Chapter 11.2, Problem 11.57P

Block B starts from rest, block A moves with a constant acceleration, and slider block C moves to the right with a constant acceleration of 75 mm/s². Knowing that at t = 2 s the velocities of B and C are 480 mm/s downward and 280 mm/s to the right, respectively, determine (a) the accelerations of A and B, (b) the initial velocities of A and C, (c) the change in position of slider block C after 3 s.

Fig. P11.57 and P11.58

To determine

The acceleration of block A $\left(a_A\right)$ and block B $\left(a_B\right)$.

Answer to Problem 11.57P

The acceleration of block A $\left(a_A\right)$ and block B $\left(a_B\right)$ are $\underset{\_}{345{\text{mm/s}}^{\text{2}}(\uparrow )}$ and $\underset{\_}{240{\text{mm/s}}^{\text{2}}(\downarrow )}$ respectively.

Explanation of Solution

Given information:

The acceleration $\left(a_C\right)$ of block C is $75{\text{mm/s}}^2$.

At time (t) 2 sec the velocity $\left(v_B\right)$ of B is $480{\text{mm/s}}^2(\downarrow )$.

At time (t) 2 sec the velocity $\left(v_C\right)$ of C is $280{\text{mm/s}}^2(\to )$.

Calculation:

Show the length and position of the cables as in Figure (1).

Choose the coordinate downward positive and right side positive.

Write the express for total lengths of cables.

$3y_A+4y_B+x_C=\text{constant}$

Differentiate the above equation with respective to time (t).

$3\frac{d{y}_A}{dt}+4\frac{d{y}_B}{dt}+\frac{d{x}_C}{dt}=\text{0}$

Denotes $\frac{d{y}_A}{dt}$ as $v_A$, $\frac{d{x}_C}{dt}$ as $v_C$ and $\frac{d{y}_B}{dt}$ as $v_B$.  Since, rate of change of any coordinate with respect to time is equal to the velocity.

$3v_A+4v_B+v_C=\text{0}$ (1)

Differentiate the equation (2) with respective to time (t).

$3\frac{d{v}_A}{dt}+4\frac{d{v}_B}{dt}+\frac{d{v}_C}{dt}=\text{0}$

Denotes $\frac{d{v}_C}{dt}$ as $a_C$, $\frac{d{v}_B}{dt}$ for $a_B$ and $\frac{d{v}_A}{dt}$ for $a_A$.

$3a_A+4a_B+a_C=\text{0}$ (2)

Calculate the velocity $\left(v_B\right)$ block B using the relation:

$v_B={\left(v_B\right)}_0+a_{B}t$

Here, ${\left(v_B\right)}_0$ is initial velocity of block B.

Substitute 0 for ${\left(v_B\right)}_0$, $480{\text{mm/s}}^2$ for $v_B$ and 2 sec for t.

$\begin{array}{l}480=0+a_B\times 2\\ a_B=\frac{480}{2}\\ a_B=240{\text{mm/s}}^{\text{2}}(\downarrow )\end{array}$

Calculate the acceleration of block A:

Substitute $240{\text{mm/s}}^{\text{2}}(\downarrow )$ for $a_B$ and $75{\text{mm/s}}^2$ for $\left(a_C\right)$.

$\begin{array}{l}3a_A+4\left(240\right)+\left(75\right)=\text{0}\\ a_A=\frac{-1035}{3}\\ a_A=-345{\text{mm/s}}^{\text{2}}\\ a_A=345{\text{mm/s}}^{\text{2}}(\uparrow )\end{array}$

Therefore, the acceleration of block A $\left(a_A\right)$ and block B $\left(a_B\right)$ are $\underset{\_}{345{\text{mm/s}}^{\text{2}}(\uparrow )}$ and $\underset{\_}{240{\text{mm/s}}^{\text{2}}(\downarrow )}$ respectively.

To determine

The initial velocities of block A ${\left(v_A\right)}_0$ and block C ${\left(v_C\right)}_0$.

