Question 15.SP.17: At the instant shown, the truck is moving forward with a spe...
At the instant shown, the truck is moving forward with a speed of 2 ft/s and is slowing down at a rate of 0.25 ft/s². The length of the boom AB is decreasing at a constant rate of 0.5 ft/s, the angular velocity of the boom is 0.1 rad/s, and the angular acceleration of the boom is 0.02 rad/s², both clockwise. Determine the velocity and acceleration of point B.
STRATEGY: Since you are not given any forces and are asked to find the velocity and acceleration of a point, use rigid-body kinematics. The boom is moving with respect to the truck, so use a rotating reference frame.
MODELING and ANALYSIS: Attach a rotating coordinate system to the boom housing with its origin at A (Fig. 1).Velocity of B. 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}_B=\mathbf{v}_A + \boldsymbol{\Omega} \times \mathbf{r}_{B / A} + \mathbf{v}_{\mathrm{rel}} (1)
where \mathbf{v}_A=(2 \mathrm{ft} / \mathrm{s}) \mathbf{i}, \mathbf{r}_{B / A}=\left(20 \cos 30^{\circ} \mathrm{ft}\right) \mathbf{i} + \left(20 \sin 30^{\circ} \mathrm{ft}\right) \mathbf{j}, and \boldsymbol{\Omega}=\left(-0.1 \mathrm{rad} / \mathrm{s}^2\right) \mathbf{k}. To find the relative velocity, ask yourself what the velocity of B would be, assuming that the rotating coordinate system is not moving. In this case, \mathbf{v}_{\text {rel }}=-\left(0.5 \cos 30^{\circ} \mathrm{ft} / \mathrm{s}\right) \mathbf{i} – \left(0.5 \sin 30^{\circ} \mathrm{ft} / \mathrm{s}\right) \mathbf{j}. Substituting into Eq. (1) gives
\mathbf{v}_B=2 \mathbf{i} + (-0.1 \mathbf{k}) \times(17.32 \mathbf{i} + 10 \mathbf{j}) – (0.433 \mathbf{i} + 0.25 \mathbf{j})
\mathbf{v}_B=(2.57 \mathrm{ft} / \mathrm{s}) \mathbf{i} – (1.982 \mathrm{ft} / \mathrm{s}) \mathbf{j}
Acceleration of B. 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}_{\mathrm{rel}} + \mathbf{a}_{\mathrm{rel}} (15.35′)
\mathbf{a}_B=\mathbf{a}_A + \dot{\mathbf{\Omega}} \times \mathbf{r}_{B / A} – \Omega^2 \mathbf{r}_{B / A} + 2 \boldsymbol{\Omega} \times \mathbf{v}_{\mathrm{rel}} + \mathbf{a}_{\mathrm{rel}} (2)
where \mathbf{a}_A=-\left(0.25 \mathrm{ft} / \mathrm{s}^2\right) \mathbf{i}, \dot{\Omega}=-\left(0.02 \mathrm{rad} / \mathrm{s}^2\right) \mathbf{k}, and \mathbf{a}_{\text {rel }} = 0. Substituting into Eq. (2) gives
\begin{aligned}\mathbf{a}_B=&-0.25 \mathbf{i} + (-0.02 \mathbf{k}) \times(17.32 \mathbf{i} + 10 \mathbf{j})-0.1^2(17.32 \mathbf{i} + 10 \mathbf{j}) \\&+2(-0.1 \mathbf{k}) \times(-0.433 \mathbf{i} – 0.25 \mathbf{j}) + 0 \\=&-0.25 \mathbf{i} + (-0.3464 \mathbf{j} + 0.2 \mathbf{i}) – (0.1732 \mathbf{i} + 0.10 \mathbf{j}) + (0.0866 \mathbf{j} – 0.05 \mathbf{i}) + 0\end{aligned}
\mathbf{a}_B=\left(-0.273 \mathrm{ft} / \mathrm{s}^2\right) \mathbf{i} – \left(0.360 \mathrm{ft} / \mathrm{s}^2\right) \mathbf{j}
REFLECT and THINK: The biggest challenge with this problem is interpreting what you are given in the problem statement. After that, it is straightforward to substitute into the governing equations. The last four terms in Eq. (2) are analogous to the polar coordinate expressions we used in Chapter 11. The following terms represent the same physical quantities: \dot{\boldsymbol{\Omega}} \times \mathbf{r}_{B / A} \rightarrow r \ddot{\theta},-\Omega^2 \mathbf{r}_{B / A} \rightarrow-r \dot{\theta}^2, 2 \boldsymbol{\Omega} \times \mathbf{v}_{\mathrm{rel}} \rightarrow 2 \dot{r} \dot{\theta}
