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.

  

To determine

(a)

The acceleration A and B.

Answer to Problem 11.57P

$a_B=240mm/s^2$ Downwards

$a_A=345mm/s^2$ Upwards

Explanation of Solution

Given information:

Block ‘B’ starts from rest

Block ‘A’ moves with constant acceleration

Block ‘C’ moves with constant acceleration of $75mm/s^2$

After, $t=2s$

$\begin{array}{l}{v}_B=480mm/s\\ v_C=280mm/s\end{array}$

In a dependent motion of particles such as the pulley system shown above,

The total length of the rope is a constant

For example,

$2x_A+2x_B+x_C=C$

$C$ - Constant

$2\frac{d{x}_A}{dt}+2\frac{d{x}_B}{dt}+\frac{d{x}_C}{dt}=0$

Calculation:

In above system the total length of rope is constant, therefore,

$3y_A+4y_B+x_C=C$

Where,

$C$ - Constant

Differentiate above equation,

$3v_A+4v_B+v_C=0$

Differentiate above equation,

$3a_A+4a_B+a_C=0$

For block ‘B’

At $t=2s$

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

Substitute,

$\begin{array}{l}480mm/s=0+a_B\left(2s\right)\\ a_B=240mm/s^2\end{array}$

Therefore,

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

Conclusion:

Acceleration of block A and B are,

$a_B=240mm/s^2$ Downwards

$a_A=345mm/s^2$ Upwards

To determine

(b)

Initial velocities of A and C.

Answer to Problem 11.57P

$\begin{array}{l}{v}_{A_0}=43.33mm/s\\ v_{C_0}=130mm/s\end{array}$

Explanation of Solution

Given information:

Block ‘B’ starts from rest

Block ‘A’ moves with constant acceleration

Block ‘C’ moves with constant acceleration of $75mm/s^2$

After, $t=2s$

$\begin{array}{l}{v}_B=480mm/s\\ v_C=280mm/s\end{array}$

For a uniformly accelerated motion,

$v=v_0+at$

In above equation,

$v$ - End velocity

$v_0$ - Start velocity

$a$ - Acceleration

$t$ - Elapsed time

Calculation:

For block ‘C’

At $t=2s$

$\begin{array}{l}{v}_C={\left(v_C\right)}_0+a_{C}t\\ 280mm/s={\left(v_C\right)}_0+\left(75mm/s^2\right)\left(2s\right)\\ {\left(v_C\right)}_0=130mm/s\\ \end{array}$

According to sub-part a,

We know that

$3v_A+4v_B+v_C=0$

Therefore,

$\begin{array}{l}3v_A+4v_B+v_C=0\\ 3v_A+0+130mm/s=0\\ v_{A_0}=43.33mm/s\end{array}$

Conclusion:

Initial velocitiesof block A and C are,

$\begin{array}{l}{v}_{A_0}=43.33mm/s\\ v_{C_0}=130mm/s\end{array}$

To determine

(c)

The change in position of block ‘C’ after $3s.$

Answer to Problem 11.57P

$x_C-x_{C_0}=727.5mm$

Explanation of Solution

Given information:

Block ‘B’ starts from rest

Block ‘A’ moves with constant acceleration

Block ‘C’ moves with constant acceleration of $75mm/s^2$

After, $t=2s$

$\begin{array}{l}{v}_B=480mm/s\\ v_C=280mm/s\end{array}$

For a uniformly accelerated motion,

$x=x_0+v_{0}t+\frac{1}{2}a{t}^2$

In above equation,

$x$ - End distance

$x_0$ - Start distance

$v_0$ - Initial velocity

$t$ - Elapsed time

$a$ -Acceleration

Calculation:

According to sub part b,

$v_{C_0}=130mm/s$

For block C,

After $3s$

$\begin{array}{l}{x}_C=x_{C_0}+v_{C_0}t+\frac{1}{2}{a}_C{t}^2\\ x_C-x_{C_0}=\left(130mm/s\right)\left(3s\right)+\frac{1}{2}\left(75mm/s^2\right){\left(3s\right)}^2\\ x_C-x_{C_0}=727.5mm\end{array}$

Conclusion:

Change in distance of block C after $3s$ is equal to $x_C-x_{C_0}=727.5mm$