Chapter 12.1, Problem 12.28P

To determine

(a)

The magnitude of force P.

Answer to Problem 12.28P

$P=10{\rm N}$

Explanation of Solution

Given information:

Mass of block A $=10kg$

Mass of block B=C $=5kg$

Time $t=2\sec$

Let the position co-ordinate be Y and it is positive towards downwards.

Now, constraint of cord AD: $y_A+y_D=$ constant

Then, $\begin{array}{l}{v}_A+v_D=0\\ \Rightarrow v_D=-v_A\_\_\_\_\_\_\_\_\_\_(1)\end{array}$ or,

$\begin{array}{l}{a}_A+a_D=0\\ \Rightarrow a_D=-a_A\_\_\_\_\_\_\_\_\_\_(2)\end{array}$

Similarly, constraint of cord BC: $\left(y_B-y_D\right)+\left(y_C-y_D\right)=$ constant

$y_B+y_C-2y_D=$ Constant

Then, $v_B+v_C-2v_D=0$ or,

$a_B+a_C-2a_D=0$

Now, from equation (1) and (2);

$\begin{array}{l}{v}_B+v_C-2\left(-v_A\right)=0\\ 2v_A+v_B+v_C=0\end{array}$ And,

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

As all of the forces are constant then, we have uniformly accelerated motions.

$y_B={\left(y_B\right)}_0+{\left(v_B\right)}_{0}t+\frac{1}{2}{a}_B{t}^2$ or,

${\left(v_B\right)}_0=0$

Then,

$\begin{array}{l}{a}_B=\frac{2\left[y_B-{\left(y_B\right)}_0\right]}{t^2}\\ a_B=\frac{2\left[3\right]}{{\left(2\right)}^2}=1.5m/s^2\end{array}$

Now, calculate all the forces for pulley D,

$\sum F_y=0$

$\begin{array}{l}{T}_{BC}+T_{BC}-T_{AD}=0\\ 2T_{BC}-T_{AD}=0or,\\ 2T_{BC}=T_{AD}\end{array}$

Now, calculate all the forces for block A, for this the draw the free body and kinetic diagram of the block A:

$\sum F_y=m{a}_y$

$\begin{array}{l}{W}_A-T_{AD}=m_A{a}_A\\ a_A=\frac{W_A-T_{AD}}{m_A}or,\\ a_A=\frac{W_A-2T_{BC}}{m_A}\_\_\_\_\_\_\_\_(4)\end{array}$

Calculate all the forces for block C, for this the draw the free body and kinetic diagram of block C:

$\sum F_y=m{a}_y$

$\begin{array}{l}{W}_C-T_{BC}=m_C{a}_C\\ a_C=\frac{W_C-T_{BC}}{m_A}\_\_\_\_\_\_\_\_(5)\end{array}$

Calculate all the forces for block B, for this the draw the free body and kinetic diagram of block B:

$\sum F_y=m{a}_y$

$\begin{array}{l}P+W_B-T_{BC}=m_B{a}_B\\ P=m_B{a}_B+T_{BC}-W_B\_\_\_\_\_\_(6)\end{array}$

Substituting the values of equation (4) and (5) into equation (3);

$\begin{array}{l}2\left(\frac{W_A-2T_{BC}}{m_A}\right)+a_B+\left(\frac{W_C-T_{BC}}{m_C}\right)=0\\ 2\left(\frac{m_{A}g-2T_{BC}}{m_A}\right)+a_B+\left(\frac{m_{C}g-T_{BC}}{m_C}\right)=0\\ 2g-\frac{4T_{BC}}{m_A}+a_B+g-\frac{T_{BC}}{m_C}=0\\ 3g+a_B=\left(\frac{4}{m_A}+\frac{1}{m_C}\right)T_{BC}\\ 3\left(9.81\right)+1.5=\left(\frac{4}{10}+\frac{1}{5}\right)T_{BC}\\ T_{BC}\left(\frac{3}{5}\right)=30.93\\ T_{BC}=51.55{\rm N}\end{array}$

Putting all the values in equation (6);

$\begin{array}{l}P=m_B{a}_B+T_{BC}-W_B\\ P=5\left(1.5\right)+51.55-5\left(9.81\right)\\ P=7.5+51.55-49.05\\ P=10{\rm N}\end{array}$

To determine

(b)

The tension in cord AD.

