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Solved: In Figure 9.21, the cg of the pole held by the

College Physics | 1st Edition | ISBN: 9781938168000 | Authors: Paul Peter Urone, Roger Hinrichs ISBN: 9781938168000 42

Solution for problem 18PE Chapter 9

College Physics | 1st Edition

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College Physics | 1st Edition | ISBN: 9781938168000 | Authors: Paul Peter Urone, Roger Hinrichs

College Physics | 1st Edition

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Problem 18PE

In Figure 9.21, the cg of the pole held by the pole vaulter is 2.00 m from the left hand, and the hands are 0.700 m apart. Calculate the force exerted by (a) his right hand and (b) his left hand. (c) If each hand supports half the weight of the pole in Figure 9.19, 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 described above.

Step-by-Step Solution:

Solution 18 PE

a.)

Step 1 of 10</p>

The force exerted by his right and left hand can be calculated by applying the second condition for equilibrium of a system ‘ the net external torque on a system must be zero’

Here we chose the centre of gravity point as the pivot around which the torque is calculated.

Step 2 of 10</p>

Sum of torque about the centre of gravity is zero.

Where, L= torque due to left hand

              R = torque due to right hand

               CG= 0

   

Where,FR is the force exerted by right hand in N

             FL is the force exerted by left hand in N

              xR is the distance between between right hand and the centre of gravity in m

             xL is the distance between between left  hand and the centre of gravity in m

Step 3 of 10</p>

Sum of forces acting in vertical direction is zero. So,

Solving for FL

Substituting FR+W  for FL in

()

Solving for FR

Substituting 5 kg for m, 9.8 m/s2, 2.00 m for xL, 2.70 m for xR

           

                                          =

                                                  FR= 140 N

Therefore, the force exerted by his right hand is 140 N.

b.)

Step 4 of 10</p>

The force exerted by his left hand can be calculated using

Where, m = mass of the pole

             g = acceleration due to gravity

Substituting 140 N for FR, 5 kg for m and 9.8 m/s2 for g

   = 140+49

FL = 189 N

Therefore, the force exerted by his left hand is 189 N.

c.)

Step 5 of 10</p>

We can show that the second condition for equilibrium (net τ = 0) is satisfied for a pivot other than the one located at the centre of gravity of the pole by picking a pivot in such a way that the net calculated torque about that pivot is equal to zero.

 

Step 6 of 9

Chapter 9, Problem 18PE is Solved
Step 7 of 9

Textbook: College Physics
Edition: 1
Author: Paul Peter Urone, Roger Hinrichs
ISBN: 9781938168000

The answer to “In Figure 9.21, the cg of the pole held by the pole vaulter is 2.00 m from the left hand, and the hands are 0.700 m apart. Calculate the force exerted by (a) his right hand and (b) his left hand. (c) If each hand supports half the weight of the pole in Figure 9.19, show that the second condition for equilibrium (net ? = 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 described above.” is broken down into a number of easy to follow steps, and 100 words. This full solution covers the following key subjects: hand, pole, left, show, figure. This expansive textbook survival guide covers 34 chapters, and 3125 solutions. Since the solution to 18PE from 9 chapter was answered, more than 1487 students have viewed the full step-by-step answer. This textbook survival guide was created for the textbook: College Physics , edition: 1. The full step-by-step solution to problem: 18PE from chapter: 9 was answered by , our top Physics solution expert on 07/07/17, 04:39PM. College Physics was written by and is associated to the ISBN: 9781938168000.

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