17. (a) Although details in Fig. 6-27 might suggest otherwise, we assume (as the problem states) that only
static friction holds block Bin place. An excellent discussion and equation development related to
this topic is given in Sample Problem 6-3. We merely quote (and apply) their main result (Eq. 6-13)
for the maximum angle for which static friction applies (in the absence of additional forces such as
the
Fof part (b) of this problem).
θmax =tan
−1µs=tan
−10.63 ≈32◦.
This is greater than the dip angle in the problem, so the block does not slide.
(b) We analyze forces in a manner similar to that shown in Sample Problem 6-3, but with the addition
of a downhill force F.
F+mg sin θ−fs,max =ma =0
N−mg cos θ=0.
Along with Eq. 6-1 (fs,max =µsN) we have enough information to solve for F.Withθ=24
◦and
m=1.8×107kg, we find
F=mg (µscos θ−sin θ)=3.0×107N.