A uniform metal beam AB is pivoted on a vertical wall at A. The beam is supported by a wire joining end B to the wall at C. The beam makes an angle of 30° with the wall and the wire makes an angle of 60° with the wall.

Question 2

(a) Define moment of a force. [1]

(b) An arrangement for lifting heavy loads is shown in Fig. 4.1.

2.8 m

Fig. 4.1

A uniform metal beam AB is pivoted on a vertical wall at A. The beam is supported by a wire joining end B to the wall at C. The beam makes an angle of 30° with the wall and the wire makes an angle of 60° with the wall.

The beam has length 2.8 m and weight of 500 N. A load of 4000 N is supported from B. The tension in the wire is T. The beam is in equilibrium.

(i) By taking moments about A, show that T is 2.1 kN. [2]

(ii) Calculate the vertical component Tv of the tension T. [1]

(iii) State and explain why Tv does not equal the sum of the load and the weight of the beam although the beam is in equilibrium. [2]

Reference: Past Exam Paper – November 2015 Paper 22 Q4

Solution:

(a) Moment of a force is defined as the product of the force and the perpendicular distance from the pivot to the line of action of the force.

(b)

(i)

{Since the beam is in equilibrium, the sum of clockwise moment is equal to the sum of anti-clockwise moment at any point.

The beam is uniform, so its weight acts at the centre, a distance of (2.8 / 2 =) 1.4 m from its ends.

Clockwise moment due to weight of beam = 500 × 1.4 × sin 30°

Clockwise moment due to 4000N load = 4000 × 2.8 × sin 30°

Clockwise moment about A = (500 × 1.4 × sin 30°) + (4000 × 2.8 × sin 30°)

Anticlockwise moment about A = T × 2.8}

 

(4000 × 2.8 × sin 30°) + (500 × 1.4 × sin 30°) = T × 2.8       

giving T = 2100 (2125) N                                                       

(ii) (Tv = 2100 cos 60° =) 1100 (1050) N

(iii)

There is an upward (vertical component of a) force at A such that

upward force at A + Tv = sum of downward forces /weight + load/4500 N

{This force could be a contact force such as friction …}