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Two ropes support a load of 478 kg. The two ropes are perpendicular to each other, and the tension in the first rope is 2.2 times that of the second rope. Find the tension in the second rope. The acceleration of gravity is 9.8 m/s 2 . Answer in units of N.

Sagot :

hamzaahmeds

Answer:

T₂ = 1937.68 N

Explanation:

First, we will calculate the weight of the object:

[tex]W = mg = (478\ kg)(9.81\ m/s^2)\\W = 4689.18\ N[/tex]

Now, we will calculate the resultant tension in the ropes. Since the ropes are perpendicular. Therefore,

[tex]T = \sqrt{T_1^2+T_2^2}\\[/tex]

where,

T = Resultant Tension

T₁ = Tension in rope 1

T₂ = Tension in rope 2

According to the given condition tension in the first rope is 2.2 times the tension in the second rope:

T₁ = 2.2 T₂

Therefore

[tex]T = \sqrt{(2.2T_2)^2 + T_2^2}\\\\T = 2.42T_2[/tex]

Now, the weight of the object must be equal to the resultant tension for equilibrium:

[tex]T = W\\2.42T_2 = 4689.18\ N\\\\[/tex]

T₂ = 1937.68 N

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