A uniform meter stick of mass $M$ has a half-filled can of fruit juice of mass $m$ attached to one end. The meter stick and the

Question

2 years ago

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A uniform meter stick of mass $M$ has a half-filled can of fruit juice of mass $m$ attached to one end. The meter stick and the

can balance at a point 21.2 cm from the end of the stick where the can is attached. When the balanced stick-can system is suspended from a scale, the reading on the scale is 2.58 N. Calculate the mass of the meter stick.

Physics

1 answer:

notka56 [123] · Answer 1 · 2022-03-01T22:20:19+03:00

Answer:

111.6 g

Explanation:

Given

m + M = 2.58 / 9.8

= 0.2632 kg

When the can of fruit juice is balance from scale , we get following relation

M x ( 50 - 21.2 ) = m x 21.2 ( balancing the torque due to weight of scale and can about the balancing point )

M x 28.8 = 21.2 m

= 21.2 ( 0.2632 - M )

= 5.58 - 21.2 M

M ( 28.8 + 21.2 ) = 5.58

M = .1116 kg

= 111.6 g