Draw a free body diagram. The string supports the weight if the meter stick and the two point masses. From vertical force equilbrium:
19.6 N = (0.7 + 0.1 + m)g
0.7 + 0.1 + m = 19.6/9.8 = 2.0 kg
m = 1.2 kg
Next, to a moment balance to find out where to place m. Let x be the unknown distance from the 0 end of the stick. Take moments about any point - I will choose the point where the string is attached. The 0.7 kg mass is 35 cm left of that point, and the Center of Mass is 10 cm to the right. The 1.2 kg mass is x-40 cm to the right of the string.
The moment balance says that
0.7*35 = 0.1*10 + 1.2*(x-40)
24.5 = 1.0 + 1.2x - 48.0
1.2x = 71.5
x = 59.6 cm
That is 59.6 - 40 = 19.6 cm to the right of the string
A .100 kg meter stick is supported at its 40.0 cm mark by a string attached to the ceiling. A .700 kg mass hangs vertically from the 5.00 cm mark. A mass, m, is attached somewhere on the meter stick to keep it horizontal and in rotational and translational equilibrium. If the tension in the string attached to the ceiling is 19.6, determine (a) the value of m and (b) its point of attachment on the stick. (answer: m=1.2 kg, d=19.6 m, and position= 59.6 cm) I don't know where to start. Any help would be helpful!
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