Home / IGCSE Physics (0625) 1.5.2 Turning effect Paper 3

IGCSE Physics (0625) 1.5.2 Turning effect Paper 3

Question

Fig. 3.1 shows the forces acting on a uniform balanced beam. The beam is pivoted at its centre.

(a) Calculate the moment of the 5.2 N force about the pivot and show that its value is close to 30 Ncm

(b) The beam is balanced. Calculate force P.
force P = ……………………………………………… N

Answer/Explanation

Answer:

(a) (moment of force =) force × (perpendicular) distance of force from pivot 
5.2 × 6.0
31.2

(b) (sum of) clockwise moment(s) = (sum of) anticlockwise moment(s)
P × 2.0 + 8.1 × 2.0 = 5.2 × 6.0 OR 31.2 OR answer from (a)
P = (31.2 – 16.2) ÷ 2.0 OR 15 ÷ 2.0
7.5 (N)

Question

A 50 cm rule is balanced at its mid-point. A force of 8.0 N acts at a distance of 10 cm from one end of the rule.

 

a) Calculate the moment of the 8.0 N force about the pivot. Give the unit.
moment = ……………………………………………………..
unit = ……………………………………………………..

b) Another force acts at a point 10 cm from the pivot. It makes the rule balance.On Fig. 2.1, draw an arrow to show the position and direction of this force.

Answer/Explanation

Answer:
(a)
moment = force × distance from pivot in any form
(distance of force from pivot = (25 – 10)=) 15 (cm)
8 × 15
120
N cm

(b) arrow giving clockwise moment
arrow drawn 10 cm from pivot by eye

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