A block of weight 1 ,N rests on an inclined p | vedclass

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(D) Given weight of the block $W = mg = 1 \,N$.When the block is about to move up the inclined plane,the forces acting on it are:$1$. The applied forc

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$\mu \cos 	heta + \sin 	heta$

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(D) Given weight of the block $W = mg = 1 \,N$.When the block is about to move up the inclined plane,the forces acting on it are:$1$. The applied force $F$ acting up the plane.$2$. The component of weight acting down the plane,$mg \sin 	heta$.$3$. The limiting friction force $f_L$ acting down the plane,which is $f_L = \mu N_{normal} = \mu mg \cos 	heta$.For the body to just move up the plane,the applied force must balance both the gravitational component and the friction force:$F = mg \sin 	heta + f_L$$F = mg \sin 	heta + \mu mg \cos 	heta$Substituting $mg = 1 \,N$:$F = 1 \cdot \sin 	heta + \mu \cdot 1 \cdot \cos 	heta$$F = \sin 	heta + \mu \cos 	heta$Thus,the minimum force required is $\mu \cos 	heta + \sin 	heta$.

$A$ block of weight $1 \,N$ rests on an inclined plane of inclination $\theta$ with the horizontal. The coefficient of friction is $\mu$. The minimum force that has to be applied parallel to the inclined plane to make the body just move up the plane is

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