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(D) To push the body up the inclined plane, the applied force $F_1$ must overcome both the component of gravitational force $mg \sin 	heta$ and the m

Vedclass · Accepted Answer

$3$

Vedclass · Answer

(D) To push the body up the inclined plane, the applied force $F_1$ must overcome both the component of gravitational force $mg \sin 	heta$ and the maximum static friction force $f_{max} = \mu mg \cos 	heta$ acting downwards. Thus, $F_1 = mg \sin 	heta + \mu mg \cos 	heta$.To prevent the body from sliding down the inclined plane, the applied force $F_2$ acts upwards along with the maximum static friction force $f_{max} = \mu mg \cos 	heta$ to balance the gravitational component $mg \sin 	heta$ acting downwards. Thus, $F_2 = mg \sin 	heta - \mu mg \cos 	heta$.Given $	an 	heta = 2\mu$, we can write $\sin 	heta = 2\mu \cos 	heta$. Substituting this into the expressions for $F_1$ and $F_2$:$F_1 = mg(2\mu \cos 	heta) + \mu mg \cos 	heta = 3\mu mg \cos 	heta$$F_2 = mg(2\mu \cos 	heta) - \mu mg \cos 	heta = \mu mg \cos 	heta$Therefore, the ratio is:$\frac{F_1}{F_2} = \frac{3\mu mg \cos 	heta}{\mu mg \cos 	heta} = 3$.

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