Chapter 2, Problem 87P

(a)

To determine

The magnitude of the force $\overrightarrow{F}$ holding the ball.

Answer to Problem 87P

The magnitude of the force $\overrightarrow{F}$ holding the ball is $34\text{ N}$ .

Explanation of Solution

The free body diagram of the system is shown in figure 1.

Since the ball is held in its position, it will be in equilibrium so that the net force on it along horizontal and vertical directions will be zero.

Write the equation for the net force along $y$ direction.

  $\Sigma F_y=0$        (I)

Here, $\Sigma F_y$ is the net force on the ball along $y$ direction.

Write the expression for $\Sigma F_y$ .

  $\Sigma F_y=T_y+W$        (II)

Here, $W$ is the gravitational force and $T_y$ is the $y$ component of the tension in the string.

Write the expression for $W$ .

  $W=-mg$

Here, $m$ is the mass of the ball and $g$ is the acceleration due to gravity.

Write the expression for $T_y$ .

  $T_y=T\sin \theta$

Here, $T$ is the tension in the string and $\theta$ is the angle the string makes with the horizontal.

Put the above two equations in equation (II).

  $\begin{array}{c}\Sigma F_y=T\sin \theta +\left(-mg\right)\\ =T\sin \theta -mg\end{array}$

Put the above equation in equation (I).

$\begin{array}{c}T\sin \theta -mg=0\\ T=\frac{mg}{\sin \theta }\end{array}$        (III)

Write the equation for the net force along $x$ direction.

  $\Sigma F_x=0$        (IV)

Here, $\Sigma F_x$ is the net force on the crate along $x$ direction.

Write the expression for $\Sigma F_x$ .

  $\Sigma F_x=F+T_x$        (V)

Here, $F$ is the magnitude of the force holding the ball and $T_x$ is the $x$ component of the tension.

Write the expression for $T_x$ .

  $T_x=-T\cos \theta$

Put the above equation in equation (V).

  $\begin{array}{c}\Sigma F_x=F+\left(-T\cos \theta \right)\\ =F-T\cos \theta \end{array}$

Put the above equation in equation (IV).

  $\begin{array}{c}F-T\cos \theta =0\\ F=T\cos \theta \end{array}$        (VI)

Conclusion:

The value of $g$ is $9.80{\text{ m/s}}^2$ .

Substitute $2.0\text{ kg}$ for $m$ , $9.80{\text{ m/s}}^2$ for $g$ and $30°$ for $\theta$ in equation (III) to find $T$ .

  $\begin{array}{c}T=\frac{\left(2.0\text{ kg}\right)\left(9.80{\text{ m/s}}^2\right)}{\sin 30°}\\ =39.2\text{ N}\end{array}$

Substitute $39.2\text{ N}$ for $T$ and $30°$ for $\theta$ in equation (VI) to find $F$ .

  $\begin{array}{c}F=\left(39.2\text{ N}\right)\cos 30°\\ =34\text{ N}\end{array}$

Therefore, the magnitude of the force $\overrightarrow{F}$ holding the ball is $34\text{ N}$ .

(b)

To determine

The tension in the string.

Answer to Problem 87P

The tension in the string is $39\text{ N}$ .

Explanation of Solution

Equation (III) can be used to determine the tension in the string.

Conclusion:

Substitute $2.0\text{ kg}$ for $m$ , $9.80{\text{ m/s}}^2$ for $g$ and $30°$ for $\theta$ in equation (III) to find $T$ .

  $\begin{array}{c}T=\frac{\left(2.0\text{ kg}\right)\left(9.80{\text{ m/s}}^2\right)}{\sin 30°}\\ =39.2\text{ N}\\ \approx \text{39 N}\end{array}$

Therefore, the tension in the string is $39\text{ N}$ .