Chapter 3, Problem 80SP

(a) What is the smallest force parallel to a $37°$ incline needed to keep a 100-N weight from sliding down the incline if the coefficients of static and kinetic friction are both 0.30? (b) What parallel force is required to keep the weight moving up the incline at constant speed? (c) If the parallel pushing force is 94 N, what will be the acceleration of the object? (d) If the object in (c) starts from rest, how far will it move in 10 s?

To determine

The magnitude of the minimum force required to keep the $100\text{ N}$ object from sliding down the $37°$ inclined plane when the external force is applied parallel to the incline plane.

Answer to Problem 80SP

Solution:

$\text{36 N}$

Explanation of Solution

Given data:

The weight of the objectis $100\text{ N}$.

Coefficient of static friction and kinetic friction between the object and $37°$ incline is $0.30$.

Formula used:

Write the expression of the coefficient of static friction:

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

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

Write the expression for the first condition of the force’s equilibrium:

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

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

Explanation:

Consider, $F$ is the smallest applied force that is required to keep the object from sliding. Since after theapplication of force $F$, the object will not slide, So, the object will remain at rest. Hence, it will have zero acceleration.

Draw the free body diagram of the object:

In the above diagram, $F_f$ is the friction force, $100\text{ N}$ is the weight of the object, $100\cos 37°$ is the component of force $100\text{ N}$ perpendicular to the plane, $100\sin 37°$ is the component of the $100\text{ N}$ forceparallel to the inclined plane, and $W$ is the weight of the object.

Recall the expression for the first condition of the force’s equilibrium:

${\displaystyle \sum F_y}=0$

Considering along the perpendicular to the inclined plane, the direction of the upward forces is positive and the direction of the downward forces is negative. Therefore,

$\begin{array}{c}{F}_N-100\cos 37°=0\\ F_N=80\text{ N}\end{array}$

Here, considering coefficient of the static friction force because the object is just sliding down.

Recall the expression of static friction force:

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

Substitute $0.20$ for ${\mu }_s$ and $80\text{ N}$ for $F_N$

$\begin{array}{c}{\mu }_s=\frac{F_f}{F_N}\\ 0.30=\frac{F_f}{80\text{ N}}\\ F_f=24\text{ N}\end{array}$

Understand that the acceleration of the object will be zero because it will be at rest after the application of force $F$.

Rewrite the expression for the first condition of the force’s equilibrium:

${\displaystyle \sum F_x}=0$

Considering along the parallel to the inclined plane, the direction of the rightward forces is positive and the direction of the leftward forces is negative. Therefore,

$F+F_f-100\sin 37°=0$

Substitute $24\text{ N}$ for $F_f$

$\begin{array}{c}F+24\text{ N}-100\sin 37°=0\\ F+24-60.18=0\\ F-36.18=0\\ F\simeq 36\text{ N}\end{array}$

Conclusion:

Therefore, the magnitude of the minimum force required to keep the object from sliding down the incline plane is $36\text{ N}$.

To determine

The magnitude of the forceapplied parallel to the $37°$ incline plane needed to keep the $100\text{ N}$ object sliding up the incline at a constant speed.

Answer to Problem 80SP

Solution:

$\text{84 N}$

Explanation of Solution

Given data:

The weight of the object is $100\text{ N}$.

Coefficient of static friction and kinetic friction between the object and $37°$ incline is $0.30$.

The acceleration of the blockis zero because its moves with constant velocity.

Formula used:

Write the expression of the coefficient of static friction:

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

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

Write the expression for the first condition of the force’s equilibrium:

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

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

Explanation:

Since the $100\text{ N}$ object is moving at constant speed, there will be no acceleration.

Draw the free body diagram of the object:

In the above diagram, $F$ is the applied force to the object, $F_f$ is the friction force, $100\text{ N}$ is the weight of the object, $100\cos 37°$ is the component of force $100\text{ N}$ perpendicular to the plane, $100\sin 37°$ is the component of the $100\text{ N}$ force parallel to the inclined plane, $W$ is the weight of the object, $m$ is the mass of the object, and $g$ is the acceleration due to gravity.

The object is moving with constant velocity.

Rewrite the expression for the first condition of the force’s equilibrium.

${\displaystyle \sum F_x}=0$

Considering along the parallel to the inclined plane, the direction of the rightward forces is positive and the direction of the leftward forces is negative. Therefore,

$F-F_f-100\sin 37°=0$

Substitute $24\text{ N}$ for $F_f$ and $10.19\text{ kg}$ for $m$

$\begin{array}{c}F-24-100\sin 37°=0\\ F-24-60.18=10.19a_y\\ F-84.18=0\\ F\simeq 84\text{ N}\end{array}$

Conclusion:

Therefore, the magnitude of the parallel force needed to keep the objectmoving up the inclineat a constant speed is $\text{84 N}$.

