G t is the tension force of the rope on the crate n f

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G T is the tension force of the rope on the crate, N F G is the normal force of the floor on the crate, mg G is the force of gravity, and G f is the force of friction. We take the + x direction to be horizontal to the right and the + y direction to be up. We assume the crate is motionless. The equations for the x and the y components of the force according to Newton’s second law are: T cos θ f = 0 sin 0 N T F mg θ + = where θ = 15° is the angle between the rope and the horizontal. The first equation gives f = T cos θ and the second gives F N = mg T sin θ . If the crate is to remain at rest, f must be less than μ s F N , or T cos θ < μ s ( mg T sin θ ). When the tension force is sufficient to just start the crate moving, we must have T cos θ = μ s ( mg T sin θ ). We solve for the tension:
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(a) If 0.600 s μ = and 0.500, k μ = then the magnitude of f G has a maximum value of ,max (0.600)( 0.500 sin 20 ) 0.497 . s s N f F mg mg mg μ = = ° = On the other hand, cos 0.500 cos20 0.470 . F mg mg θ = ° = Therefore, ,max cos s F f θ < and the block remains stationary with 0 a = . (b) If 0.400 s μ = and 0.300, k μ = then the magnitude of f G has a maximum value of ,max (0.400)( 0.500 sin 20 ) 0.332 . s s N f F mg mg mg μ = = ° = In this case, ,max cos 0.500 cos20 0.470 . s F mg mg f θ = ° = > Therefore, the acceleration of the block is ( ) [ ] 2 2 2 cos sin (0.500)(9.80 m/s ) cos20 (0.300)sin 20 (0.300)(9.80 m/s ) 2.17 m/s . k k F a g m θ μ θ μ = + = °+ ° − = 14. (a) The free-body diagram for the block is shown on the right, with F G being the force applied to the block, N F G the normal force of the floor on the block, mg G the force of gravity, and G f the force of friction. We take the + x direction to be horizontal to the right and the + y direction to be up. The equations for the x and the y components of the force according to Newton’s second law are: cos sin 0 x y N F F f ma F F F mg θ θ = = = + = Now f = μ k F N , and the second equation gives F N = mg F sin θ , which yields ( sin ) k f mg F μ θ = . This expression is substituted for f in the first equation to obtain F cos θ μ k ( mg F sin θ ) = ma , so the acceleration is ( ) cos sin k k F a g m θ μ θ μ = + .
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15. An excellent discussion and equation development related to this problem is given in Sample Problem 6-2. We merely quote (and apply) their main result: 1 1 tan tan 0.04 2 . s θ μ = = °
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16. (a) We apply Newton’s second law to the “downhill” direction: mg sin θ f = ma , where, using Eq. 6-11, f = f k = μ k F N = μ k mg cos θ . Thus, with μ k = 0.600, we have a = g sin θ μ k cos θ = –3.72 m/s 2 which means, since we have chosen the positive direction in the direction of motion (down the slope) then the acceleration vector points “uphill”; it is decelerating. With 0 18.0 m/s v = and Δ x = d = 24.0 m, Eq. 2-16 leads to 2 0 2 12.1 m/s. v v ad = + = (b) In this case, we find a = +1.1 m/s 2 , and the speed (when impact occurs) is 19.4 m/s.
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17. (a) The free-body diagram for the block is shown below.
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