Calculating Torque of a 30cm Ruler with 10cm Extension | Ruler Torque Question

[ph4all]

Homework Statement

A 30 cm ruler has a mass of 5g. Mark the pivot point and calculate the torque for the following situation:

Ruler extends 10cm out from desk

Homework Equations

T= r x F

F= |F|sinΘ

F= m Fg

The Attempt at a Solution

F= mFg
= (0.005kg) (9.8 m/s^2)
=0.049 N

T= r x F
=|r||F|sinΘ
=(0.10m)(0.049N) sin 90 degrees
=0.00049 Nm

The only thing I am not sure on is that only a third of the ruler is hanging off the desk so I am not sure if I have divide the mass.

 

[nate808]

you are correct to consider the distribution of mass along the ruler. In this situation, since only a third of the ruler is hanging off the desk, you can consider the mass to be concentrated at the center of mass of that third. This means that the distance from the pivot point to the center of mass of the hanging portion would be 10 cm/3 = 3.33 cm. Therefore, the torque would be:

T = r x F
= |r||F|sinΘ
= (0.033m)(0.049N)sin 90 degrees
= 0.00016 Nm

It is important to accurately consider the distribution of mass in order to calculate the correct torque.

 

[Architeuthis Dux]

Your calculations for the torque are correct. The mass of the ruler should not be divided as the entire ruler is still exerting a force on the pivot point, even though only a portion of it is extending out from the desk. However, it would be helpful to label the pivot point and specify the direction and magnitude of the force being exerted on the ruler in order to fully understand the situation. Additionally, it would be beneficial to include units for all values in your calculations. Overall, your approach and solution to calculating the torque of the ruler are sound.