This preview shows page 1. Sign up to view the full content.
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. 65 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....
View
Full
Document
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.
 Spring '08
 Any
 Physics, Force, Friction

Click to edit the document details