Finding Mass m2 for Equilibrium: Two Masses on a Meterstick

[k-rod AP 2010]

Two masses hang from a meterstick whose fulcrum is at 50.0cm. M1=.100kg acts at 20.0 cm, what value of mass m2 at position 70.0 puts the the system at equilibrium?

(.3m*(.1kg*9.8m/s2))=(.2m*(9.8m/s2*x kg))

.294 N*m=.2m(9.8*x N)

1.47 N=9.8*x N

x=.15 kg

Is this procedure correct? The distances in the first equation are from the center of mass, which is y they are different from the distances given in the problem.

 

[PhanthomJay]

Two masses hang from a meterstick whose fulcrum is at 50.0cm. M1=.100kg acts at 20.0 cm, what value of mass m2 at position 70.0 puts the the system at equilibrium?

(.3m*(.1kg*9.8m/s2))=(.2m*(9.8m/s2*x kg))

.294 N*m=.2m(9.8*x N)

1.47 N=9.8*x N

x=.15 kg

Is this procedure correct? The distances in the first equation are from the center of mass, which is y they are different from the distances given in the problem.

In this problem, the center of mass of the meter stick is located at the pivot point. While you can sum torques about any point, it is best to sum moments about a point where a force is unknown , since it avoids having to solve for the unknown force at that point. Your procedure is nevertheless correct ( you could have saved a step by not converting mass to weight, but in general, it is a good idea to do so ).

 

[donutz610]

Looks right

 

[Paulie323]

your process looks good to me but as PhantomJay said that is one step i would reccomend as well

 

[k-rod AP 2010]

ok that is what i was hoping to hear, thanks a lot guys. torque is confusing but once you get the center of mass stuff down its ok,,,