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## 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?