# Torques on a Ruler

## Homework Statement

A centimeter ruler, balanced at its center point, has two coins placed on it, as shown in the figure. One coin, of mass 10 g , is placed at the zero mark; the other, of unknown mass , is placed at the 4.7 cm mark. The center of the ruler is at the 3.0 cm mark. The ruler is in equilibrium; it is perfectly balanced.

Find the unknown mass.

τ = F * r

## The Attempt at a Solution

I set the zero mark to be in a center of the ruler. Then I tried to compute the counterclockwise Torque.

τ = (0.01 kg)(0.03 m)(9.8 m/s^2)

I'm really unsure about using the acceleration due to gravity. Should I have been using τ = Iα instead? I figure if I know how to properly find the counterclockwise torque, the clockwise torque will follow the same method.

#### Attachments

• RulerTorques.jpg
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No, your method is fine. The Force term is given by F=mg.

You said you put your zero point in the centre of the ruler. But the question says that the centre is at 3.0cm. Your calculation for the counterclockwise moment is still correct though, as the zero point is 3cm from the fulcrum.

Last edited:
Yeah, the problem did say to put the zero mark at the end but I thought it would be easier to put it in the middle since I wouldn't have to worry about that force. Here's what I've done so far:

counterclockwise torque = (0.01 kg)(0.03m)(9.8 m/s^2)
counterclockwise torque = 0.00294 N * M

clockwise torque = (x kg)(9.8 m/s^2)(0.017 m)
0.00294 N * M (I did this because in order for the ruler to be in equilibrium the torques must cancel)= (x kg)(9.8 m/s^2)(0.017 m)

0.00294 N * M = 0.1666 1/s^2 * (x kg)
x = 0.017 kg

They want the answer in grams though so x = 1.7 g. This doesn't make sense because the second mass is closer to the pivot point which means it must have more mass than 10 g.

1kg=1000g

Wowowowowowowowow! Haha, I think ≈ 18 g sounds a bit better.