Q5E - University Physics with Modern Physics

Question

A picture frame hung against a wall is suspended by two wires attached to its upper corners. If the two wires make the same angle with the vertical, what must this angle be if the tension in each wire is equal to 0.75 of the weight of the frame? (Ignore any friction between the wall and the picture frame.).

Step-by-Step Solution

Answer

The angle of each wire from vertical is ${48}^{°}$ .

Step 1: Angle of wire with vertical

Given Data:

• The tension in each wire is $T_1=T_2=0.75W$  

 

The angle of each wire is calculated by writing the equations for vertical equilibrium for each wire. Equate the total vertical tension of both wires to the weight of the frame.

Step 2: Determine the expression for the vertical equilibrium of the picture frame

The equation for the vertical equilibrium of the frame is given as:

$T_1\cos \theta +T_2\cos \theta =W$ 

Here, $T_1$ and $T_2$ are the tensions in the wire and its value is 0.75 W for each wire, $\theta$ is the angle from vertical, and its value is same for both wires.

Step 3: Determine the angle of each wire from vertical

Substitute all the values in the above equation, and we get,

$$\begin{gathered} \left(0.75 \mathrm{W}\right)\cos \theta +\left(0.75 \mathrm{W}\right)\cos \theta =W \\ \cos \theta =\frac{1}{1.5} \\ \theta ={48}^{°} \end{gathered}$$  

Therefore, the angle of each wire from vertical is ${48}^{°}$ .