Chapter 13.1, Problem 13.29P

A 7.5-lb collar is released from rest in the position shown, slides down the inclined rod, and compresses the spring. The direction of motion is reversed and the collar slides up the rod. Knowing that the maximum deflection of the spring is 5 in., determine (a) the coefficient of kinetic friction between the collar and the rod, (b) the maximum speed of the collar.

Fig. P13.29

To determine

Find the coefficient of kinetic friction between the collar and rod $\left(\mu \right)$.

Answer to Problem 13.29P

The coefficient of kinetic friction between the collar and rod $\left(\mu \right)$ is $\underset{\_}{0.159}$.

Explanation of Solution

Given information:

The weight of the collar $\left(W_C\right)$ is $7.5\text{lb}$.

The maximum deflection of the spring (x) is $\text{5in}\text{.}$ or $0.41667\text{ft}$.

The distance between the spring and collar (d) is $18\text{in}\text{.}$ or $\text{1}\text{.5ft}$.

The spring constant (k) is $60\text{lb}/\text{ft}$.

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

Assume the acceleration due to gravity (g) is $32.2\text{ft}/{\text{s}}^2$.

Calculation:

Show the free body diagram of the inclined rod with the forces acting as in Figure (1).

Here, the initial kinetic energy $\left(T_1\right)$ and final kinetic energy $\left(T_2\right)$ are zero respectively.

Calculate the work done ${\left(U_{1-2}\right)}_F$ due to friction using the formula:

${\left(U_{1-2}\right)}_F=-F\left(x+d\right)$

Here, F is the frictional force.

Substitute $\mu N$ for F, $0.41667\text{ft}$ for x, and $\text{1}\text{.5ft}$ for d.

$\begin{array}{c}{\left(U_{1-2}\right)}_F=-\mu N\left(0.41667+1.5\right)\\ =-1.91667\mu N\end{array}$

Substitute $W_C\cos \theta$ for N.

${\left(U_{1-2}\right)}_F=-1.91667\mu \left(W_C\cos \theta \right)$

Substitute $30°$ for $\theta$ and $7.5\text{lb}$ for $W_C$.

$\begin{array}{c}{\left(U_{1-2}\right)}_F=-1.91667\mu \left(7.5\times \cos 30°\right)\\ =-12.44914\mu \end{array}$

Calculate the work done ${\left(U_{1-2}\right)}_S$ due to spring using the formula:

${\left(U_{1-2}\right)}_S=\frac{1}{2}k{x}^2$

Substitute $0.41667\text{ft}$ for x and $60\text{lb}/\text{ft}$ for k.

$\begin{array}{c}{\left(U_{1-2}\right)}_S=\frac{1}{2}\left(60\right){\left(0.41667\right)}^2\\ =5.20842\text{lb}\end{array}$

Calculate the work done ${\left(U_{1-2}\right)}_G$ due to gravity using the formula:

${\left(U_{1-2}\right)}_G=W_C\left(x+d\right)\sin \theta$

Substitute $0.41667\text{ft}$ for x, $\text{1}\text{.5ft}$ for d, $7.5\text{lb}$ for $W_C$, and $30°$ for $\theta$.

$\begin{array}{c}{\left(U_{1-2}\right)}_G=7.5\left(0.41667+1.5\right)\sin 30°\\ =7.187513\text{lb}\end{array}$

Calculate the total work done $\left(U_{1-2}\right)$ using the relation:

$\left(U_{1-2}\right)={\left(U_{1-2}\right)}_F-{\left(U_{1-2}\right)}_S+{\left(U_{1-2}\right)}_G$

Substitute $-12.44914\mu$ for ${\left(U_{1-2}\right)}_F$, $5.20842\text{lb}$ for ${\left(U_{1-2}\right)}_S$, and $7.187513\text{lb}$ for ${\left(U_{1-2}\right)}_G$.

$\begin{array}{c}\left(U_{1-2}\right)=-12.44914\mu -5.20842\text{lb}+7.187513\text{lb}\\ =-12.44914\mu +1.979093\end{array}$

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 coefficient of kinetic friction between the collar and rod $\left(\mu \right)$:

$T_1+U_{1-2}=T_2$

Substitute 0 for $T_1$, $-12.44914\mu -1.979093$ for $U_{1-2}$, and 0 for $T_2$.

