A block of mass 1 kg is at rest on a horizontal table. The coefficient of static friction between the block and the table is 0.5. If $g = 10ms^{- 2}$, then the magnitude of the force acting upwards at an angle of 60${^\circ}$ from the horizontal that will just start the block moving is

Question

Dr. Abubacker · Accepted Answer

(b) $R + F\sin{60{^\circ} = mg}\ $ or $R = mg - \frac{\sqrt{3}F}{2}$
$F\cos 60{^\circ} = f = \mu R$
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or $\frac{F}{2} = 0.5\left\lbrack 1 	imes 10 - \frac{\sqrt{3}F}{2} ightbrack$
or $F + \frac{\sqrt{3}F}{2} = 10$ or $F = \frac{20}{2 + \sqrt{3}}$
or $F = \frac{20}{3.732}$ N= 5.36 N

Q. A block of mass 1 kg is at rest on a horizontal table. The coefficient of static friction between the block and the table is 0.5. If $g = 10ms^{- 2}$, then the magnitude of the force acting upwards at an angle of 60${^\circ}$ from the horizontal that will just start the block moving is

Explanation

Reviewed by Dr. Abubacker

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