Chapter 13.1, Problem 13.12P

A package is thrown down an incline at A with a velocity of 1 m/s. The package slides along the surface ABC to a conveyor belt that moves with a velocity of 2 m/s. Knowing that d = 6 m and μ_k = 0.2 between the package and all surfaces, determine (a) the speed of the package at C, (b) the distance the package will slide on the conveyor belt before it comes to rest relative to the belt.

To determine

Find the speed $\left(v_C\right)$ of the package at the C.

Answer to Problem 13.12P

The speed $\left(v_C\right)$ of the package at the C is $\underset{\_}{3.4643\text{m}/\text{s}}$.

Explanation of Solution

Given information:

The velocity of conveyor $\left(v_C\right)$ is $2\text{m}/\text{s}$.

The coefficient of the static friction between package with surface $\left({\mu }_k\right)$ is $0.25$.

The distance between the point C to point B $\left(d_{CB}\right)$ is $7\text{m}$.

The inclined angle of the member BA $\left(\theta \right)$ is $30°$.

The velocity at the point A $\left(v_A\right)$ is $1\text{m}/\text{s}$.

The distance between the point A to B (d) is $6\text{m}$.

Assume the acceleration due to gravity (g) is $9.81\text{m}/{\text{s}}^{\text{2}}$.

Calculation:

Calculate the weight of the conveyor (W) using the formula:

$W=mg$

Substitute $9.81\text{m}/{\text{s}}^{\text{2}}$ for g.

$\begin{array}{c}W=\left(m\right)9.81\\ =9.81m\end{array}$

Show the free body diagram of the package sliding from the corner A to B as in Figure (1).

Calculate the normal force at point AB $\left(N_{AB}\right)$ using the formula:

$N_{AB}=W\cos \theta$

Substitute $9.81m$ for W and $30°$ for $\theta$.

$\begin{array}{c}{N}_{AB}=9.81m\cos 30°\\ =8.496m\end{array}$

Calculate the force at point AB $\left(F_{AB}\right)$ using the formula:

$F_{AB}={\mu }_k{N}_{AB}$

Substitute $0.20$ for ${\mu }_k$ and $8.496m$ for $N_{AB}$.

$\begin{array}{c}{F}_{AB}=\left(0.20\right)8.496m\\ =1.6992m\end{array}$

Calculate the work done $U_{A-B}$ using the formula:

$U_{A-B}=Wd\sin 30°-F_{AB}d$

Substitute $9.81m$ for W and $1.6992m$ for $F_{AB}$.

$\begin{array}{c}{U}_{A-B}=9.81md\sin 30°-\left(1.6992md\right)\\ =md\left(9.81\sin 30°-1.6992\right)\\ =3.2058md\end{array}$

Show the free body diagram of the package sliding from corner B to C as in Figure (2).

Calculate the normal force at point BC $\left(N_{BC}\right)$ using the formula:

$N_{BC}=W$

Substitute $9.81m$ for W.

$N_{AB}=9.81m$

Calculate the force at point BC $\left(F_{BC}\right)$ using the formula:

$F_{BC}={\mu }_k{N}_{BC}$

Substitute $0.2$ for ${\mu }_k$ and $9.81m$ for $N_{BC}$.

$\begin{array}{c}{F}_{BC}=\left(0.2\right)9.81m\\ =1.962m\end{array}$

Calculate the work done $U_{B-C}$ using the formula:

$U_{B-C}=-F_{BC}{d}_{BC}$

Substitute $1.962m$ for $F_{BC}$ and $7\text{m}$ for $d_{BC}$.

$\begin{array}{c}{U}_{B-C}=-1.962m\left(7\right)\\ =-13.734m\end{array}$

Assume the corner B has no energy.

Use work and energy principle which states that kinetic energy of the particle at a displaced point can be obtained by adding the initial kinetic energy and the work done on the particle during its displacement.

Find the speed $\left(v_C\right)$ of the package at the C:

$T_A+U_{A-B}+U_{B-C}=T_C$

Substitute $\frac{1}{2}m{v}_A^2$ for $T_A$, $\frac{1}{2}m{v}_C^2$ for $T_C$, $3.2058md$ for $U_{A-B}$, and $-13.734m$ for $U_{B-C}$.

