Chapter 13.2, Problem 13.73P

A 10-lb collar is attached to a spring and slides without friction along a fixed rod in a vertical plane. The spring has an undeformed length of 14 in. and a constant k = 4 lb/in. Knowing that the collar is released from rest in the position shown, determine the force exerted by the rod on the collar at (a) point A, (b) point B. Both these points are on the curved portion of the rod.

Fig. P13.73

To determine

Find the force exerted by the rod on the collar at A $\left(N_A\right)$.

Answer to Problem 13.73P

The force exerted by the rod on the collar at A $\left(N_A\right)$ is $\underset{\_}{82.4\text{lb}}$.

Explanation of Solution

Given information:

The weight of the collar (W) is $10\text{lb}$.

The un-deformed length $\left(l_0\right)$ is $14\text{in}\text{.}$

The spring constant (k) is $4\text{lb}/\text{in}\text{.}$

The length of top support to point A $\left(l_A\right)$ is $\text{14in}\text{.}$

The length of point A to point B $\left(l_{AB}\right)$ is $\text{14in}\text{.}$

The horizontal distance from weight to point A $\left(x_1\right)$ is $\text{14in}\text{.}$

The horizontal distance from point A to point B $\left(x_2\right)$ is $\text{14in}\text{.}$

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

Calculation:

Calculate the mass of the collar (m) using the relation:

$\begin{array}{c}W=mg\\ m=\frac{W}{g}\end{array}$

Substitute $10\text{lb}$ for W and $32.2\text{ft}/{\text{s}}^2$ for g.

$\begin{array}{c}m=\frac{10}{32.2}\\ =0.31056\text{lb}\cdot \text{ft}/{\text{s}}^2\end{array}$

Consider the position 1.

Calculate the length from weight to point B $\left(l_1\right)$ using the relation:

$l_1=\sqrt{\left(x_1+x_2\right)+{\left(l_A\right)}^2}$

Substitute $\text{14in}\text{.}$ for $x_1$, $\text{14in}\text{.}$ for $x_2$, and $\text{14in}\text{.}$ for $l_1$.

$\begin{array}{c}{l}_1=\sqrt{{\left(14+14\right)}^2+\left({14}^2\right)}\\ =31.305\text{in}\text{.}\end{array}$

Calculate the stretch in rod $\left(s_1\right)$ at position 1 using the relation:

$s_1=l_1-l_0$

Substitute $31.305\text{in}\text{.}$ for $l_1$ and $\text{14in}\text{.}$ for $l_0$.

$\begin{array}{c}{s}_1=31.305-\text{14in}\text{.}\\ =17.305\text{in}\text{.}\end{array}$

Here, the kinetic energy at position 1 $\left(T_1\right)$ is zero.

Calculate the potential energy in the position 1 due to elongation of the rod ${\left(V_1\right)}_e$ using the relation:

${\left(V_1\right)}_e=\frac{1}{2}k{s}_1^2$

Substitute $4\text{lb}/\text{in}\text{.}$ for k and $17.305\text{in}\text{.}$ for $s_1$.

$\begin{array}{c}{\left(V_1\right)}_e=\frac{1}{2}\times 4\times \left({17.305}^2\right)\\ =598.9261\text{lb}\cdot \text{in}\text{.}\times \frac{1}{12}\left(\frac{\text{ft}}{\text{in}\text{.}}\right)\\ =49.910\text{lb}\cdot \text{ft}\end{array}$

Here, the potential energy in the position 1 due to gravitation of the rod ${\left(V_1\right)}_g$ is zero.

Calculate the total potential energy $\left(V_1\right)$ using the relation:

$V_1={\left(V_1\right)}_e+{\left(V_1\right)}_g$

Substitute $49.910\text{lb}\cdot \text{ft}$ for ${\left(V_1\right)}_e$ and 0 for ${\left(V_1\right)}_g$.

$\begin{array}{c}{V}_1=49.910\text{lb}\cdot \text{ft+0}\\ \text{=}49.910\text{lb}\cdot \text{ft}\end{array}$

Consider the position A.

