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• # IMAT 2013 Q54 Physics and Mathematics

• ### Nicolas Sanchez

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June 21, 2021 at 9:44 am 968 Replies 159150 XP
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A uniform beam, 3.0m long, of weight 100N has a 300N weight placed 0.50m from one end. The beam is suspended by a string 1.0m from the same end.
A diagram of the weights placed on the beam is given below:

[See diagram below]

How far from the other end must a weight of 80N be placed for the beam to be balanced?

A. 0.75m

B. 2.25m

C. 1.25m

D. 1.875m

E. 0.125m

• ### MANISH MULCHANDANI

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July 15, 2021 at 10:51 pm 984 Replies 59071 XP
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You have to use concept of center of mass. You can also solve it by considering movement of beam like clockwise and anti clockwise.

Let us consider x as distance for 80N from other end to be balanced.

Concept of center of mass stats that-

m1x1 + m2x2= m3x3

Let m1= 100N and x1= 0.5m.

The 100N weight will be balanced by string with length (1-0.5)m.

Let m2= 80 N and x2= (2-x)

The 80N weight will be balanced by (2-x)m because its placed at that end.

The 300N weight will be balanced by string with length (1-0.5)m because its placed on that end.

So, Let m3= 300N and x3= 0.5m

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