ch06-p108

# ch06-p108 - (e) The situation is similar to the one...

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(a) Interpreting "temporarily at rest" (which is not meant to be the same thing as "momentarily at rest") to mean that the box is at equilibrium, we have a = 0 and, consequently, f s = mg sin θ = 3.0 N. It is positive and therefore pointed uphill. (b) Constant speed in a one-dimensional setting implies that the velocity is constant -- thus, a = 0 again. We recover the answer f s = 3.0 N uphill, which we obtained in part (a). (c) Early in the problem, the direction of motion of the luggage was given: downhill. Thus, an increase in that speed indicates a downhill acceleration a = 0.20 m/s 2 . We now solve for the friction and obtain f s = ma + mg sin = 1.6 N, which is positive -- therefore, uphill. (d) A decrease in the (downhill) speed indicates the acceleration vector points uphill; thus, a = +0.20 m/s 2 . We solve for the friction and obtain f s = ma + mg sin = 4.4 N, which is positive -- therefore, uphill.
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Unformatted text preview: (e) The situation is similar to the one described in part (c), but with a = − 0.57 m/s 2 . Now, f s = ma + mg sin = − 1.0 N, or | | 1.0 N s f = . Since s f is negative, the direction is downhill. (f) From the above, the only case where f s is directed downhill is (e). 108. The assumption that there is no slippage indicates that we are dealing with static friction f s , and it is this force that is responsible for "pushing" the luggage along as the belt moves. Thus, Fig. 6-5 in the textbook is appropriate for this problem -- if one reverses the arrow indicating the direction of motion (and removes the word "impending"). The mass of the box is m = 69/9.8 = 7.0 kg. Applying Newton's law to the x axis leads to f s − mg sin = ma where = 2.5 ° and uphill is the positive direction....
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## This note was uploaded on 05/17/2011 for the course PHY 2049 taught by Professor Any during the Spring '08 term at University of Florida.

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