Chapter 3, Problem 81SP

A 5.0-kg block rests on a $30°$ incline. The coefficient of static friction between the

block and the incline is 0.20. How large a horizontal force must push on the block if

the block is to be on the verge of sliding (a) up the incline and (b) down the incline?

To determine

The minimumhorizontal forcerequired to push the $5\text{ kg}$ blockalong the $30°$ incline if the block is to be on the verge of sliding up the incline.

Answer to Problem 81SP

Solution:

$43.10\text{ N}$

Explanation of Solution

Given data:

The mass of the block is $5\text{ kg}$.

Coefficient of static friction between the block and incline is $0.20$.

Formula used:

The expression of the coefficient of static friction is

$\mu =\frac{F_f}{F_N}$

Here, $F_f$ is the friction force and $F_N$ is normal force.

The expression of the force’s equilibrium is

$\begin{array}{l}{\displaystyle \sum F_x}=0\\ {\displaystyle \sum F_y}=0\end{array}$

Here, ${\displaystyle \sum F_x}$ is the net force in the x- direction or horizontal direction and ${\displaystyle \sum F_y}$ is the net force in the y-direction or vertical direction.

Explanation:

Draw the free body diagram of the block representing all the forces and their components as shown in the figure below:

Since the block is considered to be on the verge of sliding, the change in velocity is zero, and therefore, the acceleration of the block is zero.

In the above diagram, $mg$ is the weight of the $5\text{ kg}$ blockand $mg\cos 30°$ and $mg\sin 30°$ are the components of the weight of $5\text{ kg}$ block normal to the incline and along the incline, respectively. $F$ is the horizontal pushing force sliding the $5\text{ kg}$ block up the incline, $F\cos 30°$ and $F\sin 30°$ are the components of the force $F$ along the incline and normal to the incline, respectively, and $F_f$ is the friction force.

Recall the expression of the force equilibrium along y- direction:

${\displaystyle \sum F_y}=0$

Consider the force normal to the incline in upward direction as positive and the force normal to the incline in the downward direction as negative. Hence,

$\begin{array}{c}{F}_N-mg\cos 30°-F\sin 30°=0\\ F_N=mg\cos 30°+F\sin 30°\end{array}$

Recall the expression of $\mu$

$\mu =\frac{F_f}{F_N}$

Substitute $0.20$ for $\mu$ and $\left(mg\cos 30°+F\sin 30°\right)$ for $F_N$

$\begin{array}{c}0.20=\frac{F_f}{mg\cos 30°+F\sin 30°}\\ F_f=0.20\left(mg\cos 30°+F\sin 30°\right)\end{array}$...............(1)

Recall the expression of the force equilibrium along x- direction:

${\displaystyle \sum F_x}=0$

Consider the force along the incline in right direction as positive and the force along the incline in the left direction as negative. Hence,

$\begin{array}{c}F\cos 30°-F_f-mg\sin 30°=0\\ F\cos 30°=mg\sin 30°+F_f\end{array}$

Substitute $\text{5 kg}$ for $m$ and $0.20\left(mg\cos 30°+F\sin 30°\right)$ for $F_f$

$F\cos 30°=5g\sin 30°+0.20\left(5g\cos 30°+F\sin 30°\right)$

Further solving the equation for $F$, we get

$\begin{array}{c}F\cos 30°=5g\sin 30°+0.20\left(5g\cos 30°\right)\text{+0}\text{.20}\left(F\sin 30°\right)=0\\ F\left(\cos 30°-0.20\sin 30°\right)=5g\left(\sin 30°+0.20\cos 30°\right)\\ F\left(0.766\right)=33.02\text{ N}\\ F=43.10\text{ N}\end{array}$

Conclusion:

The magnitude of the horizontal force required to push the block up the incline is $43.10\text{ N}$.

To determine

The minimum horizontal force required to push the $5\text{ kg}$ block along the $30°$ incline if the block is to be on the verge of sliding down the incline.

Answer to Problem 81SP

Solution:

$16.5\text{ N}$

Explanation of Solution

Given data:

Mass of the block is $5\text{ kg}$.

Coefficient of static friction between the block and incline is $0.20$.

Formula used:

The expression of the coefficient of static friction is

$\mu =\frac{F_f}{F_N}$

Here, $F_f$ is the friction force and $F_N$ is normal force.

The expression of the force’s equilibrium is

$\begin{array}{l}{\displaystyle \sum F_x}=0\\ {\displaystyle \sum F_y}=0\end{array}$

Here, ${\displaystyle \sum F_x}$ is the net force in the x- direction or horizontal direction and ${\displaystyle \sum F_y}$ is the net force in the y-direction or vertical direction.

Explanation:

Draw the free body diagram of the block representing all the forces and their components as shown in the figure below:

In the above diagram, $mg$ is the weight of the $5\text{ kg}$ blockand $mg\cos 30°$ and $mg\sin 30°$ are the components of the weight of $5\text{ kg}$ block normal to the incline and along the incline, respectively. $F$ is the horizontal pushing force sliding the $5\text{ kg}$ block down the incline, $F\cos 30°$ and $F\sin 30°$ are the components of the force $F$ along the incline and normal to the incline, respectively, and $F_f$ is the friction force.

Recall the expression of the force equilibrium along x- direction:

${\displaystyle \sum F_x}=0$

Consider the force along the incline in right direction as positive and the force along the incline in the left direction as negative. Hence,

$\begin{array}{c}F\cos 30°+F_f-mg\sin 30°=0\\ F\cos 30°=mg\sin 30°-F_f\end{array}$

Substitute $\text{5 kg}$ for $m$ and $0.20\left(mg\cos 30°+F\sin 30°\right)$ for $F_f$(using equation 1)

$\begin{array}{c}F\cos 30°=5g\sin 30°-0.20\left(5g\cos 30°+F\sin 30°\right)\\ F\cos 30°=5g\sin 30°-0.20\left(5g\cos 30°\right)+\text{0}\text{.20}\left(F\sin 30°\right)=0\end{array}$

Further solving the equation for $F$, we get

$\begin{array}{c}F\cos 30°=5g\sin 30°-0.20\left(5g\cos 30°\right)-\text{0}\text{.20}\left(F\sin 30°\right)\\ =0\end{array}$

Thus,

$\begin{array}{c}F\left(\cos 30°+0.20\sin 30°\right)=5g\left(\sin 30°-0.20\cos 30°\right)\\ F\left(0.966\right)=16.02\text{ N}\\ F=16.5\text{ N}\end{array}$

Conclusion:

The magnitude of the horizontal force required to push the block up the incline is $\text{16}\text{.5 N}$