(g) The condition for the chair to slide is ,maxwhere 0.42.sxsNsFfFμμ>==Forθ= 0°, we have ,max100 N(0.42)(245 N) 103 N=<==so the crate remains at rest. (h) For = 30.0°, we find ,max86.6 N(0.42)(195 N)81.9 N=>==so the crate slides. (i) For = 60°, we get ,max50.0 N(0.42)(158 N)66.4 N==which means the crate must remain at rest. 86. (a) Our +xdirection is horizontal and is chosen (as we also do with +y) so that the components of the 100 N force GFare non-negative. Thus, Fx= Fcos = 100 N, which the textbook denotes
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