Answer to Problem 11.57P

The initial velocities of block A ${\left(v_A\right)}_0$ and block C ${\left(v_C\right)}_0$ are $\underset{\_}{43.3\text{mm/s}(\uparrow )}$ and $\underset{\_}{130\text{mm/s}(\to )}$ respectively.

Explanation of Solution

Given information:

The acceleration $\left(a_C\right)$ of block C is $75{\text{mm/s}}^2$.

At time (t) 2 sec the velocity $\left(v_B\right)$ of B is $480{\text{mm/s}}^2(\downarrow )$.

At time (t) 2 sec the velocity $\left(v_C\right)$ of C is $280{\text{mm/s}}^2(\to )$.

Calculation:

Calculate the initial velocity ${\left(v_C\right)}_0$ of block C using the relation:

$v_C={\left(v_C\right)}_0+a_{C}t$

Substitute $75{\text{mm/s}}^{\text{2}}$ for $a_C$, 2 sec for t and $280{\text{mm/s}}^2$ for $v_C$.

$\begin{array}{l}280={\left(v_C\right)}_0+\left(75\times 2\right)\\ {\left(v_C\right)}_0=280-\left(75\times 2\right)\\ {\left(v_C\right)}_0=130\text{mm/s}(\to )\end{array}$

Calculate the initial velocity ${\left(v_A\right)}_0$ of block A:

Modify equation (1) as,

$3{\left(v_A\right)}_0+4{\left(v_B\right)}_0+{\left(v_C\right)}_0=\text{0}$

At time (t) is 0 the initial velocity ${\left(v_B\right)}_0$ is 0.

Substitute at 0 for ${\left(v_B\right)}_0$ and $130\text{mm/s}$ for ${\left(v_C\right)}_0$.

$\begin{array}{l}3{\left(v_A\right)}_0+4\left(0\right)+130=\text{0}\\ {\left(v_A\right)}_0=\frac{-130}{3}\\ {\left(v_A\right)}_0=-43.33\text{mm/s}\\ {\left(v_A\right)}_0=43.33\text{mm/s}(\uparrow )\end{array}$

Therefore, the initial velocities of block A ${\left(v_A\right)}_0$ and block C ${\left(v_C\right)}_0$ are $\underset{\_}{43.3\text{mm/s}(\uparrow )}$ and $\underset{\_}{130\text{mm/s}(\to )}$ respectively.

To determine

The change in position $\left(x_C-{\left(x_C\right)}_0\right)$ of slider block C after 3 sec.

Answer to Problem 11.57P

The change in position $\left(x_C-{\left(x_C\right)}_0\right)$ of slider block C after 3sec is $\underset{\_}{728\text{mm}(\to )}$.

Explanation of Solution

Given information:

The acceleration $\left(a_C\right)$ of block C is $75{\text{mm/s}}^2$.

At time (t) 2 sec the velocity $\left(v_B\right)$ of B is $480{\text{mm/s}}^2(\downarrow )$.

At time (t) 2 sec the velocity $\left(v_C\right)$ of C is $280{\text{mm/s}}^2(\to )$.

Calculation:

Calculate the change in positon $\left(x_C-{\left(x_C\right)}_0\right)$ of slider block C after 3 sec using the relation:

$\begin{array}{l}{x}_C={\left(x_C\right)}_0+{\left(v_C\right)}_{0}t+\frac{1}{2}{a}_C{t}^2\\ \left(x_C-{\left(x_C\right)}_0\right)={\left(v_C\right)}_{0}t+\frac{1}{2}{a}_C{t}^2\end{array}$

Here, ${\left(x_C\right)}_0$ is initial position of block C.

Substitute $130\text{mm/s}$ for ${\left(v_C\right)}_0$, 3 sec for t and $75{\text{mm/s}}^2$ for $a_C$.

$\begin{array}{c}\left(x_C-{\left(x_C\right)}_0\right)=\left(130\times 3\right)+\frac{1}{2}\times 75\times 3^2\\ =390+337.5\\ =727.5\text{mm}\approx \text{728}\text{mm}(\to )\end{array}$

Therefore, the change in position $\left(x_C-{\left(x_C\right)}_0\right)$ of slider block C after 3sec is $\underset{\_}{728\text{mm}(\to )}$.