In the figure below the cg of the pole held by the pole...

In the figure below, the CG of the pole held by the pole vaulter is 2.29 m from the left hand, and the hands are 0.573 m apart. (Assume the vaulting pole has a mass of 4.12 kg.) (a)Calculate the force (in N) exerted by his right hand. (Enter the magnitude.) (b)Calculate the force (in N) exerted by his left hand. (Enter the magnitude.) (c)If each hand supports half the weight of the pole in the figure below, show that the second condition for equilibrium (net 𝜏 = 0) is satisfied for a pivot other than the one located at the center of gravity of the pole. Explicitly show how you follow the steps in the Problem-Solving Strategy for static equilibrium.

Transcript

00:01 Okay, so we have a pole vaulter here that is holding on to the pole with both hands and they are applying a force upward with their left hand and downward with their right hand.

00:14 And the distance between the hands, i've labeled letter a, and that's 0 .573 meters.

00:20 And the distance between the center of mass and the left hand, i've labeled b, and that's 2 .29 meters.

00:27 So, picture not to scale.

00:30 To find the force applied by their right hand, we can use the net torque equation.

00:40 So, the net torque, in this case, needs to be 0.

00:44 Since we are not in motion, this pole vaulter is not moving.

00:50 And so, what we can say is that the counterclockwise torques have to balance the clockwise torques.

01:02 And i'm going to choose this position right here, the position of the left hand, to be the point of rotation.

01:11 And so, we can set the counterclockwise torques, which is just the torque from the right hand, we can set that equal to the clockwise torques, which is going to be from the weight.

01:23 So, if we do that, we have the force that the right hand is applying, times a, is going to equal the weight, which is the force of gravity on the pole, as applied to the center of gravity, times b.

01:41 We can then find that right hand force by dividing by a, and then plugging in our values here.

01:49 So, the weight is mg, so that's going to be 4 .12 times 9 .8.

01:55 Then we multiply by 2 .29 meters, and then we divide by a, which is 0 .573 meters.

02:08 And if we apply that correctly, we should get a right hand force of 161 .36 newtons.

02:21 Now, in letter b, we need to figure out the force that the left hand is applying.

02:26 And for this, we can use the net force equation.

02:28 So, the net force has to be equal to 0, which means the upward forces have to balance the downward forces.

02:37 Now, both the right hand and the weight force are acting down, and only the left hand is acting up...

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