Chapter 7, Problem 61A

Gymnast Gracie of weight mg is suspended by a pair of vertical ropes attached to the ceiling.

a. In terms of Gracie’s weight, what is the tension in each rope?

b. If Gracie’s mass is 30 kg, show that the tension in each rope is 150 N.

To determine

The tension in each rope in terms of G’s weight, $mg$ , if she is suspended by the pair of vertical ropes attached to the ceiling.

Answer to Problem 61A

Tension in the ropes will be same and will be $T=\frac{mg}{2}$ .

Explanation of Solution

Introduction:

According to Newton’s second law of motion, if the force $F$ is applied to the mass $m$ , then the acceleration produced is given by,

  $a=\frac{F}{m}$

Let the tension in the ropes be $T$ and since the ropes are identical, thus the tension in the ropes will be same.

Now, considering the Newton’s second law of motion, since there is no motion in vertical direction, thus net force on G is zero. Hence it can be written that,

  $\begin{array}{c}T+T=mg\\ T=\frac{mg}{2}\end{array}$

Conclusion:

Hence, the tension in each rope is $\frac{mg}{2}$ .

To determine

To prove: The tension in each rope is $150\text{N}$ if G’s mass is $30\text{kg}$ .

Explanation of Solution

Given:

From part (a), the tension in each rope is $\frac{mg}{2}$ .

Calculation:

It is known that $g=10{\text{m/s}}^{\text{2}}$

Then, the tension in each rope can be written as,

  $\begin{array}{c}T=\frac{\left(30\right)\left(10\right)}{2}\\ T=150\text{N}\end{array}$

Conclusion:

Hence, the tension in each rope is $150\text{N}$ when the G’s mass is $30\text{kg}$ .