Chapter+6+HW+Solutions

# Chapter+6+HW+Solutions - Chapter 6 HW Solutions 34 The...

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Unformatted text preview: Chapter 6 HW Solutions 34. The free-body diagrams for the slab and block are shown below. F is the 100 N force applied to the block, Ns F is the normal force of the floor on the slab, Nb F is the magnitude of the normal force between the slab and the block, f is 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. For both objects, we take the + x direction to be to the right and the + y direction to be up. Applying Newton’s second law for the x and y axes for (first) the slab and (second) the block results in four equations: s s Ns Ns s b b Nb b f m a F F m g f F m a F m g- =-- =- =- = from which we note that the maximum possible static friction magnitude would be 2 (0.60)(10 kg)(9.8 m/s ) 59 N . s Nb s b F m g μ μ = = = We check to see if the block slides on the slab. Assuming it does not, then a s = a b (which we denote simply as a ) and we solve for f : f m F m m s s b = + = + = (40 40 80 kg)(100 N) kg 10 kg N which is greater than f s ,max so that we conclude the block is sliding across the slab (their accelerations are different). (a) Using f = μ k Nb F the above equations yield 2 2 (0.40)(10 kg)(9.8 m/s ) 100 N 6.1 m/s . 10 kg k b b b m g F a m μ-- = = = - The negative sign means that the acceleration is leftward. That is, 2 ˆ ( 6.1 m/s )i b a = - (b) We also obtain 2 2 (0.40)(10 kg)(9.8 m/s ) 0.98 m/s . 40 kg k b s s m g a m μ = - = - = - As mentioned above, this means it accelerates to the left. That is, 2 ˆ ( 0.98 m/s )i s a = - 50. The situation is somewhat similar to that shown in the “loop-the-loop” example done in the textbook (see...
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Chapter+6+HW+Solutions - Chapter 6 HW Solutions 34 The...

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