Question 149: Laws of Motion

Question

Two blocks A and B of masses m
A
= 1 kg and m
B
= 3 kg are kept on the table as shown in figure. The coefficient of friction between A and B is 0.2 and between B and the surface of the table is also 0.2. The maximum force F that can be applied on B horizontally, so that the block A does not slide over the block B is :
[Take g = 10 m/s
2
]
![](https://imagex.cdn.examgoal.net/3rsa0w2w9jwzjluo5/a529b6a2-5f20-4153-81dc-10654e39e3e0/12ab8260-9089-11e9-86c1-73f7f991b766/file-3rsa0w2w9jwzjluo6.png)

Accepted Answer

M
A
= 1 kg, M
B
= 3 k
${\mu _{AB}} = 0.2$
${\mu _B} = 0.2$
F
max
= (M
A
+ M
B
) ÃÂ 0.2 ÃÂ 10 + (M
A
+ M
B
) ÃÂ 0.2 ÃÂ 10
= 4 ÃÂ 2 + 4 ÃÂ 2 = 16

Answer

8 N

Answer

16 N

Answer

40 N

Answer

12 N

Laws of Motion

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