Question

rope of length L is pulled by constant force f what is the tension in the Rope at a distance x from the end where the force is applied

Answer 1

Let the mass of the Rope be m then the acceleration of the rope is f/m
Let's take a part of the rope which is of x length from the point of application of force then drawing free body diagram on that part we see that the forces on that part is the tension by the remaining part and the force applied. Since the acceleration of the rope is f/m hence acceleration of that part will also be f/m. Using Newton's second law of motion the net external force applied on the body is equal to its mass multiplied bye its acceleration., so the net external force applied on the system that is the part which we have taken is f - the force applied by the remaining part of the rope which is equal to the tension let's say t hence f- t is equal to mass of that part multiplied by its acceleration now the mass of the part is (x/L) *m.
f-t=(mx/L) *(f/m)
f-t= fx/L
t =, f(L-x) /x this is the answer
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Answer 2

Answer:

Explanation:

Let m be the mass per unit length of the rope.

for part PB, f−T=mxa.....(1)

for part PA, T=m(l−x)a...(2)

(1)+(2),f=mxa+mla−mxa=mla :   a= f/ml

now using (2), T=m(l−x).f/ml

= (l-x)f/l