1. We do not consider the possibility that the bureau might tip, and treat this as a purely horizontal motion
problem (with the person’s push
Fin the +xdirection). Applying Newton’s second law to the xand y
axes, we obtain
F−fs,max =ma
N−mg =0
respectively. The second equation yields the normal force N=mg, whereupon the maximum static
friction is found to be (from Eq. 6-1) fs,max =µsmg. Thus, the first equation becomes
F−µsmg =ma =0
wherewehaveseta= 0 to be consistent with the fact that the static friction is still (just barely) able
to prevent the bureau from moving.
(a) With µs=0.45 and m= 45 kg, the equation above leads to F= 198 N. To bring the bureau into
a state of motion, the person should push with any force greater than this value. Rounding to two
significant figures, we can therefore say the minimum required push is F=2.0×102N.
(b) Replacing m= 45 kg with m= 28 kg, the reasoning above leads to roughly F=1.2×102N.