A block of mass 2 kg is pushed against a rough vertical wall with a force of 40 N, coefficient of static friction being 0.5. Another horizontal force of 15 N, is applied
on the block in a direction parallel to the wall. If the block will move, then its direction would be

15^o with 15 N force

53^o with 15 N force

45^o with 15 N force

75^o with 15N force

Solution

53^o with 15 N force

Now, When a force of 40N is applied, it will be balanced by the normal reaction from the wall

The net force = $\sqrt{20^{2} + 15^{2}}$

= 625

F = 25 N

This is greater than the 20N (friction), therefore block will move

∴ Direction

$tanθ = \frac{20}{15}$

= $\frac{4}{3}$

∴ $θ = \tan^{- 1} (\frac{4}{3})$

So, the block will move at an angle 53° with the 15 N force.

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