Class: Name: ( ) Date:
Marks: / 16
Quiz 2-5 (Lower tier) (Book 2 Chapter 5)
A Multiple-choice questions (4 marks)
1 Two forces are acting on a rod pivoted at a fixed point as shown below.
Find the net moment about the pivot.
A 8.43 N m B 26.0 N m
C 68.4 N m D 91.7 N m
2 When an object is in equilibrium,
(1) The net force acting on the object is zero.
(2) The net moment about its centre of gravity is zero.
(3) The net moment about any point on its edge is zero.
A (1) and (2) only B (1) and (3) only
C (2) and (3) only D (1), (2) and (3)
B Short questions (12 marks)
3 Describe a method to experimentally locate the centre of gravity of a plane object. (4 marks)
New Senior Secondary Physics at Work (Second Edition) 1 Chapter Quiz 2-5 (Lower tier)
Oxford University Press 2015
4 A rod PQ of length 24 cm and weight 5 N is hung by two strings A and B. Its centre of
gravity is located at the middle. A ball of weight 3 N is hung on the rod by a light string C as
shown below. The system is in equilibrium.
(a) What is the tension in string C? Explain briefly. (2 marks)
(b) Find the tension in string B. (2 marks)
(c) The horizontal and vertical components of the tension in string A are respectively Tx and
Ty. Find their values. (4 marks)
End of quiz
New Senior Secondary Physics at Work (Second Edition) 2 Chapter Quiz 2-5 (Lower tier)
Oxford University Press 2015
Solutions to Quiz 2-5 (Lower tier)
1 C
Take the clockwise direction as positive. By τ = Fl sin θ,
Net moment = –8 × 6 × sin 60° + 10 × (6 + 5) = 68.4 N m
2 D
When an object is in equilibrium, the net force acting on it is zero, and the net moment about any
point on the object is also zero.
3
Suspend the object from a point about which the object can rotate freely. 1A
Draw a vertical line passing through the point when it reaches equilibrium. 1A
Repeat the above steps with another point which is not on the line. 1A
The centre of gravity is located at the intersection point of the two lines drawn. 1A
4
(a) Since the ball is in equilibrium, the net force acting on it is zero. Hence, the
tension TC in string C is equal to the weight of the ball, 1A
i.e. 3 N. 1A
(b) Let TB be the tension in string B.
Take moment about point P.
Anticlockwise moment = clockwise moment 1M
TB × 0.24 × sin (90° – 40°) = 3 × (0.24 – 0.04) + 5 ×
TB = 6.53 N 1A
(c) Since the rod is in equilibrium, its horizontal and vertical net forces are zero.
Tx = TB sin 40° 1M
= 6.53 × sin 40°
= 4.20 N 1A
Ty = W + TC – TB cos 40° 1M
= 5 + 3 – 3.26 × cos 40°
=3N 1A
New Senior Secondary Physics at Work (Second Edition) 3 Chapter Quiz 2-5 (Lower tier)
Oxford University Press 2015
