Question 268: Laws of Motion

Question

A block is moving on an inclined plane making an angle $$45^\circ $$ with the horizontal and the coefficient of friction is $$\mu $$. The force required to just push it up the inclined plane is 3 times the force required to just
prevent it from sliding down. If we define N = 10 $$\mu $$, then N is

Accepted Answer

The pushing force $${F_1} = mg\sin 	heta  + f$$
$$	herefore$$ $${F_1} = mg\sin 	heta  + \mu mg\cos 	heta  = mg(\sin 	heta  + \mu \cos 	heta )$$
The force required to just prevent it from sliding down
$${F_2} = mg\sin 	heta  - \mu N = mg(\sin 	heta  - \mu \cos 	heta )$$
Given, $${F_1} = 3{F_2}$$
$$	herefore$$ $$\sin 	heta  + \mu \cos 	heta  = 3(\sin 	heta  - \mu \cos 	heta )$$
$$	herefore$$ $$1 + \mu  = 3(1 - \mu )$$ [$$\because$$ $$\sin 	heta  = \mu \cos 	heta $$]
$$	herefore$$ $$4\mu  = 2$$
$$	herefore$$ $$\mu  = 0.5$$
$$	herefore$$ $$N = 10\mu  = 5$$

Laws of Motion

A block is moving on an inclined plane making an angle $45^\circ with the horizontal and the coefficient of friction is \mu . The force required to just push it up the inclined plane is 3 times the force required to just prevent it from sliding down. If we define N = 10 \mu $, then N is

Integer / Numeric Answer Type