Chapter 2, Problem 39A

If a picture frame were supported by a pair of vertical wires, tension in each wire would be half the weight of the frame. When the frame is supported by wires at an angle, as shown below, how does the tension in each wire compare with that of vertical wires?

To determine

To compare: The tension in each inclined wires with that of each vertical wire.

Answer to Problem 39A

The tension in each inclined wire is greater than half the weight of picture or tension in each vertical wire.

Explanation of Solution

Given:

The given figure is shown below.

  

Formula used:

The expression for weight is,

  $W=mg$

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

Calculation:

Sketch the situation that shows the picture frame supported by vertical wires as shown below.

  

From the free-body diagram,

  $\begin{array}{l}T+T=W\\ 2T=mg\\ T=\frac{mg}{2}\end{array}$ ….. (1)

So, the tension in each vertical wire is, $T=\frac{mg}{2}$

Consider the angle of the inclined wire to the horizontal as $\alpha$ .

Sketch the situation that shows the picture frame supported by the inclined wires as shown below.

  

From the free-body diagram,

  $\begin{array}{l}T\sin \alpha +T\sin \alpha =W\\ 2T\sin \alpha =mg\\ T=\left(\frac{mg}{2}\right)\times \frac{1}{\sin \alpha }\end{array}$ ….. (2)

Tension in each inclined wire will be $T=\left(\frac{mg}{2}\right)\times \frac{1}{\sin \alpha }$

Compare equation (1) and (2).For any value of angle $\alpha$ between $0°$ and $90°$ , the value of $\sin \alpha$ is less than or equal to 1. So, the value of $\frac{1}{\sin \alpha }$ will be greater than 1. Thus, the tension in each inclined wire will be more than half the weight of picture or tension in each vertical wire.

Conclusion:

Thus, the tension in each inclined wire is greater than that of the tension in each vertical wire.