Question

A block of mass 5 kg is placed on a horizontal surface with coefficient of friction =0.2, then the maximum and minimum value of force f for which the block remain at rest

Answer 1

answer : minimum force = 9.8N

and maximum force = 10N

explanation : let a force is applied on a block at an angle x with horizontal.

see figure,

at equilibrium,

horizontal force = frictional force

or, Fcosx = μN......(1), where N is normal reaction between block and surface.

again, vertical force + normal reaction acting on body = weight of body

or, Fsinx + N = mg

or, N = mg - Fsinx .......(2)

from equations (1) and (2),

Fcosx = μ(mg - Fsinx)

or, F = μmg/(cosx + μsinx)

denominator of force = (cosx + μsinx)

we know from trigonometric functions, -√(a² + b²) ≤ asinx + bcosx ≤ √(a² + b²)

so, -√(1 + μ²) ≤ (cosx + μsinx) ≤ √(1 + μ²)

so, to get minimum force ,use (cosx + μsinx) = √(1 + μ²)

so, minimum force , F = μmg/√(1 + μ²)

now putting values of μ , m and g

so,F = 0.2 × 5kg × 10m/s²/√(1 + 0.2²)

= 10/√(1.04)

= 9.8N

now maximum force when angle made between force and horizontal line is zero.

so, Fcos0° = μmg

or, F = μmg = 0.2 × 5 × 10 = 10N