40. The situation is somewhat similar to that shown in the “loop-the-loop” example done in the textbook
(see Figure 6-10) except that, instead of a downward normal force, we are dealing with the force of the
boom
FBon the car – which is capable of pointing any direction. We will assume it to be upward as we
apply Newton’s second law to the car (of total weight 5000 N):
FB−W=ma where m=W
g,and a=−v2
r
Note that the centripetal acceleration is downward (our choice for negative direction) for a body at the
top of its circular trajectory.
(a) If r=10mandv=5.0 m/s, we obtain FB=3.7×103N=3.7kN(up).
(b) If r=10mandv= 12 m/s, we obtain FB=−2.3×103N = -2.3 kN where the minus sign indicates
that
FBpoints downward.