To determine

The magnitude of the acceleration of an object if a $94\text{ N}$ force is applied on the 100 N weightparallel to the $37°$ incline plane.

Answer to Problem 80SP

Solution:

$0.98{\text{ m/s}}^2$ up the incline

Explanation of Solution

Given data:

The weight of the object is $100\text{ N}$.

Parallel pushing force is 94N.

Coefficient of static friction and kinetic friction between the object and incline is $0.30$.

Formula used:

Write the expression of the coefficient of the kinetic friction:

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

From the Newton’s second law of the motion, the expression of the force is

${\displaystyle \sum F_x}=m{a}_x$

Here, ${\displaystyle \sum F_x}$ is the sum of forces in x- direction, $m$ is the mass of the object, and $a_x$ is acceleration in x- direction.

Write the expression for the weight of the $100\text{ N}$ object:

$W=mg$

Here, $W$ is the weight of object, $m$ is the mass of the object, and $g$ is acceleration due to gravity.

Explanation:

Since the $100\text{ N}$ object is moving, so there will be an acceleration.

Draw the free body diagram of the object:

In the above diagram, considering the objectup the incline, $100\text{ N}$ is the weight of the object and $100\cos 37°$ is the component of force $100\text{ N}$ along the perpendicular to the plane, $100\sin 37°$ is the component of the $100\text{ N}$ force along the parallel to the inclined plane, $F$ is the pushing force parallel to incline plane, and $a$ is the acceleration of an object.

Here, considering the coefficient of the kinetic friction because $94\text{ N}$ pushing force is used to move theobject $100\text{ N}$ up the incline.

Recall the expression of coefficient of kinetic friction:

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

Here, ${\mu }_k$ is the coefficient of kinetic friction.

Substitute $0.30$ for ${\mu }_s$ and $80\text{ N}$ for $F_N$

$\begin{array}{c}0.30=\frac{F_f}{80\text{ N}}\\ F_f=24\text{ N}\end{array}$

Recall the expression for the weight of the $100\text{ N}$ object:

$W=mg$

Substitute $\text{100 N}$ for $W$ and $9.81{\text{ m/s}}^2$ for $g$

$\begin{array}{c}100=m\left(9.81{\text{ m/s}}^2\right)\\ m=10.19\text{ kg}\end{array}$

Recall the expression for Newton’s second law of the motion:

${\displaystyle \sum F_x}=m{a}_x$

Considering along the parallel to the inclined plane, the direction of the rightward forces is positive and the direction of the leftward forces is negative. Therefore,

$F-F_f-100\sin 37°=ma$

Here, $a$ is the acceleration of the object.

Substitute $24\text{ N}$ for $F_f$, $94\text{ N}$ for $F$, and $10.19\text{ kg}$ for $m$

$\begin{array}{c}94-24-100\sin 37°=10.19a\\ 9.82=10.19a\\ a=\frac{9.82}{10.19}\\ =0.963{\text{ m/s}}^2\\ \simeq 0.98{\text{ m/s}}^2\end{array}$

Conclusion:

Therefore, the magnitude of the acceleration of the object is $0.98{\text{ m/s}}^2$.

To determine

Thedistance coveredby the object after $10\text{ s}$ due to the application of a parallel pushing force of 94 N if itstarts from rest .

Answer to Problem 80SP

Solution:

$49\text{ m}$

Explanation of Solution

Given data:

The weight of the object is $100\text{ N}$.

Coefficient of static friction and kinetic friction between the object and incline is $0.30$.

Time to considered is $10\text{ s}$.

Initial velocity is $0\text{ m/s}$.

Formula used:

Write the expression of the displacement:

$s=v_{i}t+\frac{1}{2}a{t}^2$

Here, $v_i$ is the initial velocity of the object, $a$ is the acceleration of the object, $t$ is the time, and $s$ is the displacement of the object.

Explanation:

Recall the expression of the displacement:

$s=v_{i}t+\frac{1}{2}a{t}^2$

Substitute $\text{0}{\text{.98 m/s}}^2$ for $a$, $10\text{ s}$ for $t$, and $0\text{ m/s}$ for $v_i$

$\begin{array}{c}s=\left(0\right)t+\frac{1}{2}\left(\text{0}\text{.98}\right){\left(10\right)}^2\\ s=\frac{98}{2}\\ =49\text{ m}\end{array}$

Conclusion:

The displacement of the object is $49\text{ m}$.