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A sign is held in equilbrium by 7 vertically hanging ropes attached to the ceiling. If each rope has an equal tension of 53 Newtons, what is the mass of the sign in kg?
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A sign is held in equilbrium by 7 vertically hanging ropes attached to the ceiling. If each rope has an equal tension of 53 Newtons, what is the mass of the sign in kg?

User Guillaume Darmont
by
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1 Answer

3 votes

Answer:

37.86 kg

Step-by-step explanation:

The weight of sign board is equally divided on each rope. It means the tension in all the ropes is equal to the weight of the sign board in equilibrium condition.

Tension in each rope = 53 N

Tension in 7 ropes = 7 x 53 N = 371 N

Thus, The weight of sign = 371 N

Now, weight = m g

where m is the mass of sign.

m = 371 / 9.8 = 37.86 kg

User Peeyush Kumar
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