Question

a uniform rope of length 5 metres is on a smooth horizontal surface .it is being pulled by a horizontal force of 20 Newtons at one end the ratio of tension at the distance of 2 metres from force end to tension at a distance of 2 metres from free end is​

Answer 1

Answer:T1/T2 = 3/2

Explanation:

You can use general formula

Tension = F(L-x)/L

where x is length of rope from force end

Answer 2

Answer:

T1/T2 = 3/2 .

Explanation:

Since, the length of the rope is L let at point x a small tension is given in a direction opposite to the acceleration of the rope.

For x length of rope = Mx/L.

Again, F-T=(M/L - x)(F/M).

F-T=Fx/L.

So, tension T = F-Fx/L.

Tension = F(L-x)/L .

Now, putting the values we will get that T1 = 20(5-2)/5 = 12N.

Again T2 will be F(L-x1)/L  = 20(5-(5-2))/5 = 20(5-3)/5 = 8N.

So, the ratio will be T1/T2 = 12/8 = 3/2.