Question

In the figure shown, a force F is applied at the top
of a disc of mass 4 kg and radius 0.25 m. Find
maximum value of F for no slipping.​

Answer 1

Maximum value of F for no slipping is 72 N.

Given-

• Mass of the disc = 4 kg
• Radius of the disc = 0.25 m
• Coefficient of friction = 0.6

From the attached figure we can write that-

F + f = ma ----------(i)

Torque equation is -

FR-fR = Iα -----------(ii)

R (F-f) = mR²α/2

F-f = mRα/2

This is pure rolling so no slipping takes place.

αR = a

So,

F-f = ma/2 -------(iii)

By adding (i) and (ii) we get

2F = 3ma/2

F = 3ma/4

when f will be maximum then we get the maximum value of F.

μ = 0.6

f max = 0.6 × N

f max = 0.6 × mg

F + 0.6 mg = ma [from equation (i)]

a = F + 0.6mg/m

By substituting the value of a we get

F = 3ma/4

F = 3m/4 ( F+0.6mg/m)

F = 3F/4 + 1.8mg/4

F = 1.8 mg

F = 1.8 × 4 × 10 = 72 N