20. If the block is sliding then we compute the kinetic friction from Eq. 6-2; if it is not sliding, then we
determine the extent of static friction from applying Newton’s law, with zero acceleration, to the xaxis
(which is parallel to the incline surface). The question of whether or not it is sliding is therefore crucial,
and depends on the maximum static friction force, as calculated from Eq. 6-1. The forces are resolved
in the incline plane coordinate system in Figure 6-5 in the textbook. The acceleration, if there is any, is
along the xaxis, and we are taking uphill as +x. The net force along the yaxis, then, is certainly zero,
which provides the following relationship:
Fy=0 =⇒N=Wcos θ
where W= 45 N is the weight of the block, and θ=15
◦is the incline angle. Thus, N=43.5N,which
implies that the maximum static friction force should be fs, max =(0.50)(43.5) = 21.7N.
(a) For
P=5.0 N downhill, Newton’s second law, applied to the xaxis becomes
f−P−Wsin θ=ma where m=W
g.
Here we are assuming
fis pointing uphill, as shown in Figure 6-5, and if it turns out that it points
downhill (which is a possibility), then the result for fswill be negative. If f=fsthen a=0,we
obtain fs= 17 N, which is clearly allowed since it is less than fs, max.
(b) For
P=8.0 N downhill, we obtain (from the same equation) fs= 20 N, which is still allowed since
it is less than fs, max.
(c) But for
P= 15 N downhill, we obtain (from the same equation) fs= 27 N, which is not allowed
since it is larger than fs, max. Thus, we conclude that it is the kinetic friction, not the static friction,
that is relevant in this case. We compute the result fk=(0.34)(43.5) = 15 N. Here, as in the other
parts of this problem, the friction is directed uphill.