http://www.ck12.org Chapter 8. Angular Motion and Statics
FIGURE 8.12
Once again, there is only one unknown in the problem- distance. Therefore, we will use the equilibrium condition
∑τ=0. We’ll take the pivotPas our point of rotation. The clockwise torque due to the weight of the beam must
balance the counterclockwise torque due to the weight of the daredevil. The angle betweenrandFis 90◦throughout
the problem so write the torques as justrF.
We must determine the distance,xbfrom the center of mass of the beam to the pivot,P.
Since the location of the center of mass is^12 Land the location of the pivot is^23 L, (seeFigure8.12) the distance
between the center of mass and the pivot isxb=^23 L−^12 L=^16 L. The condition for rotational equilibrium gives
∑τ=^0 →mbgL 6 −mdgxd=0.
After cancelinggand substituting the givens we have
∑τ=^0 →(^270 kg)
( 1
6 L
)
−( 60 kg)xd= 0 →xd=^34 LWell, just how much of a daredevil is he?