Chapter 2.3, Problem 2.65P

A cable loop of length 1.5 m is placed around a crate. Knowing that the mass of the crate is 300 kg, determine the tension in the cable for each of the arrangements shown.

To determine

c

The tension in the cable for the given arrangement of the cable loop.

Answer to Problem 2.65P

The tension in the cable $T_{AE}$ is $\underset{\_}{\text{2450}\text{N}}$.

Explanation of Solution

The length of the cable loop is $1.5\text{m}$, the mass of the crate is $300\text{kg}$.

Above figure represents the free body diagram of the cable loop which is placed around a crate. The forces $T_{AB}$ and $T_{BE}$ are of same magnitude.

Write the expression for the weight acting on the system.

$W=mg$ (I)

Here, $W$ is the weight, $m$ is the mass, and $g$ is the gravitational acceleration.

Figure 2 represents the force triangle of the given free body diagram to find the tension.

Refer from the figure P2.65And Write the expression for finding the cosine angle.

$\begin{array}{c}\cos \alpha =\frac{\text{Adjacent side }}{\text{Hypotenuse}}\\ =\frac{CD/2}{EB}\end{array}$ (II)

Write the expression for the total length of loop.

$L=BE+AC+CD+DB+EA$ (III)

From the force triangle it is shown that $T_{AE}=T_{BE}$

Refer the figure 2.

According to the law of sines,

$T_{AE}\sin \alpha =\frac{1}{2}2943\text{N}$ (IV)

Conclusion:

Substitute $300\text{Kg}$ for $m$ and $9.8\text{m}/{\text{s}}^2$ for $g$ in equation (I) to find $W$.

$\begin{array}{c}W=\left(300\text{kg}\right)\left(9.8\text{m}/{\text{s}}^2\right)\\ =2943.0\text{N}\end{array}$

From the figure given in the question the length $BE$ is same as $EA$.rearrange equation $\text{III}$ to find $EB$ and substitute $1500\text{ mm}$ for $L$, $300\text{mm}$ for $AC$, $400\text{mm}$ for $CD$ , and $300\text{mm}$ for $DB$ to find $EB$.

$L=AB+AC+CD+DB+2EB$

Simplify to find $EB$.

$\begin{array}{c}EB=\frac{1}{2}\left(L-AC-CD-DB\right)\\ =\frac{1}{2}\left(1500\text{mm}-400\text{mm}-300\text{mm}-\text{300}\text{mm}\right)\\ =250\text{mm}\end{array}$

Substitute $250\text{mm}$ for $EB$ in the equation II to find $\alpha$.

$\begin{array}{c}\cos \alpha =\frac{200\text{mm}}{250\text{mm}}\\ =0.8\alpha \\ {\cos }^{-1}\left(0.8\right)=36.87°\\ \alpha =36.87°\end{array}$

Substitute $36.87°$ for $\alpha$ in the equation IV to find $T_{AE}$.

$\begin{array}{c}{T}_{AE}\sin \left(36.87°\right)=\frac{1}{2}\left(2943.0\text{N}\right)\\ T_{AE}=2452.5\text{N}\\ \approx \text{2450}\text{N}\end{array}$

Therefore, the tension in the cable,$T_{AE}$ is $\underset{\_}{\text{2450}\text{N}}$.

To determine

The tension in the cable for the given arrangement of the cable loop.

Answer to Problem 2.65P

The tension in the cable $T_{AE}$ is $\underset{\_}{\text{2220}\text{N}}$.

Explanation of Solution

The length of the cable loop is $1.5\text{m}$, the mass of the crate is $300\text{kg}$.

Figure represents the free body diagram of the cable loop which is placed around a crate. The forces $T_{AB}$ and $T_{BE}$ are of same magnitude.

Write the expression for the weight acting on the system.

$W=mg$ (V)

Here, $W$ is the weight, $m$ is the mass, $g$ is the gravitational acceleration.

Figure2 represents the force triangle of the given free body diagram to find the tension.

Write the expression for finding the cosine angle.

$\cos \alpha =\frac{\text{Adjacent side }}{\text{Hypotenuse}}$ (VI)

Write the expression for the total length.

$L=AB+AC+CD+DB+2EB$

From the force triangle it is shown that $T_{AE}=T_{BE}$

From the figure 2a

According to the law of sines,

$T_A\sin \alpha =\frac{1}{2}2943\text{N}$ (IV)

Conclusion:

Substitute $300\text{Kg}$ for $m$ and $9.8\text{m}/{\text{s}}^2$ for $g$ in equation (I) to find $W$.

$\begin{array}{c}W=\left(300\text{kg}\right)\left(9.8\text{m}/{\text{s}}^2\right)\\ =2943.0\text{N}\end{array}$

From the figure given in the question the length $EC$ is same as $EA$.rearrange equation $III$ to find $EA$ and substitute $1500\text{ mm}$ for $L$, $300\text{mm}$ for $AC$, $400\text{mm}$ for $CD$ , and $300\text{mm}$ for $DB$ to find $EB$.

$\begin{array}{c}EA=\frac{1}{2}\left(L-CD-DB-AB\right)\\ =\frac{1}{2}\left(1500\text{mm}-400\text{mm}-300\text{mm}-4\text{00}\text{mm}\right)\\ =200\text{mm}\end{array}$

Substitute $200\text{mm}$ for $EB$ and $150$ for $1/2\left(DB\right)$ for adjacent side in the equation II to find $\alpha$.

$\begin{array}{c}\cos \alpha =\frac{150\text{mm}}{200\text{mm}}\\ =0.75\\ \alpha ={\cos }^{-1}\left(0.75\right)\\ =41.4096°\end{array}$

Substitute $41.4096°$ for $\alpha$ in the equation IV to find $T_{AE}$.

$\begin{array}{c}{T}_{AE}\sin \left(41.4096°\right)=\frac{1}{2}\left(2943.0\text{N}\right)\\ T_{AE}=2224.7\text{N}\\ \approx \text{2220}\text{N}\end{array}$

Therefore, the tension in the cable,$T_{AE}$ is $\underset{\_}{\text{2220}\text{N}}$.