Q63P - University Physics with Modern Physics

Question

In a repair shop a truck engine that has mass 409 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 722 N, what are the tensions in cables B and C?

                                   

Step-by-Step Solution

Answer

The tensions in cables B and C are 4963 N and 1197 N respectively.

Step 1: Identification of given data

The given data can be listed below as:

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

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 below:

Here, $T_A$ is the tension of cable A, $T_B$ is the tension of cable B, and $T_c$ is the tension of cable C.

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

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

$$\begin{gathered} T_A=T_c\sin \theta \\ {T}_c=\frac{T_A}{\sin \theta } \end{gathered}$$ 

Here, $T_A$ is the tension of cable A, $T_C$ is the tension of cable C, and $\theta$ is the angle made by cable C

Substitute the values in the above equation.

$$\begin{gathered} T_C=\frac{722 N}{\sin 37.1°} \\ =\frac{722 N}{0.603} \\ =1197N \end{gathered}$$

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

$T_B=T_{C}cos\theta +T_D$ 

Here, $T_D$ 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:

$T_B=T_{C}cos\theta +mg$

Here, m is the mass of the truck engine and g is the acceleration due to gravity.

Substitute the values in the above equation.

$$\begin{gathered} T_B=\left(1197 N\right)\cos 37.1°+\left(409 kg\right)\left(9.8 m/s^2\right) \\ =\left(1197 N\right)\left(0.797\right)+4008.2 kg·m/s^2 \\ =\left(954.009 N\right)+4008.2 kg·m/s^2\times \frac{1N}{1kg·m/s^2} \\ =4963 N \end{gathered}$$

Thus, the tensions in cables B and C are 4963 N and 1197 Nrespectively.