A block of mass 10, kg is kept on a rough inc | vedclass

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Question

(A) Given: Mass $m = 10\, kg$,angle $	heta = 45^\circ$,coefficient of static friction $\mu = 0.6$,downward force $F_{down} = 3\, N$,$g = 10\, ms^{-2}

Vedclass · Accepted Answer

$32$

Vedclass · Answer

(A) Given: Mass $m = 10\, kg$,angle $	heta = 45^\circ$,coefficient of static friction $\mu = 0.6$,downward force $F_{down} = 3\, N$,$g = 10\, ms^{-2}$.For the block not to move downward,the upward force $P$ must balance the downward components of gravity and the applied force,minus the maximum static friction force acting upwards.The downward force component due to gravity is $mg \sin 	heta = 10 	imes 10 	imes \sin 45^\circ = 100 	imes \frac{1}{\sqrt{2}} \approx 70.71\, N$.The normal force is $N = mg \cos 	heta = 10 	imes 10 	imes \cos 45^\circ = 100 	imes \frac{1}{\sqrt{2}} \approx 70.71\, N$.The maximum static friction force is $f_{max} = \mu N = 0.6 	imes 70.71 \approx 42.43\, N$.To prevent downward motion,the minimum upward force $P$ must satisfy: $P + f_{max} \geq mg \sin 	heta + 3\, N$.$P + 42.43 \geq 70.71 + 3$.$P \geq 73.71 - 42.43 = 31.28\, N$.Rounding to the nearest integer,the minimum force $P$ is $32\, N$.

Similar Questions

$A$ body takes $1\frac{1}{3}$ times as much time to slide down a rough inclined plane as it takes to slide down an identical but smooth inclined plane. If the angle of the inclined plane is $45^{\circ}$,the coefficient of friction is:

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