Calculate the length at point A $\left(l_A\right)$ using the relation:

$l_A=\sqrt{\left(x_1\right)+{\left(l_A\right)}^2}$

Substitute $\text{14in}\text{.}$ for $x_1$ and $\text{14in}\text{.}$ for $l_A$.

$\begin{array}{c}{l}_A=\sqrt{{\left(14\right)}^2+\left({14}^2\right)}\\ =19.799\text{in}\text{.}\end{array}$

Calculate the stretch in rod $\left(s_A\right)$ at position A using the relation:

$s_A=l_A-l_0$

Substitute $19.799\text{in}\text{.}$ for $l_A$ and $\text{14in}\text{.}$ for $l_0$.

$\begin{array}{c}{s}_A=19.799-\text{14in}\text{.}\\ =5.799\text{in}\text{.}\end{array}$

Calculate the kinetic energy at position A $\left(T_A\right)$ using the formula:

$T_A=\frac{1}{2}m{v}_A^2$

Here, $v_A$ is the velocity in the position A.

Substitute $0.31056\text{lb}\cdot \text{ft}/{\text{s}}^2$ for m.

$\begin{array}{c}{T}_A=\frac{1}{2}\left(0.31056\text{lb}\cdot \text{ft}/{\text{s}}^2\right)v_A^2\\ =0.15528v_A^2\end{array}$

Calculate the potential energy in the position A due to elongation of the rod ${\left(V_A\right)}_e$ using the relation:

${\left(V_A\right)}_e=\frac{1}{2}k{s}_A^2$

Substitute $4\text{lb}/\text{in}\text{.}$ for k and $5.799\text{in}\text{.}$ for $s_A$.

$\begin{array}{c}{\left(V_A\right)}_e=\frac{1}{2}\times 4\times \left({5.799}^2\right)\\ =67.257\text{lb}\cdot \text{in}\text{.}\times \frac{1}{12}\left(\frac{\text{ft}}{\text{in}\text{.}}\right)\\ =5.605\text{lb}\cdot \text{ft}\end{array}$

Here, the potential energy in the position 1 due to gravitation of the rod ${\left(V_1\right)}_g$ is zero in the datum at level A.

Calculate the total potential energy $\left(V_A\right)$ in position A using the relation:

$V_A={\left(V_A\right)}_e+{\left(V_A\right)}_g$

Substitute $5.605\text{lb}\cdot \text{ft}$ for ${\left(V_A\right)}_e$ and 0 for ${\left(V_A\right)}_g$.

$\begin{array}{c}{V}_A=5.605\text{lb}\cdot \text{ft+0}\\ \text{=}5.605\text{lb}\cdot \text{ft}\end{array}$

The expression for principle for conservation of energy as follows;

$T_1+V_1=T_A+V_A$

Substitute 0 for $T_1$, $49.910\text{lb}\cdot \text{ft}$ for $V_1$, $0.15528v_A^2$ for $T_A$ and $5.605\text{lb}\cdot \text{ft}$ for $V_A$.

$\begin{array}{l}0+49.910\text{lb}\cdot \text{ft}=0.15528v_A^2+5.605\text{lb}\cdot \text{ft}\\ 0.15528v_A^2=49.910-5.605\\ 0.15528v_A^2=44.305\\ v_A^2=\frac{44.305}{0.15528}\\ v_A^2=285.3233\text{ft}/\text{s}\\ v_A=\sqrt{285.3233}\\ v_A=16.892\text{ft}/\text{s}\end{array}$

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

Calculate the normal acceleration at position A ${\left(a_A\right)}_n$ using the formula:

${\left(a_A\right)}_n=\frac{v_A^2}{\rho }$

Substitute $285.3233\text{ft}/\text{s}$ for $v_A^2$ and $14\text{in}\text{.}$ for $\rho$.

$\begin{array}{c}{\left(a_A\right)}_n=\frac{285.3233}{14\text{in}\text{.}\times \frac{1}{12}\frac{\text{ft}}{\text{in}\text{.}}}\\ =244.56\text{ft}/{\text{s}}^2\end{array}$

Calculate the spring force at position A $\left(F_A\right)$ using the formula:

$F_A=k{s}_A$

Substitute $4\text{lb}/\text{in}\text{.}$ for k and $5.799\text{in}\text{.}$ for $s_A$.

