Question 15.SP.18: In a can crusher, bar AB has a length of 30 in. and slides i...
In a can crusher, bar AB has a length of 30 in. and slides inside a collar located at point P. This collar is attached to plunger DP, which is constrained to move vertically. At the instant shown, the velocity of point B is a constant 4 ft/s perpendicular to the bar. Determine the velocity and acceleration of the plunger D.
STRATEGY: You are not given any forces and are asked to find the velocity and acceleration of a point, so use rigid-body kinematics. Since the collar is moving with respect to the bar, use a rotating reference frame.
MODELING and ANALYSIS: Attach a rotating coordinate system to the bar with its origin at A (Fig. 1).Angular Velocity of AB. Rod AB is undergoing fixed-axis rotation, so
\omega_{A B}=\frac{v_B}{r_{B / A}}=\frac{48 \mathrm{in} . / \mathrm{s}}{30 \mathrm{in} .}=1.60 \mathrm{rad} / \mathrm{s} \circlearrowright
Velocity of P. Points D and P have the same velocity and acceleration because the plunger is constrained to translate only. From Eq. (15.32′) you know
\mathbf{v}_P=\mathbf{v}_O + \boldsymbol{\Omega} \times \mathbf{r} + \mathbf{v}_{\mathrm{rel}} (15.32′)
\mathbf{v}_P=\mathbf{v}_A + \boldsymbol{\Omega} \times \mathbf{r}_{P / A} + \mathbf{v}_{\mathrm{rel}} (1)
where \mathbf{v}_A=0, \mathbf{r}_{P / A}=(20 \mathrm{in} .) \mathbf{i}+(12 \mathrm{in} .) \mathbf{j} \text {, and } \boldsymbol{\Omega}=-(1.6 \mathrm{rad} / \mathrm{s}) \mathbf{k}. To find the relative velocity, ask yourself what the velocity of P would be assuming that the rotating coordinate system is not moving. In this case, \mathbf{v}_{\text {rel }}=v_{\text {rel }} \cos \theta \mathbf{i}+v_{\text {rel }} \sin \theta \mathbf{j} \text { where } \theta=\tan ^{-1}(12 /20)=30.96^{\circ}. Substituting into Eq. (1) gives
\mathbf{v}_P \mathbf{j}=0+(-1.6\mathbf{k}) \times (20\mathbf{i} + 12\mathbf{j}) + (v_{\text{rel}} \cos \theta \mathbf{i} + v_{\text{rel}} \sin \theta \mathbf{j}) \\ =-32\mathbf{j} + 19.2\mathbf{i} + 0.8575 v_{\text{rel}} \mathbf{i}+ 0.5145 v_{\text{rel}}\mathbf{j}
Equating components allows you to solve for the unknown velocities:
\begin{array}{llrl}\mathrm{i}: & 0=19.2 + 0.8575 v_{\text {rel }} & \longrightarrow & v_{\text {rel }}=-22.39 \mathrm{in} . / \mathrm{s} \\ \mathrm{j}: & -v_P=-32 + 0.5145 v_{\text {rel }} & \longrightarrow & v_P=43.53 \mathrm{in} . / \mathrm{s}\end{array}\mathbf{v}_P=43.53 \text { in./s } \downarrow
Acceleration of P. From Eq. (15.35′), you know
\mathbf{a}_P=\mathbf{a}_O + \dot{\boldsymbol{\Omega}} \times \mathbf{r} – \Omega^2 \mathbf{r} + 2 \boldsymbol{\Omega} \times \mathbf{v}_{\text {rel }} + \mathbf{a}_{\mathrm{rel}} (15.35′)
\mathbf{a}_P=\mathbf{a}_A + \dot{\boldsymbol{\Omega}} \times \mathbf{r}_{P / A} – \Omega^2 \mathbf{r}_{P / A} + 2 \boldsymbol{\Omega} \times \mathbf{v}_{\mathrm{rel}} + \mathbf{a}_{\mathrm{rel}} (2)
where \mathrm{a}_A=0, \dot{\boldsymbol{\Omega}}=0, \mathbf{a}_{\mathrm{rel}}=a_{\mathrm{rel}} \cos \theta \mathbf{i}+a_{\mathrm{rel}} \sin \theta \mathbf{j}. Substituting into Eq. (2) gives
\begin{aligned} – a_P \mathbf{j}=& 0 + 0-1.6^2(20 \mathbf{i} + 12 \mathbf{j}) + 2(-1.6 \mathbf{k}) \times \\&(-22.39 \cos \theta \mathbf{i} – 22.39 \sin \theta \mathbf{j}) + \left(a_{\text {rel }} \cos \theta \mathbf{i} + a_{\text {rel }} \sin \theta \mathbf{j}\right) \\=&(-51.2 \mathbf{i} – 30.72 \mathbf{j}) + (61.44 \mathbf{j} – 36.86 \mathbf{i}) + \left(0.8575 a_{\text {rel }} \mathbf{i} + 0.5145 a_{\text {rel }} \mathbf{j}\right)\end{aligned}
Equating components allows you to solve for the unknown accelerations:
\begin{aligned}& \mathbf{j}: \quad 0=-51.2-36.86+0.8575 a_{\text {rel }} \quad \longrightarrow \quad a_{\text {rel }}=102.7 \text { in./s }{ }^2 \\&\mathbf{j}: \quad-a_P=-30.72+61.44+0.5145 a_{\text {rel }} \quad \longrightarrow \quad a_P=-83.56 \text { in./s }{ }^2 \\&\end{aligned}a_P=-83.6 \text { in./s }{ }^2 \downarrow
REFLECT and THINK: You used the same strategy for the telescoping boom in Sample Prob. 15.17 as you did for the sliding collar in this problem. For each case, the point of interest was moving with respect to a coordinate frame attached to a rigid body. The same strategy is used in problems where pins move within slotted bodies (such as the Geneva mechanism in Sample Prob. 15.19).