Answer to Problem 12.28P

$T_{AD}=103.1{\rm N}$

Explanation of Solution

Given information:

Mass of block A $=10kg$

Mass of block B=C $=5kg$

Time $t=2\sec$

Let the position co-ordinate be Y and it is positive towards downwards.

Now, constraint of cord AD: $y_A+y_D=$ constant

Then, $\begin{array}{l}{v}_A+v_D=0\\ \Rightarrow v_D=-v_A\_\_\_\_\_\_\_\_\_\_(1)\end{array}$ or,

$\begin{array}{l}{a}_A+a_D=0\\ \Rightarrow a_D=-a_A\_\_\_\_\_\_\_\_\_\_(2)\end{array}$

Similarly, constraint of cord BC: $\left(y_B-y_D\right)+\left(y_C-y_D\right)=$ constant

$y_B+y_C-2y_D=$ Constant

Then, $v_B+v_C-2v_D=0$ or,

$a_B+a_C-2a_D=0$

Now, from equation (1) and (2);

$\begin{array}{l}{v}_B+v_C-2\left(-v_A\right)=0\\ 2v_A+v_B+v_C=0\end{array}$ And,

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

As all of the forces are constant then we have uniformly accelerated motions.

$y_B={\left(y_B\right)}_0+{\left(v_B\right)}_{0}t+\frac{1}{2}{a}_B{t}^2$ or,

${\left(v_B\right)}_0=0$

Then,

$\begin{array}{l}{a}_B=\frac{2\left[y_B-{\left(y_B\right)}_0\right]}{t^2}\\ a_B=\frac{2\left[3\right]}{{\left(2\right)}^2}=1.5m/s^2\end{array}$

Now, calculate all the forces for pulley D,

$\sum F_y=0$

$\begin{array}{l}{T}_{BC}+T_{BC}-T_{AD}=0\\ 2T_{BC}-T_{AD}=0or,\\ 2T_{BC}=T_{AD}\end{array}$

Now, calculate all the forces for block A, for this the draw the free body and kinetic diagram of the block A:

$\sum F_y=m{a}_y$

$\begin{array}{l}{W}_A-T_{AD}=m_A{a}_A\\ a_A=\frac{W_A-T_{AD}}{m_A}or,\\ a_A=\frac{W_A-2T_{BC}}{m_A}\_\_\_\_\_\_\_\_(4)\end{array}$

Calculate all the forces for block C, for this the draw the free body and kinetic diagram of the block C:

$\sum F_y=m{a}_y$

$\begin{array}{l}{W}_C-T_{BC}=m_C{a}_C\\ a_C=\frac{W_C-T_{BC}}{m_A}\_\_\_\_\_\_\_\_(5)\end{array}$

Substituting the values of equation (4) and (5) into equation (3);

$\begin{array}{l}2\left(\frac{W_A-2T_{BC}}{m_A}\right)+a_B+\left(\frac{W_C-T_{BC}}{m_C}\right)=0\\ 2\left(\frac{m_{A}g-2T_{BC}}{m_A}\right)+a_B+\left(\frac{m_{C}g-T_{BC}}{m_C}\right)=0\\ 2g-\frac{4T_{BC}}{m_A}+a_B+g-\frac{T_{BC}}{m_C}=0\\ 3g+a_B=\left(\frac{4}{m_A}+\frac{1}{m_C}\right)T_{BC}\\ 3\left(9.81\right)+1.5=\left(\frac{4}{10}+\frac{1}{5}\right)T_{BC}\\ T_{BC}\left(\frac{3}{5}\right)=30.93\\ T_{BC}=51.55{\rm N}\end{array}$

Now, tension in cord AD; $T_{AD}=2T_{BC}$

$\begin{array}{l}{T}_{AD}=2\left(51.55\right)\\ T_{AD}=103.1{\rm N}\end{array}$