A block of mass 50 , kg can slide on a rough | vedclass

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(D) Given: Mass $m = 50 \, kg$,coefficient of friction $\mu = 0.6$,angle $	heta = 30^\circ$ with the vertical.For the block to just slide,the horizon

Vedclass · Accepted Answer

$294.3$

Vedclass · Answer

(D) Given: Mass $m = 50 \, kg$,coefficient of friction $\mu = 0.6$,angle $	heta = 30^\circ$ with the vertical.For the block to just slide,the horizontal component of the force must equal the limiting friction:$F \sin 	heta = f_L = \mu R$Here,$R$ is the normal reaction. Balancing vertical forces:$R + F \cos 	heta = mg$$R = mg - F \cos 	heta$Substituting $R$ into the friction equation:$F \sin 	heta = \mu (mg - F \cos 	heta)$$F \sin 	heta = \mu mg - \mu F \cos 	heta$$F (\sin 	heta + \mu \cos 	heta) = \mu mg$$F = \frac{\mu mg}{\sin 	heta + \mu \cos 	heta}$Substituting values $(g = 9.81 \, m/s^2)$:$F = \frac{0.6 	imes 50 	imes 9.81}{\sin 30^\circ + 0.6 \cos 30^\circ} = \frac{294.3}{0.5 + 0.6 	imes 0.866} = \frac{294.3}{0.5 + 0.5196} = \frac{294.3}{1.0196} \approx 288.6 \, N$.Note: If we use $g = 10 \, m/s^2$,$F = \frac{300}{1.0196} \approx 294.2 \, N$. Thus,option $D$ is the correct choice.

$A$ block of mass $50 \, kg$ can slide on a rough horizontal surface. The coefficient of friction between the block and the surface is $0.6$. The least force of pull acting at an angle of $30^\circ$ to the upward drawn vertical which causes the block to just slide is ........ $N$.

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