A box of mass 2 kg is placed on an inclined | vedclass

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(C) The block has a tendency to slide downwards,so the friction force $f$ acts in the upward direction along the incline.For the box not to slide,the

Vedclass · Accepted Answer

$32.7$

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(C) The block has a tendency to slide downwards,so the friction force $f$ acts in the upward direction along the incline.For the box not to slide,the maximum static friction must balance the component of weight acting downwards along the incline.$f_{max} \geq mg \sin 	heta$Since $f_{max} = \mu N$,where $N$ is the normal reaction force.The force $F$ is applied perpendicular to the incline,so the normal reaction $N$ is:$N = mg \cos 	heta + F$Substituting $N$ into the friction equation:$\mu(mg \cos 	heta + F) \geq mg \sin 	heta$$\mu mg \cos 	heta + \mu F \geq mg \sin 	heta$$\mu F \geq mg \sin 	heta - \mu mg \cos 	heta$$F \geq \frac{mg(\sin 	heta - \mu \cos 	heta)}{\mu}$Given: $m = 2 \ kg$,$g = 10 \ ms^{-2}$,$	heta = 30^{\circ}$,$\mu = 0.2$.$F \geq \frac{2 	imes 10 	imes (\sin 30^{\circ} - 0.2 	imes \cos 30^{\circ})}{0.2}$$F \geq \frac{20 	imes (0.5 - 0.2 	imes 0.866)}{0.2}$$F \geq \frac{20 	imes (0.5 - 0.1732)}{0.2}$$F \geq \frac{20 	imes 0.3268}{0.2}$$F \geq 100 	imes 0.3268 = 32.68 \ N$Rounding to one decimal place,the minimum value of $F$ is $32.7 \ N$.

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