Question

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

Answer 1

Answer: 2g/5

Explanation:

T = mg - ma

T= breaking force= 60×mg/100

m(g-a) = 60×mg /100

a= 10g - 6g /10

a=4g/10

a= 2g/5

Answer 2

Answer:

The minimum acceleration is $\frac{2}{5}g$.

Explanation:

Given that :

Inertia force $(ma)=$ minimum

Force due to gravity $(mg)=$$\frac{3}{5}$

$T-mg+ma=0\\ma=mg-T$

$ma=mg-\frac{3}{5}mg$

Dividing throughout with $m$ : $a=g-\frac{3}{5}g$

$a=\frac{5g-3g}{5}\\ a=\frac{2g}{5}$

D'Alembert's concept is used in this procedure.

• We can solve this utilising dynamic equilibrium if we add the inertia force to the rope. Because he is accelerating downward, the inertia force acts upward $F = ma$ (opposite to direction of acceleration). The human is still under the influence of tension and gravity.

• If $a=0$, the rope tension is equal to the person's weight.

• NOTE: If he accelerates up the rope, the inertia force acts downward, causing the rope tension to exceed the person's weight.