A meter stick is supported by a pivot at its center of mass. Assume that the met

Question

A meter stick is supported by a pivot at its center of mass. Assume that the meter stick is uniform and that the center of mass is at the 50 cm mark. a) If a mass m1 = 80 g is suspended at the 30 cm mark, at which cm mark would a mass m2 = 110 g need to be suspended for the system to be in equilibrium?b) If a mass m1=80g is suspended at the 25cm mark,and a mass m2 =110g is suspended at the 60 cm mark, from what cm mark would a mass m3 = 45 g need to be suspended for the system to be in equilibrium?

Explanation / Answer

m1 = 80g
m2 = 110 g

Let m2 be suspended at x m from the Center of the mass
Now For Equilibrium,
m1 * (50cm - 30cm ) = m2 * x
80 * 20 = 110 * x
x = 14.55 cm

cm Mark at which m2 need to be suspended = 50 cm + x
= 50 + 14.55 cm
= 64.55 cm

(b)
m1 = 80 g
m2 = 110 g
m3 = 45 g

Let m3 be suspended at x m from the Center of the mass -
For Equilibrium ==>
m1 * (50 - 25) - m2 * (60 - 50) = m3 *x
80 * 25 - 110 * 10 = 45 * x
x = 20 cm

cm Mark at which m3 need to be suspended = 50 cm + x
= 50 + 20 cm
= 70 cm

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