Tw masses, (m1 = 600g and m2 = 150g) are hanging from a meter stick. m1 is locat

Question

Tw masses, (m1 = 600g and m2 = 150g) are hanging from a meter stick. m1 is located at the 10cm mark and m2 is at the 7cm mark. If the center-of-mass of the combination (made up of the meter stick and the two point masses) is at the 30-cm mark of the meter stick, what is the mass of the meter stick?

I don't understand how to solve this... I've tried a number of different methods, and all return innacurate ansers. Note: The solution is 300g

Here is what I have tried:

ΣFy=0

-Mrg-m1g-2g+N=0

(Where Mr is the mass of the meter stick and N is a normal force pointed in the posetive-y direction at the same x-coordinate as the center of mass)

ΣFτ=0

0-(20cm)m1g-(40cm)m2g+0=0

(Where the first 0 represents the torque from gravity and the last 0 represents the torque from the normal force)

I feel so stuck... I'd love some advice on this one!

Explanation / Answer

Let us say the mass of stick = M, and position of COM = 50 .
Then COM equation.
m1*x1 + m2*x2 + M*50 = (m1 + m2 + M)*x(com)
m1 =600
m2 =150
x1= 10
x2= 7
x(com) =30
putting it we have
M = 772.5 g

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