Chapter 2, Problem 24A

Percy does gymnastics, suspended by one rope in A and by two ropes in positions B, C, and D. Rank the tensions in the ropes from greatest to least.

To determine

To rank:The tensions in the ropes from greatest to least.

Answer to Problem 24A

The rank of the tensions in the ropes from greatest to least is $T_{\text{A}}=T_{\text{D}}>T_{\text{C}}>T_{\text{B}}$ .

Explanation of Solution

Given:

The given figure is shown below.

  

Formula used:

The expression for the weight is,

  $W=mg$

Here, $m$ is the mass and $g$ is the acceleration due to gravity.

Calculation:

Consider the weight of gymnastic is denoted by $W$ .

Consider the tension in each rope is denoted by $T$ .

Sketch the free body diagram of gymnastics for each case as shown below.

  

Consider the angle of the rope to the horizontal for case C as ${\theta }_{\text{C}}=60°$ .

Consider the angle of the rope to the horizontal for case D as ${\theta }_{\text{D}}=30°$ .

The tension in the rope for case A is,

  $\begin{array}{l}T=W\\ T=mg\end{array}$

The tension in the rope for case B is,

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

The tension in the rope for case C is,

  $\begin{array}{l}T\sin {\theta }_{\text{C}}+T\sin {\theta }_{\text{C}}=W\\ T\sin 60°+T\sin 60°=mg\\ 2T\sin 60°=mg\\ T\left(\frac{\sqrt{3}}{2}\right)=\frac{mg}{2}\\ T=\frac{mg}{\sqrt{3}}\end{array}$

The tension in the rope for case D is,

  $\begin{array}{l}T\sin {\theta }_{\text{D}}+T\sin {\theta }_{\text{D}}=W\\ T\sin 30°+T\sin 30°=mg\\ 2T\sin 30°=mg\\ T\left(\frac{1}{2}\right)=\frac{mg}{2}\\ T=mg\end{array}$

By comparing the values of the tension in the ropes $T_{\text{A}}=T_{\text{D}}>T_{\text{C}}>T_{\text{B}}$ .

Conclusion:

Thus, the rank of the tensions in the ropes from greatest to least is $T_{\text{A}}=T_{\text{D}}>T_{\text{C}}>T_{\text{B}}$ .