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Grade 12Mechanics

A rope of mass 5kg is moving vertically in vertical vertical position with an upward force of 100N acting at the upper end and adownward force of 70N.The tension at the midpmidpoint of the rope is?Do explain the concept also.

Profile image of agam  goel
11 Years agoGrade 12
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4 Answers

Profile image of Sumit Majumdar
11 Years ago
Dear student,
The tension at the midpoint would be given by:
100 N – 70 N = 30 N
Which would be acting at the centre of mass/mid point of the rope, assuming the mass is uniformly distributed. The 100 N is acting at the centre of mass of the rope, in the upward direction and 70 N is acting at the centre of mass in the downward direction.
Regards
Sumit
Profile image of Shivang
8 Years ago
​​When you are given one upward and one downward force on a rope of some mass, and you have to find the tension at the mid- point, then you can simply average the 2 forces. So, the correct answer is (100+70)/3 = 85N
Profile image of Shivang
8 Years ago
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(100+70)/2* 3 was by mistake
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Profile image of Nishant gautam
7 Years ago

The acceleration of the rope is (100-70)/5= 6m/s.s

For the lower half part of the rope(of mass 2.5 kg), two forces act, Tension(T) and 70N.

Therefore, T-70 = 2.5 x 6, [ F=m.a] , giving T=85 N.