Two horizontal and opposite forces, each of magnitude
1.2 N, act at points A and B on the edge of the disc. A force of 6.0 N, at an
angle θ below the horizontal, acts on the midpoint C of a radial line
of the disc, as shown in Fig. 2.1. The disc has negligible weight and is in
equilibrium.
(i) State an expression, in terms of r,
for the torque of the couple due to the forces at A and B acting on the disc.
(ii) Friction between the disc and the pin is
negligible. Determine the angle θ.
(iii) State the magnitude of the force of the pin on the disc.
Solution:
(i) Torque = force × perpendicular
distance between the 2 forces
Torque = 1.2 × 2r=2.4r
Also Torque =(1.2r + 1.2r)
(ii) Since the disc is in equilibrium, So
Anticlockwise moment = Clockwise moment ……..(i)
Clockwise moment (due to forces at A and B) =
(1.2 × r) + (1.2 × r) = 2.4r……..(ii)
We Know,
Moment = force × perpendicular distance
The force at C acts at a distance r/2 from the pin.
Perpendicular component of force = 6.0 sinθ
Anticlockwise moment = 6.0 × (r / 2) × sinθ……..(iii)
From (i), (ii), (iii)
Anticlockwise moment = Clockwise moment
6.0 × (r / 2) × sinθ =
2.4r
θ= 53°