Physics Question on Torque

This forum made possible through the generous support of SDN members, donors, and sponsors. Thank you.

ieatshrimp24

Full Member
10+ Year Member
Joined
Feb 18, 2013
Messages
186
Reaction score
30
This question is from Nova Physics book Chapter 7 Question 13.

A meter stick of mass 0.6 kg sits on a fulcrum located at the 0.3-m mark at equilibrium. At the 0.0-m mark hangs a mass m. What is m?

The solution states that m(10 m/s^2)(0.3 m) - (0.6 kg)(10m/s^2)(0.2 m) = 0. My question is where did the 0.2 m come from? Thanks.
 
That is the distance from the fulcrum to center of mass of the meter stick.

That solution is the calculation of net torque in equilibrium - with torque being Distance x Force.
 
To supplement Cawolf's comments. Torque exerted by each object is = mgL where L can be thought as the distance between the center of mass of the object (under study) and the pivot point. It is important to first find a pivot point. Here it is given at 0.3 m mark. The center of mass of the meter stick is at 0.5m and L = 0.2m (0.5-0.3). The center of mass of the unknown mass is at 0.0m and L = 0.3m (0.3-0).
 
Top