Question 12
(a)
Define the moment of
a force. [1]
(b)
A thin disc of radius r is supported at its
centre O by a pin. The disc is supported so that it is vertical. Three forces
act in the plane of the disc, as shown in Fig. 2.1.
Fig. 2.1
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.
[1]
(ii)
Friction between the disc and the pin is negligible.
Determine the angle θ. [2]
(iii)
State the magnitude of the force of the pin on the disc. [1]
[Total: 5]
Reference: Past Exam Paper – November 2017 Paper 22 Q2
Solution:
(a)
The moment of a force is defined as the product of the
force and the perpendicular of the line of action of the force to a
point.
(b)
(i)
{Torque = force × perpendicular distance between the 2 forces
Torque = 1.2 × 2r}
Torque = 2.4r or (1.2 × 2r) or (1.2r + 1.2r)
(ii)
{Since the disc is in equilibrium,
anticlockwise moment = clockwise moment
Clockwise moment (due to forces at A and B) = (1.2
× r) + (1.2 × r) = 2.4r
Recall: 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
{Anticlockwise moment = clockwise moment}
6.0 × r / 2 × sinθ = 2.4r
θ =
53°
(iii)
{For equilibrium,
Resultant torque / moment = 0 (rotational equilibrium,) – this was
already seen in the previous parts
Resultant force = 0 (translational equilibrium)
The 1.2 N forces cancel each other. So, we need
another 6.0 N force for equilibrium.}
Force = 6.0 N
Please explain how you got 6*r/2*sin(theta) in b ii).
ReplyDeleteIt's already explained above. Re-read the 'recall' part in (b)(ii)
DeleteI would like to thank you for all your work
ReplyDeleteHow is the answer of b(iii) 6N??
ReplyDeleteAs explained above, the resultant force should be zero.
Deletethe 12N forces already cancel out each other.
For cancel the 6N force present, we need another 6.0 N