EK General Physics Book #60

This forum made possible through the generous support of SDN members, donors, and sponsors. Thank you.
Get help with your application

Use all the free resources available to you from SDN: articles, guides, expert advising, forums discussions, and school research.

desi0chick

Full Member
10+ Year Member
Advertisement - Members don't see this ad
I'm not sure if there's a typo for this question but for number 60, regarding torques how is the solution 3kg? The key states the equation setup as: (0.2)(3)=(0.3)(x), I just don't get how they got 0.3m, if it's a meter stick shouldn't the other end be 0.8? 😕
 
1zuq1l.png


here's the problem, the answer is B.
 
Advertisement - Members don't see this ad
First of all, recognize that the pivot point or fulcrum is at 0.2m. Since the board is 1m long, the center of mass of the board is at 0.5m. So you can imagine that there's a weight at 0.5m, which is 0.3m away from the pivot point (this is the moment arm for the board's mass). Now all you have to do is balance the left torques with the right torques and you get 0.2(3) = 0.3(x). x = 2
 
Gravity always acts on the centre of mass. You can think of the plank as weightless with all of its weight concentrated as a block hanging off its centre.