Chapter 2, Problem 22A

In the diagram below, identical blocks are suspended by ropes, each rope having a scale to measure the tension (stretching force) in the rope. Rank the scale readings from greatest to least.

To determine

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

Answer to Problem 22A

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

Explanation of Solution

Given:

The given diagram showing the identical blocks suspended by ropes is,

  

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 four blocks A, B, C and D. The mass of each of the four block is denoted by $m$ .

Sketch the free body diagram of identical blocksas shown below.

  

Find the tension in the rope for case A .

For Equilibrium of forces,

  $\begin{array}{c}\text{Tension}\text{in}\text{the ropes}=\text{Weight}\text{of}\text{the}\text{block}\text{A}\\ T+T+T=mg\\ 3T=mg\\ T_{\text{A}}=\frac{mg}{3}\end{array}$

Find the tension in the rope for case B.

For Equilibrium of forces,

  $\begin{array}{c}\text{Tension}\text{in}\text{the ropes}=\text{Weight}\text{of}\text{the}\text{block}\text{B}\\ T_{\text{B}}=mg\end{array}$

Find the tension in the rope for case C.

For Equilibrium of forces,

  $\begin{array}{c}\text{Tension}\text{in}\text{the ropes}=\text{Weight}\text{of}\text{the}\text{block}\text{C}\\ T+T=mg\\ 2T=mg\\ T_{\text{C}}=\frac{mg}{2}\end{array}$

Find the tension in the rope for caseD.

For Equilibrium of forces,

  $\begin{array}{c}\text{Tension}\text{in}\text{the ropes}=\text{Weight}\text{of}\text{the}\text{block}\text{D}\\ T_{\text{D}}=mg\end{array}$

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

Conclusion:

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