Question

Find the range of the force applied for which the body remains at rest.
[REFER TO THE ATTACHMENT FOR YOUR ANSWER]

Kindly help me with this.
Thanks =]​

Answer 1

Force by gravity

= mgsin30°

= 10 x 10 x 1/2

= 50 N

Force by friction

= N x ųs

= mg x 0.6

= 100 x 0.6

= 60 N

= 50 N (Given: Static friction)

As the force by friction is greater than the force of gravity, here the force of friction is acting on the object.

Range = 0N to 50 N

Answer 2

Heya, here's your answer!

To find the force at which it remains at rest, we will first need to find the components of the forces being applied.

Consider the rough surface as a plane.

The forces acting parallel to the plane are mg cos 30° to the left, F cos 30° to the right, and frictional force f to the right.

The forces acting perpendicular to the plane are mg sin 30° downwards, F sin 30° downwards and normal reaction N upwards.

At equilibrium,

N = F sin 30° + mg sin 30°

= F/2 + 50

and,

mg cos 30° = F cos 30° + uN (for max frictional force)

or,

mg cos 30° = F cos 30° (for minimum frictional force)

50 root 3 = F root 3 + 0.6(F/2 + 50)

50 root 3 = F root 3 + 0.3F + 30

50×1.7 = 1.7F + 0.3F + 30 (root 3 ≈ 1.7)

85 = 2F + 30

F = 55/2 = 27.5 N

50 root 3 = F root 3

F = 50 N

So the range of F is [27.5, 50]