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Numerical Problem-A casting weighing 9 kN hangs freely from a rope which makes 2.5 turns ............

Question: 

A casting weighing 9 kN hangs freely from a rope which makes 2.5 turns round a drum of 300 mm diameter revolving at 20 rpm. The other end of the rope is pulled by a man. Taking μ = 0.25, determine (i) the force required by the man (ii) the power to raise the casting.

 

Answer: 

Given: W= T1= 9 kN =9000N, d= 0.3 m, N = 20 rpm , µ= 0.25

(i) Force reqd. by a man 

T2- force reqd. by man

As rope makes 2.5 turns,

Therefore angle of contact ,

θ =2.5x2π = 5 π rad.

We know that,

2.3 log {T1/T2} = µ θ = 0.25 x 5 π = 3.9275

log {T1/T2} = 3.9275/2.3 = 1.71 or T1/T2 = 51

T2 = 9000/51 = 176.47 N

(ii) Power to raise casting : 

As velocity of rope, v = πdN/60 = 3.14x0.3x20/60 = 0.3142 m/s

Power to raise casting = (T1-T2) x v = (9000-176.47) x 0.3142

                                     = 2.772 kW.