$\begin{array}{l}0+\left(-12.44914\mu +1.979093\right)=0\\ 12.44914\mu =1.979093\\ \mu =0.159\end{array}$

Therefore, the coefficient of kinetic friction between the collar and rod $\left(\mu \right)$ is $\underset{\_}{0.159}$.

To determine

Find the maximum speed $\left(v_{\max }\right)$ of the collar.

Answer to Problem 13.29P

The maximum speed $\left(v_{\max }\right)$ of the collar is $\underset{\_}{5.915\text{ft}/\text{s}}$

Explanation of Solution

Given information:

The weight of the collar $\left(W_C\right)$ is $7.5\text{lb}$.

The maximum deflection of the spring (x) is $\text{5in}\text{.}$ or $0.41667\text{ft}$.

The distance between the spring and collar (d) is $18\text{in}\text{.}$ or $\text{1}\text{.5ft}$.

The spring constant (k) is $60\text{lb}/\text{ft}$.

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

Assume the acceleration due to gravity (g) is $32.2\text{ft}/{\text{s}}^2$.

Calculation:

Calculate the kinetic energy $\left(T_3\right)$ using the formula:

$T_3=\frac{1}{2}\left(\frac{W_C}{g}\right)v_{\max }^2$

Substitute $7.5\text{lb}$ for $W_C$ and $32.2\text{ft}/{\text{s}}^2$ for g.

$\begin{array}{c}{T}_3=\frac{1}{2}\left(\frac{7.5}{32.2}\right)v_{\max }^2\\ =0.1165v_{\max }^2\end{array}$

Calculate the work done ${\left(U_{1-3}\right)}_F$ due to friction using the formula:

${\left(U_{1-3}\right)}_F=-F\left(d\right)$

Here, F is the frictional force.

Substitute $\mu N$ for F and $\text{1}\text{.5ft}$ for d.

$\begin{array}{c}{\left(U_{1-3}\right)}_F=-\mu N\left(1.5\right)\\ =-1.5\mu N\end{array}$

Substitute $W_C\cos \theta$ for N.

${\left(U_{1-3}\right)}_F=-1.5\mu \left(W_C\cos \theta \right)$

Substitute $30°$ for $\theta$ and $7.5\text{lb}$ for $W_C$.

$\begin{array}{c}{\left(U_{1-3}\right)}_F=-1.5\mu \left(7.5\times \cos 30°\right)\\ =-9.743\mu \end{array}$

Calculate the work done ${\left(U_{1-3}\right)}_G$ due to gravity using the formula:

${\left(U_{1-3}\right)}_G=W_C\left(d\right)\sin \theta$

Substitute $\text{1}\text{.5ft}$ for d, $7.5\text{lb}$ for $W_C$, and $30°$ for $\theta$.

$\begin{array}{c}{\left(U_{1-3}\right)}_G=7.5\left(1.5\right)\sin 30°\\ =5.625\text{lb}\end{array}$

Calculate the total work done $\left(U_{1-3}\right)$ using the relation:

$\left(U_{1-3}\right)={\left(U_{1-3}\right)}_F+{\left(U_{1-3}\right)}_G$

Substitute $-9.743\mu$ for ${\left(U_{1-3}\right)}_F$, and $5.625\text{lb}$ for ${\left(U_{1-3}\right)}_G$.

$\left(U_{1-3}\right)=-9.743\mu +5.625\text{lb}$

Substitute 0.159 for $\mu$.

$\begin{array}{c}\left(U_{1-3}\right)=-9.743\left(0.159\right)+5.625\text{lb}\\ =4.075863\end{array}$

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 maximum speed $\left(v_{\max }\right)$ of the collar:

$T_1+U_{1-3}=T_3$

Substitute 0 for $T_1$, 4.075863 for $U_{1-3}$, and $0.1165v_{\max }^2$ for $T_3$.

$\begin{array}{l}0+\left(4.075863\right)=0.1165v_{\max }^2\\ v_{\max }^2=\frac{4.075863}{0.1165}\\ v_{\max }=5.915\text{ft}/\text{s}\end{array}$

Therefore, the maximum speed $\left(v_{\max }\right)$ of the collar is $\underset{\_}{5.915\text{ft}/\text{s}}$