Question 155: Laws of Motion

Question

A block of mass 10 kg is kept on a rough inclined plane as shown in the figure. A force of 3 N is applied on the block. The coefficient of static friction between the plane and the block is 0.6. What should be the minimum value of force P, such that the block does not move downward? (take g = 10 ms
$$-$$2
)

Accepted Answer

At equilibrium,
N = mgcos45
o
. . . . . . (1)
3 + mgsin45
o
= P + $$\mu $$N
$$ \Rightarrow $$   3 + $${{mg} \over {\sqrt 2 }}$$  =  P + 0.6 $$\left( {{{mg} \over {\sqrt 2 }}} ight)$$
$$ \Rightarrow $$   3 + $${{100} \over {\sqrt 2 }}$$ = P + 0.6 $$ 	imes $$ $${{100} \over {\sqrt 2 }}$$
$$ \Rightarrow $$    P = 3 + $${{40} \over {\sqrt 2 }}$$
$$ \Rightarrow $$   P = 3 + 20 $${\sqrt 2 }$$
$$ \Rightarrow $$   P = 31.28 $$ \simeq $$ 32 N

Answer

32 N

Answer

18 N

Answer

23 N

Answer

25 N

Laws of Motion

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