11.2 Applications of statics, including problem-solving strategies  (Page 2/3)

What force is needed to support a weight held near its cg?

For the situation shown in [link] , calculate: (a) $F_R$ , the force exerted by the right hand, and (b) $F_L$ , the force exerted by the left hand. The hands are 0.900 m apart, and the cg of the pole is 0.600 m from the left hand.

Strategy

[link] includes a free body diagram for the pole, the system of interest. There is not enough information to use the first condition for equilibrium $\text{(net}F=0$ ), since two of the three forces are unknown and the hand forces cannot be assumed to be equal in this case. There is enough information to use the second condition for equilibrium $\text{(net}\tau =0\text{)}$ if the pivot point is chosen to be at either hand, thereby making the torque from that hand zero. We choose to locate the pivot at the left hand in this part of the problem, to eliminate the torque from the left hand.

Solution for (a)

There are now only two nonzero torques, those from the gravitational force ( ${\tau }_{\text{w}}$ ) and from the push or pull of the right hand ( ${\tau }_R$ ). Stating the second condition in terms of clockwise and counterclockwise torques,

or the algebraic sum of the torques is zero.

Here this is

since the weight of the pole creates a counterclockwise torque and the right hand counters with a clockwise torque. Using the definition of torque, $\tau =\text{rF}\text{sin}\theta$ , noting that $\theta =\mathrm{90º}$ , and substituting known values, we obtain

Thus,

Solution for (b)

The first condition for equilibrium is based on the free body diagram in the figure. This implies that by Newton’s second law:

From this we can conclude:

Solving for $F_L$ , we obtain

Discussion

$F_L$ is seen to be exactly half of $F_R$ , as we might have guessed, since $F_L$ is applied twice as far from the cg as $F_R$ .