Q63P - University Physics with Modern Physics

Question

In a repair shop a truck engine that has mass $\mathbf{409}\mathbf{ }\mathbf{\text{kg}}$ is held in place by four light cables (Fig. P5.63). Cable A is horizontal, cables B and D are vertical, and cable C makes an angle of $\mathbf{37}\mathbf{.}\mathbf{1}\mathbf{°}$ with a vertical wall. If the tension in cable A is $\mathbf{722}\mathbf{ }\mathbf{\text{N}}$, what are the tensions in cables B and C?

Step-by-Step Solution

Answer

The tensions in cables B and C are $4963 \text{N}$ and $1197 \text{N}$respectively.

Step 1: Identification of given data

The given data can be listed below as:

• The mass of the truck engine is $\text{m}=409 \text{kg}$.
• The angle made by cable C is $\theta =37.1°$.
• The tension in cable A is ${\text{T}}_{\text{A}}=722 \text{N}$.

Step 2: Significance of the tension

The tension is described as the pulling force that is transmitted in an axial direction with the help of a string or pulley. It is also referred to as the pair of the action and the reaction force.

Step 3: Determination of the tensions in cables B and C

The free body diagram of the system has been drawn belo  

Here, ${\text{T}}_{\text{A}}$is the tension of cable A, ${\text{T}}_{\text{B}}$  is the tension of cable B, and ${\text{T}}_{\text{C}}$ is the tension of cable C.

All the forces in the and in the direction is zero.

The equation of the forces acting in the x-direction is expressed as:

$$\begin{gathered} {\text{T}}_{\text{A}}={\text{T}}_{\text{C}}\text{sinθ} \\ {\text{T}}_{\text{C}}=\frac{{\text{T}}_{\text{A}}}{\text{sinθ}} \end{gathered}$$

Here, ${\text{T}}_{\text{A}}$ is the tension of cable A, ${\text{T}}_{\text{C}}$ is the tension of cable C, and $\mathrm{\theta }$ is the angle made by cable C.

Substitute the values in the above equation.

$\begin{array}{rcl}{\text{T}}_{\text{C}}& =& \frac{722 \text{N}}{\sin 37.1°}\\ & =& \frac{722 \text{N}}{0.603}\\ & =& 1197 \text{N}\\ & & \end{array}$

The equation of the forces acting in the y-direction is expressed as:

${\text{T}}_{\text{B}}={\text{T}}_{\text{C}}\text{cosθ}+{\text{T}}_{\text{D}}$

Here, ${\text{T}}_{\text{D}}$ is the tension of cable D.

As the weight of the engine is the tension of the cable D, then the above equation can be expressed as:

${\text{T}}_{\text{B}}={\text{T}}_{\text{C}}\text{cosθ}+\text{mg}$

Here, $\mathrm{m}$ is the mass of the truck engine and $\mathrm{g}$ is the acceleration due to gravity.

Substitute the values in the above equation.

$\begin{array}{rcl}{\text{T}}_{\text{B}}& =& \left(1197 \text{N}\right)\cos 37.1°+\left(409 \text{kg}\right)\left(9.8 {\text{m/s}}^{\text{2}}\right)\\ & =& \left(1197 \text{N}\right)\left(0.797\right)+4008.2 \text{kg}·{\text{m/s}}^{\text{2}}\\ & =& \left(954.009 \text{N}\right)+4008.2 \text{kg}·{\text{m/s}}^{\text{2}}\times \frac{1 \text{N}}{1 \text{kg}·{\text{m/s}}^{\text{2}}}\\ & =& 4963 \text{N}\\ & & \end{array}$

Thus, the tensions in cables B and C are $4963 \text{N}$ and $1197 \text{N}$respectively.