; where a b is the acceleration of the block. First check to see if the block slides on the slab. Assume it does not. Then a s = a b . Use a to denote both these accelerations. Use f = m s a to eliminate a from F f = m b a , then solve for f . The result is f = m s F m s + m b = ( kg)( N) kg + kg N : According to the last of the second law equations N b = m b g , so s N b = s m b g = ( : )( kg)( : m/s ) = N : Since f > s N b the block slides on the slab. They have di erent accelerations. Carefully note that N b , not N s is used to compute the upper limit of the static frictional force. The force of friction is exerted by the slab and block on each other so the normal force these objects exert on each other is used to compute the upper limit. Since the block slides on the slab, the magnitude of the frictional force is given by f = k N b . The x component of Newton's second law for the slab is now k N b = m s a s , the x component of Newton's second law for the block is F k N b = m b a b , and the y component is N b m b g = . The last of these gives N b = m b g . Substitute into the rst equation and solve for a s : a s = k m b g m s = ( : )( kg)( : m/s ) kg : m/s : Substitute N b = m b g into the second equation and solve for a b : a b = F k m b g m b = N ( : )( kg)( : m/s ) kg : m/s : P Each side of the trough exerts a normal force on the crate. The rst diagram in the next page shows the view looking in toward a cross section. The resultant is along the dotted line.
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m g N r θ f N N CHAPTER 6 FORCE AND MOTION – II 165 Since each of the normal forces makes an an- gle of with the dotted line the magnitude of the resultant is given by N r = N cos = p N , where cos = p was used. The second diagram is the free-body diagram for the crate. The force of gravity has magni- tude mg , where m is the mass of the crate, and the magnitude of the force of friction is denoted by f . Take the x axis to be down the incline and the y axis to be in the direc- tion of N r . Then the x component of New- ton's second law is mg sin f = ma and the y component is N r mg cos = , where is the angle of the incline. Since the crate is moving each side of the trough exerts a force of kinetic friction, so the total frictional force has magnitude f k N k N r = p = p k N r . Substitute this expression and N r = mg cos , from the y component of the second law, into the x component equation to obtain mg sin p mg cos = ma . Finally, solve for a : a = g (sin p k cos ).
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