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

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Aicharya · Accepted Answer

$ \begin{aligned} & F_1=m g \sin 	heta+\mu m g \cos 	heta \ & F_2=m g \sin 	heta-\mu m g \cos 	heta \end{aligned} $ Given that, $F_1=3 F_2$ or $\left(\sin 45^{\circ}+\mu \cos 45^{\circ}ight)$ $ =3\left(\sin 45^{\circ}-\mu \cos 45^{\circ}ight) $ On solving, we get $\mu=0.5$ $ 	herefore \quad N=10 \mu=5 $ $	herefore$ Answer is 5 . Analysis of Question Question is simple. Only one has to take care of direction and magnitude of friction. !

A block is moving on an inclined plane making an angle $45^{\circ}$ with the horizontal and the coefficient…

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