$\begin{array}{c}{F}_A=4\left(5.799\right)\\ =23.196\text{lb}\end{array}$

Calculate the angle $\left(\alpha \right)$ using the relation:

$\tan \alpha =\frac{l_A}{x_1}$

Substitute $14\text{in}\text{.}$ for $x_1$ and $14\text{in}\text{.}$ for $l_A$.

$\begin{array}{l}\tan \alpha =\frac{14}{14}\\ \alpha ={\tan }^{-1}\left(\frac{14}{14}\right)\\ \alpha =45°\end{array}$

Calculate the force exerted by the rod on the collar in the point A $\left(N_A\right)$ by considering the Newton’s second law:

$\begin{array}{l}{\displaystyle \sum F}=m{\left(a_A\right)}_n\\ N_A+W-F_A\sin \alpha =m{\left(a_A\right)}_n\end{array}$

Substitute $10\text{lb}$ for W, $23.196\text{lb}$ for $F_A$, $45°$ for $\alpha$, $0.31056\text{lb}\cdot \text{ft}/{\text{s}}^2$ for m and $244.56\text{ft}/{\text{s}}^2$ for ${\left(a_A\right)}_n$.

$\begin{array}{l}{N}_A+10-\left(23.196\times \sin 45°\right)=\left(0.31056\times 244.56\right)\\ N_A+10-16.40205=75.951\\ N_A=75.951-10+16.40205\\ N_A=82.4\text{lb}\end{array}$

Therefore, the force exerted by the rod on the collar at A $\left(N_A\right)$ is $\underset{\_}{82.4\text{lb}}$.

To determine

Find the force exerted by the rod on the collar at B $\left(N_B\right)$.

Answer to Problem 13.73P

The force exerted by the rod on the collar at B $\left(N_B\right)$ is $\underset{\_}{49.6\text{lb}}$.

Explanation of Solution

Given information:

The weight of the collar (W) is $10\text{lb}$.

The un-deformed length $\left(l_0\right)$ is $14\text{in}\text{.}$

The spring constant (k) is $4\text{lb}/\text{in}\text{.}$

The length of top support to point A $\left(l_A\right)$ is $\text{14in}\text{.}$

The length of point A to point B $\left(l_{AB}\right)$ is $\text{14in}\text{.}$

The horizontal distance from weight to point A $\left(x_1\right)$ is $\text{14in}\text{.}$

The horizontal distance from point A to point B $\left(x_2\right)$ is $\text{14in}\text{.}$

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

Calculation:

Consider the position B.

Calculate the length at point B $\left(l_B\right)$ using the relation:

$l_B=x_1+x_2$

Substitute $\text{14in}\text{.}$ for $x_1$ and $\text{14in}\text{.}$ for $x_1$.

$\begin{array}{c}{l}_B=14+14\\ =28\text{in}\text{.}\end{array}$

Calculate the stretch in rod $\left(s_B\right)$ at position B using the relation:

$s_B=l_B-l_0$

Substitute $\text{28in}\text{.}$ for $l_B$ and $\text{14in}\text{.}$ for $l_0$.

$\begin{array}{c}{s}_B=28-\text{14in}\text{.}\\ =14\text{in}\text{.}\end{array}$

Calculate the kinetic energy at position B $\left(T_B\right)$ using the formula:

$T_B=\frac{1}{2}m{v}_B^2$

Here, $v_B$ is the velocity in the position B.

Substitute $0.31056\text{lb}\cdot \text{ft}/{\text{s}}^2$ for m.

$\begin{array}{c}{T}_B=\frac{1}{2}\left(0.31056\text{lb}\cdot \text{ft}/{\text{s}}^2\right)v_B^2\\ =0.15528v_B^2\end{array}$

Calculate the potential energy in the position B due to elongation of the rod ${\left(V_B\right)}_e$ using the relation:

${\left(V_B\right)}_e=\frac{1}{2}k{s}_B^2$

Substitute $4\text{lb}/\text{in}\text{.}$ for k and $14\text{in}\text{.}$ for $s_B$.

