Question

A uniform metre scale of weight 50 gf is balanced at 30 cm mark when weights of 80 gf and 60 gf act at 5 cm and 45 cm mark respectively. What force must be applied at 20 cm mark to balance the metre scale?

Answer 1

Answer:

Let a wt. of 80gf be placed at a distance x from 40cm mark. Clock wise moments = Anticlock wise moment (80 × x) + 50 × (50 – 40) = 100 × (40 – 5) 80x +500 = 3500 80x = 3500 – 500 = 3000 x = 3000/80 = 37.5cm ∴ A is at 40 + 37.5 = 77.5cm

Explanation:

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Answer 2

Given :

Weight of scale = 50gf

Length of scale = 100cm

The scale is balanced at = 30 cm

80gf weight is placed at = 5cm

60gf weight is placed at = 45cm

Distance of weight 60gf from pivot = (45-30)=15

Distance of weight 80gf from pivot = (35-5)=25

Distance of weight that is to be placed at 20cm = (30-20)=10

To find : Force to be applied at 20cm mark

Solution :

We know,

Clockwise moment = anticlockwise moment

Therefore, 60 x 15 + 50 x 20 = 80 x 25 + 8 x 10

=> 900 + 1000 = 2000 + 10F

=> 1900 = 2000 + 10F

=>F = 10

Therefore the force that is to be kept at 20cm mark is 10gf

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