Question

The minimum acceleration with which a person can slide down a rope whose breaking strength is 60percent of his weight

Answer 1

ANSWER:

Given: Tension(T) = ? ; Inertia Force(ma) = minimum ; \text { $Gravity Force }(\mathrm{mg})=\frac{3}{5}$

Sol: T-mg+ma=0

        ma= mg-T

$\mathrm{ma}=\mathrm{mg}_{-} \frac{3}{5} m g$

$\frac{m a=m g-3 / 5 m g}{m}$

$\mathrm{a}=\mathrm{g}_{-} \frac{3}{5} g$

        $\mathrm{a}=\frac{5 g-3 g}{5}$

$\mathrm{a}=\frac{2}{5} g$

EXPLANATION:

This method works with D’Alembert’s principle. If we add the inertia force to rope and we can solve using dynamic equilibrium. Since he is accelerating downward, inertia force acts F = ma upward(opposite to direction of acceleration). Tension and gravity force remains act on the person. If a=0, Rope tension is equal to weight of the person. NOTE: If he accelerate up the rope, Inertia force acts towards downward and rope tension will goes higher than the weight of the person.

And if there is no acceleration, Newton’s second law can helps to drive the problem Therefore acceleration is Zero. $\Sigma F_{y}=m a_{y}$

  T-mg=m(-a)

  T-mg+ma=0