Question

With what minimum acceleration can an athlete climb a rope whose breaking strength is one third of its weight

Answer 1

Answer:

Explanation:

acc. to question mg>t

mg = weight of the athelete

acc. to laws of motion : mg - t =ma

we also know t= 1/3mg

so ,

mg-mg/3=ma

a= 3mg-mg/3m (using L.C.M)

a=2mg/3m

a=2g/3

Answer 2

Answer:

Explanation:

Sum of the forces acting = ma. These types of questions can also be easily solved by drawing a free body diagram as shown below. If we add the inertia force (ma) to the FBD then we can solve it using dynamic equilibrium (all forces balance).

Edit: This method is D’Alembert’s principle.

Since he is accelerating downward, there is an inertia force F = ma upward (opposite the direction of acceleration). The only other two forces acting on the fireman are the rope tension upward and his weight = mg downward.