Solve the following :


Repeat part (a) of the problem 6 if the push is applied horizontally and not parallel to the incline.


Since the block is just to move up the incline so frictional force will act in downward direction.

Since, block is in equilibrium.

$N=m g \cos 30^{\circ}+F \sin 30^{\circ}$   .........(perpendicular to incline)

$F \cos 30^{\circ}=m g \sin 30^{\circ}+f f$  ..........(along the incline)


$F \cos 30^{\circ}=m g \sin 30^{\circ}+\mu N$

$F \cos 30^{\circ}=m g \sin 30^{\circ}+\mu\left(m g \cos 30^{\circ}+F \sin 30\right)$

$F \frac{\sqrt{3}}{2}=2 \times 10 \times \frac{1}{2}+0.2\left(2 \times 10 \times \frac{\sqrt{3}}{2}+\frac{F}{2}\right)$

$F=17.5 \mathrm{~N}$

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