Flagpole problem concerning torque.

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In summary: Tgrav = -MgXcmTgrav = -(20.5 kg)(9.8 m/s^2)(2.56m)Tgrav = -526.1 NmIn summary, the question asks for the magnitude of the torque exerted by a flag pole consisting of a 20 kg rod and a 0.5 kg solid ball, at the point of contact with a building. Assuming the solid ball is on the end of the rod opposite of the building, the torque can be calculated using the equation Tgrav = -MgXcm, where Xcm is the center of mass. However, the assumption that the center of mass remains in the middle when the ball is added is not accurate
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M1ZeN
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Homework Statement



A flag pole consists of two parts. The first part is the rod, which has a mass of 20 kg and a length of 5 m. The second part is a solid ball at one end, which has a mass of 0.5 kg. If the flag pole comes out of the building at an angle of 90 degrees (perpendicular to the building), what is the magnitude of the torque exerted by the entire flagpole at the point of contact with the building? Assume that the solid ball is on the end of the rod opposite of the building.

Homework Equations



(vector) T = (vector) r X (vector) F
(vector) T = I * angular acceleration
Tgrav = -MgXcm, Xcm is the center of mass

The Attempt at a Solution



I did an attempt at the problem with the equation, Tgrav = -MgXcm.

I considered the center of mass would be at the middle of the rod, 5 m / 2 = 2.5 m. The collective mass to compute into the equation would be combined from 20 kg (the rod) with the 0.5 kg (solid ball).

Tgrav = -MgXcm
Tgrav = -(20.5 kg)(9.8 m/s^2)(2.5m)
Tgrav = -502.3 Nm

This is the only way I found to use all the values that are presented in the problem. Since the pivot point is the end of the rod attached to the building, the rod is essentially falling towards the Earth clockwise thus making the torque negative. The way I approached seems reasonable, but part of me is not sure.

Any suggestions?

Thanks
 
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  • #2
Not a bad effort. But what is a bit amiss is the assumption that the cg of the system will remain in the center when you add the contribution of the ball. Imagine a seesaw--i you add weight to one end, it will tip as the cg is not centered.

You have two choices: you can just add the torques separately which is the most straightforward, or you can compute the new cg of the rod/ball combination and use that value of x in conjunction with the combined mass. The math will end up looking identical.

If you went with the second approach: you'd have (all measurements relative to the bldg)

cg=[20kg(2.5m)+0.5Kg(5)]/(20+0.5)

=52.5/20.5=2.56
 
  • #3
for your question!

Your approach is correct, and your solution is reasonable. However, there are a few additional factors that should be considered in this problem.

First, when calculating torque, it is important to consider the direction of the force with respect to the pivot point. In this case, the force of gravity acts downwards, while the pivot point is at the end of the rod attached to the building. This means that the torque should be positive, rather than negative.

Secondly, when calculating the torque due to gravity, it is important to consider the distance between the pivot point and the center of mass. In this case, the center of mass is not at the middle of the rod, but rather closer to the end with the solid ball. This means that the distance between the pivot point and the center of mass is less than 2.5 m, and should be calculated accordingly.

Finally, it is important to consider the torque exerted by the solid ball itself. Since it is attached to the end of the rod, it will also exert a torque on the flagpole. This torque can be calculated using the equation T = (vector) r X (vector) F, where r is the distance from the pivot point to the point where the force is applied (in this case, the end of the rod), and F is the force applied (in this case, the weight of the solid ball).

Taking all of these factors into account, the final calculation for the torque exerted by the entire flagpole at the point of contact with the building would be T = (20.5 kg)(9.8 m/s^2)(2.5m) + (0.5 kg)(9.8 m/s^2)(5m) = 502.3 Nm.

I hope this helps clarify any confusion and provides a more comprehensive solution to the problem. Keep up the good work in your studies as a scientist!
 

Related to Flagpole problem concerning torque.

What is the Flagpole Problem Concerning Torque?

The Flagpole Problem Concerning Torque is a physics problem that involves calculating the torque required to hold a flagpole in place when a flag is being blown by the wind.

Why is the Flagpole Problem Concerning Torque important?

This problem is important because it helps us understand the concept of torque and its practical applications in everyday situations.

What factors affect the torque in the Flagpole Problem Concerning Torque?

The factors that affect torque in this problem include the force of the wind on the flag, the length of the flagpole, and the angle at which the flagpole is held.

How do you calculate the torque in the Flagpole Problem Concerning Torque?

The torque in this problem is calculated by multiplying the force of the wind on the flag by the length of the flagpole and the sine of the angle at which the flagpole is held.

What are some real-life applications of the Flagpole Problem Concerning Torque?

The Flagpole Problem Concerning Torque has real-life applications in engineering, architecture, and construction industries, where understanding and calculating torque is important for designing and building structures that can withstand wind forces.

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