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.

Question

Sumit Majumdar · Answer

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

Shivang · Answer

​​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

Shivang · Answer

….............................................................. (100+70)/2 * 3 was by mistake ..................................................................

Nishant gautam · Answer

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.

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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.

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.

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