A store sign of mass 4.25 kg is hung by two wires that each make an angle of θ = 42.4° with the ceiling.What is the tension in each Wire?
Think:
Resolving the tension in each wire into horizontal and vertical components and we calculate the net force along x axis and y axis using Newton’s second law to obtain the tension in each wire.
Research:
The weight of the sign board will be directed downwards, tension in the wire will be resolved in to \(T \sin \theta\) and \(T \cos \theta\). Equating the forces along \(x\) axis and \(y\) axis we can evaluate the tension value.
Sketch:
The free body diagram for the system is shown below,
Simplify:
Given data
Mass of the sign board is, \(m=4.25 \mathrm{~kg}\)
Angle made by the each wire with the ceiling is, \(\theta=42.4^{\circ}\)
Let the tensions in the two wires be \(T_{1}, T_{2}\).
Using the free body diagram, we can calculate the net force along \(x\) axis as,
$$ \begin{aligned} \sum F_{x} &=m a_{x} \\ T_{1} \cos \theta-T_{2} \cos \theta &=0 \\ \left(T_{1}-T_{2}\right) \cos \theta &=0 \\ T_{1}-T_{2} &=0 \\ T_{1}=T_{2} \cdots \cdots(1) \end{aligned} $$
Similarly, we calculate the net force along \(y\) axis as,
$$ \begin{gathered} \sum F_{y}=m a_{y} \\ T_{1} \sin \theta+T_{2} \sin \theta-m g=0 \\ \sin \theta\left(T_{1}+T_{2}\right)=m g \\ T_{1}+T_{2}=\frac{m g}{\sin \theta} \ldots \ldots \text { (2) } \end{gathered} $$
Calculate:
Using the relation in equation (1), it will be easier to simplify the problem and the value of tension can be obtained.
Taking equation (2) for further simplification,
$$ \begin{aligned} T_{1}+T_{2} &=\frac{m g}{\sin \theta} \\ 2 T_{1} &=\frac{m g}{\sin \theta} \quad\left[\text { Since }, T_{1}=T_{2}\right] \\ T_{1} &=\frac{(4.25 \mathrm{~kg})\left(9.81 \mathrm{~m} / \mathrm{s}^{2}\right)}{2\left(\sin 42.4^{\circ}\right)} \\ &=30.9 \mathrm{~N} \end{aligned} $$
Thus, the tension in each wire is, \(30.9 \mathrm{~N}\)