JEE MAIN - Physics (2015 (Offline) - No. 23)
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Given in the figure are two blocks $$A$$ and $$B$$ of weight 20 N and 100 N, respectively. These are being pressed against a wall by a force $$F$$ as shown. If the coefficient of friction between the blocks is 0.1 and between block $$B$$ and the wall is 0.15, the frictional force applied by the wall on block $$B$$ is :
$$120$$ $$N$$
$$150$$ $$N$$
$$100$$ $$N$$
$$80$$ $$N$$
Explanation
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Assuming both the blocks are stationary
From the F.B.D of both the blocks we can find,
$$N=F$$
f1 = 20 N
f2 = 100 + 20 = 120 N
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Considering those two blocks as one system and due to equilibrium $$f=120N$$
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