Question

A horizontal force applied on a body on a rough horizontal surface produces an acceleration a if cofficient of friction between the body snd surface is m reduced to m/3 the acceleration increase by 2 times. Find the value of m

Answer 1

As per given condition, a2=2(a1); μ1=m; μ2=m/3

Normal reaction = R = mg

frictional force = f = μR = μmg

FBD in first case,

F - f1 = ma1

F - μ1mg = ma1

F - m(mg) = ma1

F - m²g = ma1

F = m²g+ma1 ...(i)

FBD in second case,

F - f2 = ma2

F - μ2(mg) = ma2

F - m²g/3 = ma2

F - m²g/3 = m(2a1)

F - m²g/3 = 2ma1

F = m²g/3+2ma1 ...(ii)

From (i) & (ii) we get,

m²g+ma1 = m²g/3+2ma1

3m²g+3ma1 = m²g+6ma1

2m²g = 3ma1

2mg = 3a

m = 3a/2g