Chapter 5, Problem 12P

A block of mass 3.00 kg is pushed up against a wall by a force $\stackrel{\to }{P}$ that makes an angle of θ = 50.0° with the horizontal as shown in Figure P5.12. The coefficient of static friction between the block and the wall is 0.250. (a) Determine the possible values for the magnitude of $\stackrel{\to }{P}$ that allow the block to remain stationary. (b) Describe what happens if $\left|\stackrel{\to }{P}\right|$ has a larger value and what happens if it is smaller. (c) Repeat parts (a) and (b), assuming the force makes an angle of θ = 13.0° with the horizontal.

Figure P5.12

To determine

The possible value for the magnitude of force $\overrightarrow{P}$ that allow the block to remain stationary.

Answer to Problem 12P

The possible value for the magnitude of force $\overrightarrow{P}$ that allow the block to remain stationary is $\underset{\_}{31.7\text{N}}$.

Explanation of Solution

Assume that, the block is moving in upward direction to get maximum possible value of $P$. The free body diagram of the block is shown in Figure 1.

Assume that, the net force along $x$ direction is zero.

    ${\displaystyle \sum F_x}=0$        (I)

Here, $F_x$ is the net force along $x$ axis.

From the free body diagram, write the expression for net force along $x$ direction.

    $P\cos 50.0°-n=0$        (II)

Here, $P$ is the force, and $n$ is the normal force.

Write the expression for frictional force for maximum $P$.

    $f_{\text{s,max}}={\mu }_{\text{S}}n$        (III)

Here, $f_{\text{s,max}}$ is the static frictional force for maximum $P$, and ${\mu }_{\text{s}}$ is the coefficient of static friction.

Solve equation (II) for $n$.

    $n=P\cos 50.0°$        (IV)

Use equation (IV) in (III).

    $f_{\text{s,max}}={\mu }_{\text{S}}P\cos 50.0°$        (V)

Assume that, the net force along $y$ direction is zero.

    ${\displaystyle \sum F_y}=0$        (VI)

Here, $F_y$ is the net force along $y$ axis.

From the free body diagram, write the expression for net force along $y$ direction.

    $P\sin 50.0°-f_{s,\max }-mg=0$        (VII)

Solve equation (VII) for $P_{\text{max}}$.

    $P_{\text{max}}=\frac{f_{s,\max }+mg}{\sin 50.0°}$        (VIII)

Assume that, the block is moving in downward motion to get minimum possible value of $P$. The free body diagram of the block is shown in Figure 2.

Assume that, the net force along $y$ direction is zero.

    ${\displaystyle \sum F_y}=0$

From the free body diagram, write the expression for net force along $y$ direction.

    $P\sin 50.0°+f_{s,\max }-mg=0$        (IX)

Solve equation (VII) for $P_{\text{min}}$.

    $P_{\text{min}}=\frac{-f_{s,\max }+mg}{\sin 50.0°}$        (X)

Conclusion:

Substitute, $0.250$ for ${\mu }_{\text{s}}$ in the equation (V), to find $f_{\text{s,max}}$.

    $\begin{array}{c}{f}_{\text{s,max}}=\left(0.250\right)P\cos 50.0°\\ =0.161P\end{array}$

Substitute, $0.161P$ for $f_{\text{s,max}}$, $3.00\text{kg}$ for $m$, and $9.80\text{m}/{\text{s}}^2$ for $g$ in the equation (VIII), to find $P_{\text{max}}$.

    $\begin{array}{c}{P}_{\text{max}}=\frac{0.161P+\left(3.00\text{kg}\right)\left(9.80\text{m}/{\text{s}}^2\right)}{\sin 50.0°}\\ =48.6\text{N}\end{array}$

Substitute, $0.161P$ for $f_{\text{s,max}}$, $3.00\text{kg}$ for $m$, and $9.80\text{m}/{\text{s}}^2$ for $g$ in the equation (X), to find $P_{\text{min}}$.

    $\begin{array}{c}{P}_{\text{min}}=\frac{-\left(0.161P\right)+\left(3.00\text{kg}\right)\left(9.80\text{m}/{\text{s}}^2\right)}{\sin 50.0°}\\ =31.7\text{N}\end{array}$

Therefore, The possible value for the magnitude of force $\overrightarrow{p}$ that allow the block to remain stationary is $\underset{\_}{31.7\text{N}}$.

To determine

The result if $\left|\overrightarrow{P}\right|$ has a larger and then smaller value.

Answer to Problem 12P

When $P$ has larger value, the block will slides up the wall, and $P$ has smaller value the block will slides down the wall.

Explanation of Solution

Maximum value of force $P$ is $48.6\text{N}$. If the value of $P$ is larger than the $48.6\text{N}$, then the block will slides up the wall. And the minimum value of force $P$ is $31.7\text{N}$. If the value of $P$ is smaller than $31.7\text{N}$, then the block will slides down the wall.

Conclusion:

Therefore, $P$ has larger value, the block will slides up the wall, and $P$ has smaller value the block will slides down the wall.

To determine

The possible value for the magnitude of force $\overrightarrow{P}$ that allow the block to remain stationary for the angle $13.0°$ with the horizontal, and the result happens if $\left|\overrightarrow{P}\right|$ has a larger and smaller value.

Answer to Problem 12P

The possible value for the magnitude of force $\overrightarrow{P}$ that allow the block to remain stationary for the angle $13.0°$ is $\underset{\_}{62.7\text{N}}$.

Explanation of Solution

Rewrite the equation (V) for angle $13.0°$.

    $f_{\text{s,max}}={\mu }_{\text{S}}P\cos 13.0°$        (XI)

Rewrite the equation (VIII) for angle $13.0°$.

    $P_{\text{max}}=\frac{f_{s,\max }+mg}{\sin 13.0°}$        (XII)

Rewrite the equation (X) for angle $13.0°$.

    $P_{\text{min}}=\frac{-f_{s,\max }+mg}{\sin 13.0°}$        (XIII)

Conclusion:

Substitute, $0.250$ for ${\mu }_{\text{s}}$ in the equation (XI), to find $f_{\text{s,max}}$.

    $\begin{array}{c}{f}_{\text{s,max}}=\left(0.250\right)P\cos 13.0°\\ =0.244P\end{array}$

Substitute, $0.244P$ for $f_{\text{s,max}}$, $3.00\text{kg}$ for $m$, and $9.80\text{m}/{\text{s}}^2$ for $g$ in the equation (XII), to find $P_{\text{max}}$.

    $\begin{array}{c}{P}_{\text{max}}=\frac{0.244P+\left(3.00\text{kg}\right)\left(9.80\text{m}/{\text{s}}^2\right)}{\sin 13.0°}\\ =-1580\text{N}\end{array}$

It is not possible to be negative, so however large or small it is, it cannot produce upward motion

Substitute, $0.244P$ for $f_{\text{s,max}}$, $3.00\text{kg}$ for $m$, and $9.80\text{m}/{\text{s}}^2$ for $g$ in the equation (XIII), to find $P_{\text{min}}$.

    $\begin{array}{c}{P}_{\text{min}}=\frac{-\left(0.244P\right)+\left(3.00\text{kg}\right)\left(9.80\text{m}/{\text{s}}^2\right)}{\sin 13.0°}\\ =62.7\text{N}\end{array}$

Therefore, The possible value for the magnitude of force $\overrightarrow{p}$ that allow the block to remain stationary for the angle $13.0°$ is $\underset{\_}{62.7\text{N}}$. And $P$ has larger value, the block won’t slides up the wall, and $P$ has smaller value the block will slides down the wall.