Chapter 13, Problem 1FP

To determine

The tension developed in the cable.

Answer to Problem 1FP

The tension developed in the cable is $176\text{N}$.

Explanation of Solution

Given:

The mass of the crate $m$ is $20\text{kg}$.

The coefficient of kinetic friction ${\mu }_k$ is $0.3$.

Angle $\theta$ is $30°$.

Initial velocity ${\upsilon }_0$ is $0$.

Initial distance $s_0$ is $0$

Final distance $s$ is $6\text{m}$.

Time $t$ is $3\text{s}$.

Write the expression for final distance in $y$ direction.

$s=s_0+{\upsilon }_{0}t+\frac{1}{2}a{t}^2$ (I).

Here, Final distance is $s$ , initial distance is $s_0$, velocity is $\upsilon$ , initial velocity is ${\upsilon }_0$ time is $t$ and acceleration is $a$.

Draw the free body diagram of the crate as shown in Figure (1).

Refer Figure (1) write the formula to calculate horizontal force components:

$\sum F_x=m{a}_x$ (II).

Here, summation of total force acting on horizontal direction is $\sum F_x$, the mass of the object is $m$ and acceleration is $a$.

Refer Figure (1) and resolve the forces along $x-\text{axis}$.

${\displaystyle \sum F_x=T-m\left(9.81{\text{m/s}}^2\right)\sin 30°-{\mu }_{k}N}$ (III).

Refer Figure (1) and write the formula to calculate vertical force components:

$\begin{array}{c}\sum F_y=m\left(0\right)\\ =0\end{array}$

Here, summation of total force acting on vertical direction is $\sum F_y$.

Refer Figure (1) and resolve the forces along $y-\text{axis}$.

${\displaystyle \sum F_y=N-m\left(9.81{\text{m/s}}^2\right)\cos 30°}$ (IV).

Conclusion:

Substitute $\text{3s}$ for $t$, $0$ for ${\upsilon }_0$, $6\text{m}$ for $s$ and $0$ for $s_0$ in Equation (I).

$\begin{array}{l}6=0+0\left(3\right)+\frac{1}{2}a{\left(3\right)}^2\\ a=1.333{\text{m/s}}^2\end{array}$

Substitute $20\text{kg}$ for $m$ and $1.333{\text{m/s}}^2$ for $a$ in Equation (II).

$\begin{array}{c}\sum F_x=ma\\ =\left(20\right)1.333\\ =26.66\end{array}$

Substitute $0$ for $\sum F_y$, and $20\text{kg}$ for $m$ Equation (IV).

$\begin{array}{l}{\displaystyle \sum F_y=N-m\left(9.81{\text{m/s}}^2\right)\cos 30°}\\ 0=N-20\left(9.81{\text{m/s}}^2\right)\cos 30°\\ N=169.9\text{N}\end{array}$

Substitute $26.66$ for $\sum F_x$, 0.3 for ${\mu }_k$ and $169.9\text{N}$ for $N$ in Equation (III).

$\begin{array}{l}{\displaystyle \sum F_x=T-m\left(9.81{\text{m/s}}^2\right)\sin 30°-{\mu }_{k}N}\\ 26.66=T-20\left(9.81{\text{m/s}}^2\right)\sin 30°-0.3\left(169.9\right)\\ T=176\text{N}\end{array}$

Thus, the tension developed in the cable is $176\text{N}$.