Chapter 2, Problem 76P

(a)

To determine

The normal force applied on the crate by the ramp.

Answer to Problem 76P

The normal force applied on the crate by the ramp is $\text{75}\text{.2N}$

Explanation of Solution

The weight of the crate is $\text{80}\text{.0N}$. The angle of inclination of the ramp is $\text{20}\text{.0}°$. Consider the ramp to be $\text{x}$ axis.

The following pic shows the FBD of the system.

The net force along the y-axis is zero as there is no motion in y-axis.

  $N-mg\cos \theta =0$                                                                

Here, $N$ is the normal force, $m$ is the mass, $g$ is the acceleration due to gravity, and $\theta$ is the angle of inclination of the ramp.

Re-write the above expression to get an expression for $N$.

  $N=mg\cos \theta$        (I)

Conclusion:

Substitute $80.0\text{N}$ for $mg$, $20.0°$ for $\theta$ in equation (I)

  $\begin{array}{c}N=\left(80.0\text{N}\right)\cos \left(20.0°\right)\\ =75.2\text{N}\end{array}$

The normal force applied on the crate by the ramp is $\text{75}\text{.2N}$

(b)

To determine

The magnitude and direction of the interaction partner of the normal force.

Answer to Problem 76P

The interaction partner of the normal force will have magnitude of $\text{75}\text{.2N}$ and applied in opposite direction to the normal force.

The normal force on the crate has magnitude of $\text{75}\text{.2N}$. Thus the interaction partner of the normal force will have magnitude of $\text{75}\text{.2N}$ and applied in opposite direction to the normal force.

Explanation of Solution

According to the Newton’s third law the interaction partners have equal magnitude and opposite.

The normal force on the crate has magnitude of $\text{75}\text{.2N}$. Thus the interaction partner of the normal force will have magnitude of $\text{75}\text{.2N}$ and applied in opposite direction to the normal force.

Conclusion:

The interaction partner of the normal force will have magnitude of $\text{75}\text{.2N}$ and applied in opposite direction to the normal force.

(c)

To determine

The static friction force between the crate and the ramp.

Answer to Problem 76P

The static friction force between the crate and the ramp is $\text{27}\text{.4N}$

Explanation of Solution

The weight of crate is $\text{80}\text{.0N}$. The angle of inclination of the ramp is $\text{20}\text{.0}°$. Consider the ramp to be  $\text{x}$-axis.

The following pic shows the FBD of the system.

When the crate is static, the force along the $\text{x}$-axis is zero.

  $f_s-mg\sin \theta =0$        (I)

Here, $f_s$ is the friction force, $m$ is the mass, $g$ is the acceleration due to gravity, $\theta$ is the angle of inclination of the ramp.

Re-write the above expression to get an expression for $f_s$.

  $f_s=mg\sin \theta$        (II)

Conclusion:

Substitute $80.0\text{N}$ for $mg$, $20.0°$ for $\theta$ in equation (I)

  $\begin{array}{c}{f}_s=\left(80.0\text{N}\right)\sin \left(20.0°\right)\\ =27.4\text{N}\end{array}$

The static friction force between the crate and the ramp is $\text{27}\text{.4N}$

(d)

To determine

The minimum possible value for the coefficient of static friction.

Answer to Problem 76P

The minimum possible value for the coefficient of static friction is $0.364$

Explanation of Solution

 The following pic show the FBD of the system,

The minimum coefficient of static friction is just enough to make the friction force equalize the component of weight in the direction of the crate’s motion.

  $f_s=mg\sin \theta$        (I)

Here, $f_s$ is the friction force, $m$ is the mass, $g$ is the acceleration due to gravity, $\theta$ is the angle of inclination of the ramp.

Write the formal for the coefficient of static friction.

  ${\mu }_s=\frac{f_s}{N}$        (II)

Here, ${\mu }_s$ is the coefficient of static friction, $N$ is the normal force.

Write the formula for the normal force.

  $N-mg\cos \theta =0$        (III)

Substitute expression (I) and (III) in expression (II),

  $\begin{array}{c}{\mu }_s=\frac{mg\sin \theta }{mg\cos \theta }\\ =\tan \theta \end{array}$        (IV)

Conclusion:

Substitute  $20.0°$ for $\theta$ in equation (IV)

  $\begin{array}{c}{\mu }_s=\tan \left(20.0°\right)\\ =0.364\end{array}$

The minimum possible value for the coefficient of static friction is $0.364$

(e)

To determine

The magnitude and direction of the contact force exerted on the crate by the ramp.

Answer to Problem 76P

The magnitude of the contact force is $80.0\text{N}$ and it makes $70.0°$ with the $\text{x}$ axis

Explanation of Solution

The weight of the crate is $80.0\text{N}$. The angle of inclination of the ramp is $20.0°$. Consider the ramp to be $\text{x}$ axis.

The friction force and the normal force are the component of the contact force.

Write the formula for the magnitude of the contact force.

  $\left|F\right|=\sqrt{f_s{}^2+N^2}$        (I)

Here $F$ is the contact force, $f_s$ is the friction force, $N$ is the normal force.

Write the direction of the contact force.

  $\theta ={\tan }^{-1}\left(\frac{N}{f_s}\right)$        (II)

Here, $\theta$ is the angle made by the contact force with the positive $\text{x}$ axis.

Conclusion:

Substitute $\text{75}\text{.2N}$ for $N$,  $\text{27}\text{.4N}$ for $f_s$ in equation (I)

  $\begin{array}{c}\left|F\right|=\sqrt{{\left(\text{27}\text{.4N}\right)}^2+{\left(\text{75}\text{.2N}\right)}^2}\\ =80.0\text{N}\end{array}$

Substitute $\text{75}\text{.2N}$ for $N$,  $\text{27}\text{.4N}$ for $f_s$ in equation (I)

  $\begin{array}{c}\theta ={\tan }^{-1}\left(\frac{\text{75}\text{.2N}}{\text{27}\text{.4N}}\right)\\ =70.0°\end{array}$

The magnitude of the contact force is $80.0\text{N}$ and it makes $70.0°$ with $\text{x}$ axis.