Question

10. The diagram below (Fig. 1.32) shows a uniform bar
supported at the middle point 0. A weight of 40 gf is
placed at a distance 40 cm to the left of point O.
How can you balance the bar with a weight of
80 gf?
40 gf
O
1
19
50 40
30 20
10
10 20 30
40
50 cm
Fig. 1.32
>​

Answer 1

Answer:

Given that 40 gf is applied at a distance of 40 cm from point O. Hence moment of force is 40×40=1600gfcm. Hence this moment of force must be balanced by 80gf of force. After equating we get 1600gfcm=x×80gfcm

x=1600/80cm

x=20cm.

Hence 80 gf must be applied at a distance of 20 cm. This is all because of clockwise and anticlockwise moments are equal

Answer 2

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