Chapter 5, Problem 36P

(a)

To determine

The tension in the three strands of the cord.

Answer to Problem 36P

The tensions in the three strands of the cord are $T_1=31.5\text{N}$, $T_2=37.5\text{N}$ and $T_3=49\text{N}$.

Explanation of Solution

The tension forces acting on the wires due to the load hanging from the wires is as shown below.

Write the expression for the tension $T_3$ in the wire connected to load as.

  $T_3=m_{1}g$                                                                                                         (I)

Here, $T_3$ is the tension, $m_1$ is the mass of the first load and $g$ is the acceleration due to gravity.

The tension force $T{1}_{}$ and $T_2$ are resolved in the horizontal and vertical direction as.

In static equilibrium the forces acting on the particular point are balanced.

Write the expression for the force balancing in the X-direction as.

  $T_1\cos {\theta }_1=T_2\cos {\theta }_2$

Rearrange the above equation for $T_1$ as.

  $T_1=\frac{T_2\cos {\theta }_2}{\cos {\theta }_1}$                                                                                                (II)

Here, $T_1$ is the tension in wire $1$, ${\theta }_1$ is the angle made by wire 1 with the roof, ${\theta }_2$ is the angle made by wire 2 and the roof and $T_2$ is the tension in wire $2$.

Write the expression for the force balancing in Y-direction as.

  $T_1\sin {\theta }_1+T_2\sin {\theta }_2=T_3$                                                                                 (III)

Substitute $\left(\frac{T_2\cos {\theta }_2}{\cos {\theta }_1}\right)$ for $T_1$ in equation (II).

  $\left(\frac{T_2\cos {\theta }_2}{\cos {\theta }_1}\right)\sin {\theta }_1+T_2\sin {\theta }_2=T_3$

Rearrange the above equation for $T_2$ as.

  $T_2\cos {\theta }_2\tan {\theta }_1+T_2\sin {\theta }_2=T_3$                                                               (IV)

Conclusion:

Substitute $5.00\text{kg}$ for $m_1$ and $9.80\text{m}/{\text{s}}^{\text{2}}$ for $g$ in equation (I).

  $\begin{array}{c}{T}_3=\left(5.00\text{kg}\right)\left(9.80\text{m}/{\text{s}}^{\text{2}}\right)\\ =49\text{kg}.\text{m}/{\text{s}}^{\text{2}}\\ =49\text{N}\end{array}$

Substitute $\left(49\text{N}\right)$ for $T_3$, $50°$ for ${\theta }_2$ and $40°$ for ${\theta }_1$ in equation (IV).

  $\begin{array}{c}{T}_2\cos 50°\tan 40°+T_2\sin 50°=\left(49\text{N}\right)\\ T_2\left(0.643\right)\left(0.84\right)+T_2\left(0.766\right)=49\text{N}\\ T_2=\frac{49\text{N}}{1.306}\\ =37.5\text{N}\end{array}$

Substitute $37.5\text{N}$ for $T_2$, $50°$ for ${\theta }_2$ and $40°$ for ${\theta }_1$ in equation (II).

  $\begin{array}{c}{T}_1=\frac{\left(37.5\text{N}\right)\cos 50°}{\cos 40°}\\ =\frac{\left(37.5\text{N}\right)\left(0.643\right)}{\left(0.766\right)}\\ =31.478\text{N}\\ \approx 31.5\text{N}\end{array}$

Thus, the tensions in the three strands of the cord are $T_1=31.5\text{N}$, $T_2=37.5\text{N}$ and $T_3=49\text{N}$.

(b)

To determine

The tension in the three strands of the cord.

Answer to Problem 36P

The tensions in the three strands of the cord are $T_1=113\text{N}$, $T_2=56.5\text{N}$ and $T_3=98\text{N}$.

Explanation of Solution

The tension forces acting on the wires due to the load hanging from the wires is as shown below.

Write the expression for the tension $T_3$ in the wire connected to load as.

  $T_3=m_{2}g$                                                                                                       (V)

Here, $T_3$ is the tension, $m_2$ is the mass of the second load and $g$ is the acceleration due to gravity.

The tension force $T{1}_{}$ and $T_2$ are resolved in the horizontal and vertical direction as.

In static equilibrium the forces acting on the particular point are balanced.

Write the expression for the force balancing in the X-direction as.

  $T_1\cos \varphi =T_2$                                                                                                 (VI)

Here, $T_1$ is the tension in wire $1$ and $T_2$ is the tension in wire $2$.

Write the expression for the force balancing in Y-direction as.

  $T_1\sin \varphi =T_3$                                                                                                (VII)

Conclusion:

Substitute $10.0\text{kg}$ for $m_2$ and $9.80\text{m}/{\text{s}}^{\text{2}}$ for $g$ in equation (V).

  $\begin{array}{c}{T}_3=\left(10.0\text{kg}\right)\left(9.80\text{m}/{\text{s}}^{\text{2}}\right)\\ =98\text{kg}.\text{m}/{\text{s}}^{\text{2}}\\ =98\text{N}\end{array}$

Substitute $\left(98\text{N}\right)$ for $T_3$ and $60°$ for $\varphi$ in equation (VII).

  $\begin{array}{c}{T}_1\sin 60°=\left(98\text{N}\right)\\ T_1=\frac{\left(98\text{N}\right)}{0.866}\\ =113.1\text{N}\\ \approx 113\text{N}\end{array}$

Substitute $\left(113\text{N}\right)$ for $T_1$ and $60°$ for $\varphi$ in equation (VI).

  $\begin{array}{c}\left(113\text{N}\right)\cos 60°=T_2\\ T_2=56.5\text{N}\end{array}$

Thus, the tensions in the three strands of the cord are $T_1=113\text{N}$, $T_2=56.5\text{N}$ and $T_3=98\text{N}$.