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A uniform metre rule of weight 0.9 N is suspended horizontally by two vertical loops of thread A and B placed at 20cm and 30cm from its ends respectively. Find the distances from the centre of the rule at which a 2N weight must be suspended;
(a) to make loop A become slack
(b) to make loop B slack​

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Respuesta :

MathPhys

Answer:

(a) 29 cm

(b) 43.5 cm

Explanation:

(a) when loop A is slack, there are three forces acting on the metre rule.

-0.9 N at 50 cm mark

T at 70 cm mark

-2 N at x

Taking the sum of the torques about B:

βˆ‘Ο„ = IΞ±

(-0.9 N) (50 cm βˆ’ 70 cm) + (-2 N) (x βˆ’ 70 cm) = 0

18 Ncm βˆ’ 2 N (x βˆ’ 70 cm) = 0

2 N (x βˆ’ 70 cm) = 18 Ncm

x βˆ’ 70 cm = 9 cm

x = 79 cm

The distance from the center is |50 cm βˆ’ 79 cm| = 29 cm.

(b) when loop B is slack, there are three forces acting on the metre rule.

-0.9 N at 50 cm mark

T at 20 cm mark

-2 N at x

Taking the sum of the torques about A:

βˆ‘Ο„ = IΞ±

(-0.9 N) (50 cm βˆ’ 20 cm) + (-2 N) (x βˆ’ 20 cm) = 0

-27 Ncm βˆ’ 2 N (x βˆ’ 20 cm) = 0

2 N (x βˆ’ 20 cm) = -27 Ncm

x βˆ’ 20 cm = -13.5 cm

x = 6.5 cm

The distance from the center is |50 cm βˆ’ 6.5 cm| = 43.5 cm

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