Question

17. What is the frictional torque (MR) in Nm, if normal force 1.2 KN, diameter of shaft (d) is
60 mm and the coefficient of friction () is 0.03?
WS84A062
a) 1.08 Nm
c) 1.02 Nm
b) 1.06 Nm
d) 1.00 Nm​

Answer 1

Answer:

Option (a) 1.08 Nm

Step-by-step explanation:

Given

Normal force N = 1.2 KN = 1200 N

Frictional force F = μN

                             = 0.03 × 1200 N

                             = 36 N

Diameter of the shaft = 60 mm = 0.06 m

Radius of the shaft = 0.06/2 = 0.03 m

Torque due to frictional force

τ = rF

  = 0.03 × 36 Nm

Thus the frictional torque is 1.08 Nm

  = 1.08 Nm