# Why isn't gravity included in this calculation for torque?

In this textbook problem which was assigned for homework, I found a flaw in my understanding of torque.

A $2.4\ \mathrm{kg}$ block rests on a slope and is attached by a string of negligible mass to a solid drum of mass $0.85\ \mathrm{kg}$ and radius $5.0\ \mathrm{cm}$. When released, the block accelerates down the slope at $1.6\ \mathrm{m/s^2}$.

Find the coefficient of friction between the block and slope. My initial thought process is to consider the torque because one of the forces applied contributes to torque, and that force makes the "drum" rotate.

The applied forces that cause rotation, from my point of view, is the tension force and the force of gravity.

However, upon looking at the solution, I noticed that the force of gravity isn't included in the calculation for torque.

Why is that? I understand that the drum is "attached" to this triangular platform which might be why the force of gravity isn't considered, but when I was trying to answer this question and making my free body diagram, I included the force of gravity.

The force of gravity is applied at the center of mass of the drum. As a result, it does not apply any torque to the drum, since the lever arm is zero. A good way to see this is notice that the drum does not turn under the force of gravity alone.

The single remaining torque is due to the tension of the rope, so that's all you have to consider for the rotation of the drum.

You would have to consider the torque only in the case that the drum were unbalanced: for instance if it were not rotating around the center.

• How can I visualize the "lever arm" here? Would it be the distance from the axis of rotation (the center of the "circle") along the length of the radius? – DeepLearner Nov 30 '17 at 5:04
• Force is applied at a point. The lever arm is the component of that distance from the axis of rotation to that point that is perpendicular to the force. – Chris Nov 30 '17 at 5:07
• Would the force of gravity be applied to the center of mass, which is the axis of rotation in this problem? – DeepLearner Nov 30 '17 at 5:08
• @TheChemistryGuy Exactly. The force of gravity can always be considered to be applied at the center of mass. – Chris Nov 30 '17 at 5:09
• I finally get it now! Thanks so much for your help. – DeepLearner Nov 30 '17 at 5:11

Gravitational Force is acting at the center of the drum which is intersecting the axis of rotation (i.e. $\vec r$ = 0), causing value of torque zero.
Remember, Torque $\vec\tau$ = $\vec r \times \vec F$.