Question

A block of mass 10 kg is held at rest against a rough vertical wall (μ= 0.5 ) under the action a force F as shown in figure. The minimum value of F required for it is (g=10 ms-2)

Answer 1

Answer:The minimum value should be : 89.6 N.

Explanation : If u left the box then it would tend to slide downward so the friction force should be in upward direction.

Now.for minimum value of F the block should box 's weight should be just balanced .

Let F be required value so , At equilibrium -

Net upward force = Net downward force

( 0.5 × F Sin30° ) + F Cos30° = 100

On putting values u will get answer

Answer 2

Answer:

plzz refer to the attachment for answer and explanation