Question

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✍ QUESTION OF THE DAY ✍


➡ Three balls A,B and C are kept in a straight line. The separation between A and C is 1m,and B is placed at the midpoint between them.The masses of A,B and C are 100 g, 200 g and 300g respectively. Find the net gravitational force on (a) A ,(b) B and (c) C.


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Answer 1

Answer:

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Hey mate plzz refer to the attachment

Answer 2

Answer:

A ------ 0.50 m -------- B -------- 0.50

-------- C

Ma = 0.100 kg

Mc = 0.300 kg

Mb = 0.200 kg

Net gravitational force Fa on A is

towards right side.

Fa= G Ma Mb/ 0.502 + G Ma Mc/

1.02

= 6.67 * 10-11 * 0.100

[ 0.200/ 0.502 + 0.300 /1] N

= 7.34 * 10-12 N.

Net gravitational force Fc on C is

towards B, left side and its

magnitude is equal to Fa ie., 7.34 * 10-12

Net gravitational force on B is Fb & is

towards C

Fb = G Mb [Mc- Ma 1/ 0.502

= 6.67 * 10-11 * 0.200 * 0.200 * 4 =

1.07 * 10-11N

4.0 (78)

0

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