Cylinder Equilibrium (2012 F = ma #12)

[Jzhang27143]

Homework Statement

The problem is #12 found here: https://www.aapt.org/physicsteam/2013/upload/exam1-2012-unlocked-solutions.pdf.

A uniform cylinder of radius a originally has a weight of 80 N. After an off-axis cylinder hole at 2a/5 was drilled
through it, it weighs 65 N. The axes of the two cylinders are parallel and their centers are at the same height.
a 2a/5
A force T is applied to the top of the cylinder horizontally. In order to keep the cylinder is at rest, the magnitude
of the force is closest to:
(A) 6 N← CORRECT
(B) 10 N
(C) 15 N
(D) 30 N
(E) 38 N

Homework Equations

Tau = r x F
M_total x_cm = sum m_i x_i

The Attempt at a Solution

This problem has been bugging me because I keep getting the wrong answer.
So i found the center of mass of the new cylinder to be at -6a/65 with the origin being at the center. (0 = 65x + 15*2a/5 -> 65x = -6a -> x = -6a/65.)
The torque due to gravity must be -6/65a*65 = -6a relative to the point of contact. Therefore, the torque due to the horizontal force T must be 6a. 6a = 2aT -> T = 3N which does not agree with the answer of 6N. What have I done incorrectly?

 

[TSny]

Your work looks correct to me. You have wisely taken the torques about the point of contact. I agree with your answer. Hopefully, others will confirm this or else point out where we are overlooking something.

(I think the answer of 6 N would be correct if the cylinder were mounted on a horizontal, frictionless axle running along the central axis of the cylinder.)

 

[Chestermiller]

I wasn't able to follow what you did, but here is what I did. Drilling out the hole centered at x=2a/5 is equivalent to putting an upward force of 15 N at x = 2a/5 in the original undrilled cylinder. The moment of this force about the center of the cylinder is 6a N-m. So the force to balance this moment at r = a has to be 6 N.

Chet

 

[haruspex]

I agree with 3N, and there's an easier way to get it.
Think of the drilling the hole as, instead, adding a negative weight (80N-65N=15N) at the hole's mass centre.
This has torque 15N*(2a/5) = 6aN about the point of contact with the ground. The countering force is 2a perpendicularly from that point.

 

[haruspex]

I wasn't able to follow what you did, but here is what I did. Drilling out the hole centered at x=2a/5 is equivalent to putting an upward force of 15 N at x = 2a/5 in the original undrilled cylinder. The moment of this force about the center of the cylinder is 6a N-m. So the force to balance this moment at r = a has to be 6 N.

Chet

Did you check the diagram? No axle is indicated for the cylinder. Rather, it is shown lying sideways on a surface.

 

[Chestermiller]

Did you check the diagram? No axle is indicated for the cylinder. Rather, it is shown lying sideways on a surface.

You're right. The key words in the problem statement are "at rest." The force at the top has to be 3N, and a static friction force of 3N has to develop at the table top to balance the horizontal force (to prevent linear acceleration) and to provide the required moment (couple). In this way, the moment about the center of mass and the moment about the contact point at the table top are the same. Very cute.

Now, back to the problem statement. They asked which answer the force is closest to. 3N is closest to the 6N choice. I'm guessing they were allowing people who made my misinterpretation to get credit for a "correct" answer.

Chet

 

[TSny]

Now, back to the problem statement. They asked which answer the force is closest to. 3N is closest to the 6N choice. I'm guessing they were allowing people who made my misinterpretation to get credit for a "correct" answer.
Chet

Could be. But they could have achieved the same objective by using the correct answer of 3 N instead of 6 N.