$\begin{array}{l}\frac{1}{2}m{v}_A^2+3.2058md-13.734m=\frac{1}{2}m{v}_C^2\\ m\left(\frac{v_A^2}{2}+3.2058d-13.734\right)=\frac{1}{2}m{v}_C^2\\ \frac{v_A^2}{2}+3.2058d-13.734=\frac{v_C^2}{2}\end{array}$

Substitute $6\text{m}$ for d and $1\text{m}/\text{s}$ for $v_A$.

$\begin{array}{l}\frac{{\left(1\right)}^2}{2}+3.2058\left(6\right)-13.734=\frac{v_C^2}{2}\\ 6.0008=2\\ v_C=\sqrt{6.0008\left(2\right)}\\ v_C=3.4643\text{m}/\text{s}\end{array}$

Therefore, the speed $\left(v_C\right)$ of the package at the C is $\underset{\_}{3.4643\text{m}/\text{s}}$.

To determine

Find the distance $\left(x_{belt}\right)$ of the belt before it comes to rest.

Answer to Problem 13.12P

The distance $\left(x_{belt}\right)$ of the belt before it comes to rest is $\underset{\_}{2.0391\text{m}}$.

Explanation of Solution

Given information:

The velocity of conveyor $\left(v_C\right)$ is $2\text{m}/\text{s}$.

The coefficient of the static friction between package with surface $\left({\mu }_k\right)$ is $0.25$.

The distance between the point C to point B $\left(d_{CB}\right)$ is $7\text{m}$.

The inclined angle of the member BA $\left(\theta \right)$ is $30°$.

The velocity at the point A $\left(v_A\right)$ is $1\text{m}/\text{s}$.

The distance between the point A to B (d) is $6\text{m}$.

Assume the acceleration due to gravity (g) is $9.81\text{m}/{\text{s}}^{\text{2}}$.

Calculation:

Calculate the force (F) using the formula:

$F={\mu }_{k}N$

Substitute $0.2$ for ${\mu }_k$ and $9.81m$ for $N$.

$\begin{array}{c}F=\left(0.2\right)9.81m\\ =1.962m\end{array}$

Calculate the work done $U_{belt}$ using the formula:

$U_{belt}=-F{x}_{belt}$

Here, $x_{belt}$ is the distance moves by package as it slides on the belt.

Substitute $1.962m$ for $F$.

$U_{belt}=-1.962m\left(x_{belt}\right)$

Use work and energy principle which states that kinetic energy of the particle at a displaced point can be obtained by adding the initial kinetic energy and the work done on the particle during its displacement.

Find the distance $\left(x_{belt}\right)$ of the belt before it comes to rest:

$T_C+U_{belt}=T_{belt}$

Substitute $\frac{1}{2}m{v}_C^2$ for $T_C$, $-1.962m\left(x_{belt}\right)$ for $U_{belt}$ and $\frac{1}{2}m{v}_{belt}^2$ for $T_{belt}$.

$\begin{array}{l}\frac{1}{2}m{v}_C^2-1.962m\left(x_{belt}\right)=\frac{1}{2}m{v}_{belt}^2\\ m\left(\frac{v_C^2}{2}-1.962x_{belt}\right)=\frac{1}{2}m{v}_{belt}^2\\ \frac{v_C^2}{2}-1.962x_{belt}=\frac{v_{belt}^2}{2}\end{array}$

Substitute $3.4643\text{m}/\text{s}$ for $v_C$ and $2\text{m}/\text{s}$ for $v_{belt}$.

$\begin{array}{l}\frac{\left({3.4643}^2\right)}{2}-1.962x_{belt}=\frac{{\left(2\right)}^2}{2}\\ 1.962x_{belt}=\frac{\left({3.46}^2\right)}{2}-\frac{{\left(2\right)}^2}{2}\\ x_{belt}=\frac{4.000687}{1.962}\\ x_{belt}=2.0391\text{m}\end{array}$

Therefore, the distance $\left(x_{belt}\right)$ of the belt before it comes to rest is $\underset{\_}{2.0391\text{m}}$.