1. In procedure 3: suppose the ruler is asymmetrical, balancing at the 56.5 cm m

Question

1. In procedure 3: suppose the ruler is asymmetrical, balancing at the 56.5 cm mark. The ruler is now supported at the 42.4 cm mark, and a mass of 680 g is placed at the 11.4 cm mark. If the system is now in equilibrium, find the mass of the ruler.
mruler = g

2. In procedure 4: suppose a symmetrical meter stick of mass 218 g is free to rotate about a fulcrum at the 45.9 cm mark. A mass of 621 g is hanging at the 63.8 cm mark. Mass M is placed at the 18.0 cm mark, and the ruler balances. Find M.
M = g

Explanation / Answer

1. We know the center of gravity:

hence,

C.G. = {m*56.5 +680*11.4} /(m+680) = 42.4

m*56.5 +7752 = 42.4m +28832

m=1495 g.

2.

Moment by both the masses will cancel each other out.

621*(63.8 -45.9) =M*(45.9-18)

M = 398.19 g.

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