As shown in the figure,a block of mass 10 , k | vedclass

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(B) The forces acting on the block are the gravitational force $Mg$ downwards,the normal force $N$ upwards,the applied force $F$ at an angle $30^{\cir

Vedclass · Accepted Answer

$25.2$

Vedclass · Answer

(B) The forces acting on the block are the gravitational force $Mg$ downwards,the normal force $N$ upwards,the applied force $F$ at an angle $30^{\circ}$,and the frictional force $f$ opposing the motion.Resolving the force $F$ into components,we have $F \cos 30^{\circ}$ horizontally and $F \sin 30^{\circ}$ vertically upwards.The vertical equilibrium condition is $N + F \sin 30^{\circ} = Mg$,so $N = Mg - F \sin 30^{\circ}$.Given $M = 10 \, kg$ and $g = 10 \, ms^{-2}$,$N = 100 - F \sin 30^{\circ} = 100 - \frac{F}{2}$.The block just starts to move when the horizontal component of the applied force equals the limiting friction,$F \cos 30^{\circ} = \mu_{s} N$.Substituting the values,$F \cos 30^{\circ} = 0.25 	imes (100 - \frac{F}{2})$.$F \frac{\sqrt{3}}{2} = 0.25 	imes (100 - 0.5F)$.$0.866F = 25 - 0.125F$.$0.866F + 0.125F = 25$.$0.991F = 25$.$F = \frac{25}{0.991} \approx 25.22 \, N$.

As shown in the figure,a block of mass $10 \, kg$ lying on a horizontal surface is pulled by a force $F$ acting at an angle of $30^{\circ}$ with the horizontal. For $\mu_{s} = 0.25$,the block will just start to move for the value of $F$ in $N$ is: [Given $g = 10 \, ms^{-2}$]

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