Chapter 11.2, Problem 11.60P

The system shown starts from rest, and the length of the upper cord is adjusted so that A, B, and C are initially at the same level. Each component moves with a constant acceleration, and after 2 s the relative change in position of block C with respect to block A is 280 mm upward. Knowing that when the relative velocity of collar B with respect to block A is 80 mm/s downward, the displacements of A and B are 160 mm downward and 320 mm downward, respectively, determine (a) the accelerations of A and B if $a_B>10$ mm/s² (b) the change in position of block D when the velocity of block C is 600 mm/s upward.

  

To determine

(a)

The acceleration of ‘A’ and ‘B’ if $a_B>10mm/s^2$.

Answer to Problem 11.60P

$\begin{array}{l}{a}_A=20mm/s^2\\ a_B=40mm/s^2\end{array}$

Explanation of Solution

Given information:

At $t=0$

$y_{A_0}=y_{B_0}=y_{C_0}$

Each component moves in constant acceleration

After $t=2s$

The relative change in position of block ‘C’ with respect to block ‘A’ is $y_{C/A}=280mm$ in upward direction.

The velocity of collar ‘B’ with respect to ‘A’ is $80mm/s$ in downward direction.

The displacement of block ‘A’ is $160mm$ downwards.

The displacement of collar ‘B is $320mm$ downwards.

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 length of each cable is constant

For cable 1,

$2y_A+2y_B+y_C=C$

Differentiate above equation twice,

$\begin{array}{l}2v_A+2v_B+v_C=0\\ 2a_A+2a_B+a_C=0\end{array}$

For cable 2,

$\begin{array}{l}\left(y_D-y_A\right)+\left(y_D-y_B\right)=C\\ 2y_D-y_A-y_B=C\end{array}$

Differentiate above equation twice,

$\begin{array}{l}2v_D-v_A-v_B=0\\ 2a_D-a_A-a_B=0\end{array}$

For block ‘A’

$\begin{array}{l}{y}_A=y_{A_0}+v_{A_0}t+\frac{1}{2}{a}_A{t}^2\\ y_A=y_{A_0}+0+\frac{1}{2}{a}_A{t}^2\end{array}$

For block ‘C’

$\begin{array}{l}{y}_C=y_{C_0}+v_{C_0}t+\frac{1}{2}{a}_C{t}^2\\ y_C=y_{C_0}+0+\frac{1}{2}{a}_C{t}^2\end{array}$

Therefore,

$y_{C/A}=y_C-y_A=\frac{1}{2}{t}^2\left(a_C-a_A\right)$

At, $t=2s$

$\begin{array}{l}-280mm=\frac{1}{2}{\left(2s\right)}^2\left(a_C-a_A\right)\\ a_C=a_A-140\end{array}$

But we know that,

$2a_A+2a_B+a_C=0$

Then,

$\begin{array}{l}2a_A+2a_B+a_C=0\\ 2a_A+2a_B+\left(a_A-140\right)=0\\ a_A=\frac{\left(140-2a_B\right)}{3}\end{array}$

For block ‘A’

$\begin{array}{l}{y}_A=y_{A_0}+v_{A_0}t+\frac{1}{2}{a}_A{t}^2\\ y_A-y_{A_0}=0+\frac{1}{2}{a}_A{t}^2\\ 160mm=\frac{1}{2}{a}_A{t}^2\end{array}$

For block ‘B’

$\begin{array}{l}{y}_B=y_{B_0}+v_{B_0}t+\frac{1}{2}{a}_B{t}^2\\ y_B-y_{B_0}=0+\frac{1}{2}{a}_B{t}^2\\ 320mm=\frac{1}{2}{a}_B{t}^2\end{array}$

Then,

$160mm=\frac{1}{2}\left(a_B-a_A\right)t^2$

But we have already found,

$80mm=\left(a_B-a_A\right)t$

Therefore,

$t=4s$

Then, for block ‘A’

$\begin{array}{l}160mm=\frac{1}{2}{a}_A{\left(4s\right)}^2\\ a_A=20mm/s^2\end{array}$

Then, for collar ‘B’

$\begin{array}{l}320mm=\frac{1}{2}{a}_B{\left(4s\right)}^2\\ a_B=40mm/s^2\end{array}$

Conclusion:

The acceleration of ‘A’ and ‘B’ are

$\begin{array}{l}{a}_A=20mm/s^2\\ a_B=40mm/s^2\end{array}$

To determine

(b)

The change in position of block ‘D’ when the velocity of block ‘C’ is $600mm/s$

Answer to Problem 11.60P

$y_D-y_{D_0}=375mm$ Downwards

Explanation of Solution

Given information:

At $t=0$

$y_{A_0}=y_{B_0}=y_{C_0}$

Each component moves in constant acceleration

After $t=2s$

The relative change in position of block ‘C’ with respect to block ‘A’ is $y_{C/A}=280mm$ in upward direction.

The velocity of collar ‘B’ with respect to ‘A’ is $80mm/s$ in downward direction.

The displacement of block ‘A’ is $160mm$ downwards.

The displacement of collar ‘B is $320mm$ downwards.

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 a,

We have found

$a_C=a_A-140$

$2a_D-a_A-a_B=0$

Bu we know that,

$a_A=20mm/s$

Therefore,

$a_C=-120mm/s^2$

$\begin{array}{l}2a_D-a_A-a_B=0\\ 2a_D-20mm/s^2-40mm/s^2=0\\ a_D=30mm/s^2\end{array}$

When,

$v_C=-600mm/s$

Therefore,

$\begin{array}{l}{v}_C=v_{C_0}+a_{C}t\\ -600mm/s=0+\left(-120mm/s^2\right)t\\ t=5s\end{array}$

To find the change in distance of block ‘D’

$\begin{array}{l}{y}_D=y_{D_0}+v_{D_0}t+\frac{1}{2}{a}_D{t}^2\\ y_D-y_{D_0}=0+\frac{1}{2}\left(30mm/s^2\right){\left(5s\right)}^2\\ y_D-y_{D_0}=375mm\end{array}$

Conclusion:

The change in distance of block ‘D’ is equal to $y_D-y_{D_0}=375mm$ in downward direction