Question

A homogeneous cylinder 3 m diameter and weighing 400 N is resting on two rough inclined surface’s shown.If the angle of friction is 15o.Find couple C applied to the cylinder that will start it rotating clockwise.

Answer 1

Given : Angle of friction is 15o
μ = tan 15 = 0.2679
Radius = 1.5 m
To find : Couple C
Solution: 

F₁=μN₁=0.2679N₁ ………………….(1)

F₂=μN₂=0.2679N₂ …………..……….(2)
Assuming the body is in equilibrium
ΣFx=0
F₁cos40+N₁sin40+F₂cos60-N₂sin60=0
N₁(0.2679cos40 + sin40)+N₂(0.2679cos60-sin60)=0 …………(3)

ΣFy=0

-F₁sin40+N₁cos40+F₂sin60+N₂cos60-400=0
N₁(-0.2679sin40+cos40)+N₂(0.2679sin60+cos60)=400 ………..(4)
Solving (3) and (4)
N₁=277.4197 N and N₂=321.3785 N

Substituting N1 and N2 in (1 and 2)
F₁=0.2679 x 277.4197 = 74.3344 N
F₂=0.2679 x 321.3785 = 86.1131 N …….(5)
C is the couple required to rotate the cylinder clockwise
C=F₁ x r + F₂ x r
= 240.6712 Nm(clockwise) (r=1.5 m)(From 5)
The couple C required to rotate the cylinder clockwise is 240.6712 Nm(clockwise)