$\begin{array}{c}{\left(V_B\right)}_e=\frac{1}{2}\times 4\times \left({14}^2\right)\\ =6392\text{lb}\cdot \text{in}\text{.}\times \frac{1}{12}\left(\frac{\text{ft}}{\text{in}\text{.}}\right)\\ =32.667\text{lb}\cdot \text{ft}\end{array}$

Calculate the potential energy in the position B due to gravitation of the rod ${\left(V_B\right)}_g$ using the relation:

${\left(V_B\right)}_g=W{l}_0$

Substitute $10\text{lb}$ for W and $-14\text{in}\text{.}$ for $l_0$.

$\begin{array}{c}{\left(V_B\right)}_g=10\left(-14\right)\\ =-140\text{lb}\cdot \text{in}\text{.}\times \frac{1}{12}\left(\frac{\text{ft}}{\text{in}\text{.}}\right)\\ =-11.667\text{lb}\cdot \text{ft}\end{array}$

Calculate the total potential energy $\left(V_B\right)$ in the position B using the relation:

$V_B={\left(V_B\right)}_e+{\left(V_B\right)}_g$

Substitute $32.667\text{lb}\cdot \text{ft}$ for ${\left(V_B\right)}_e$ and $-11.667\text{lb}\cdot \text{ft}$ for ${\left(V_B\right)}_g$.

$\begin{array}{c}{V}_B=32.667\text{lb}\cdot \text{ft}-11.667\text{lb}\cdot \text{ft}\\ \text{=21}\text{.0lb}\cdot \text{ft}\end{array}$

The expression for principle for conservation of energy as follows;

$T_1+V_1=T_B+V_B$

Substitute 0 for $T_1$, $49.910\text{lb}\cdot \text{ft}$ for $V_1$, $0.15528v_B^2$ for $T_B$ and $\text{21}\text{.0lb}\cdot \text{ft}$ for $V_B$.

$\begin{array}{l}0+49.910\text{lb}\cdot \text{ft}=0.15528v_B^2+\text{21}\text{.0lb}\cdot \text{ft}\\ 0.15528v_B^2=49.910-21\\ 0.15528v_B^2=28.91\\ v_B^2=\frac{28.91}{0.15528}\\ v_B^2=186.18\text{ft}/\text{s}\\ v_B=\sqrt{186.18}\\ v_B=13.645\text{ft}/\text{s}\end{array}$

Show the free body diagram of the point B with the forces acting as in Figure (2).

Calculate the normal acceleration at position b ${\left(a_B\right)}_n$ using the formula:

${\left(a_B\right)}_n=\frac{v_B^2}{\rho }$

Substitute $186.18\text{ft}/\text{s}$ for $v_B^2$ and $14\text{in}\text{.}$ for $\rho$.

$\begin{array}{c}{\left(a_B\right)}_n=\frac{186.18}{14\text{in}\text{.}\times \frac{1}{12}\frac{\text{ft}}{\text{in}\text{.}}}\\ =159.583\text{ft}/{\text{s}}^2\end{array}$

Calculate the spring force at position B $\left(F_B\right)$ using the formula:

$F_B=k{s}_B$

Substitute $4\text{lb}/\text{in}\text{.}$ for k and $14\text{in}\text{.}$ for $s_B$.

$\begin{array}{c}{F}_B=4\left(14\right)\\ =56\text{lb}\end{array}$

Calculate the force exerted by the rod on the collar in the point B $\left(N_B\right)$ by considering the Newton’s second law:

$\begin{array}{l}{\displaystyle \sum F}=m{\left(a_B\right)}_n\\ N_B=m{\left(a_B\right)}_n\end{array}$

Substitute $0.31056\text{lb}\cdot \text{ft}/{\text{s}}^2$ for m and $159.583\text{ft}/{\text{s}}^2$ for ${\left(a_B\right)}_n$.

$\begin{array}{c}{N}_B=\left(0.31056\times 159.583\right)\\ =49.6\text{lb}\end{array}$

Therefore, the force exerted by the rod on the collar at B $\left(N_B\right)$ is $\underset{\_}{49.6\text{lb}}$.