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A uniform meter rule of weight 0.9N is suspended horizontally by two vertical loops of thread A and B placed 20cm and 30cm from its ends respectively. find the distances from the centre of the rule at which a 2N weight must be suspend a

A=0.09m​

Respuesta :

onyebuchinnaji

The  distance from the centre of the rule at which a 2N weight must be suspend from A is 29.3 cm.

Distance from the center of the meter rule

The distance from the centre of the rule at which a 2N weight must be suspend from A is calculated as follows;

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  20 A  (30 - x)↓      x         ↓      20 cm  B 30 cm

                       2N              0.9N

Let the center of the meter rule = 50 cm

take moment about the center;

2(30 - x) + 0.9(x)(30 - x) = 0.9(20)

(30 - x)(2 + 0.9x) = 18

60 + 27x - 2x - 0.9x² = 18

60 + 25x - 0.9x² = 18

0.9x² - 25x - 42 = 0

x = 29.3 cm

Thus, the  distance from the centre of the rule at which a 2N weight must be suspend from A is 29.3 cm.

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snehashish65

Answer:

The answer is 29 cm and  43.5 cm

Explanation:

The 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:

By using formula:

∑τ = Iα

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

18 Ncm − 2 N (x − 70 cm) = 0

2 N (x − 70 cm) = 18 Ncm

x − 70 cm = 9 cm

x = 79 cm

therefore ,the distance from the center is |50 cm − 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:

∑τ = Iα

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

-27 Ncm − 2 N (x − 20 cm) = 0

2 N (x − 20 cm) = -27 Ncm

x − 20 cm = -13.5 cm

x = 6.5 cm

Therefore the distance from the center is |50 cm − 6.5 cm

i.e. 43.5 cm

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