Solve the following :

Question:

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

Solution:

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)

Now,

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

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

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

$F=17.5 \mathrm{~N}$

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