Solved Problem on Dynamics

Three bodies A, B, and C are suspended by inextensible cords as shown in the figure. Body B is simultaneously suspended by two cords, one connected to body A and another to body C. Determine:
a) Acceleration and direction of motion if all masses are equal to m;
b) Acceleration and direction of motion, if the masses A and C are equal to m and mass B equal to 3m;
c) If the masses A and C are equal to m, what should be the value of mass B for the movement to give upwards with an acceleration equal to 0.5g?

Problem diagram:

We randomly choose a direction for acceleration, positive for bodies A and C descending and body B rising. Drawing a free-body diagram, we apply Newton's Second Law

\[ \begin{gather} \bbox[#99CCFF,10px] {\vec{F}=m\vec{a}} \tag{I} \end{gather} \]

Figure 1

Solution

Body A:

\( {\vec F}_{gA} \): gravitational force of body A;
\( \vec{T} \): tension force on the cord.

In the vertical direction, we have the acceleration in the direction of the gravitational force of the body A, this will be positive, the tension is in the opposite direction and will be negative, applying the expression (I)

\[ \begin{gather} F_{gA}-T=m_{A}a \tag{II} \end{gather} \]

Body B:

\( {\vec F}_{gB} \): gravitational force of body B;
\( \vec{T} \): tension force on the cord.

In this body, the acceleration is in the same direction of the force of tensions, these will be positive, and the gravitational force of the body B is in the opposite direction and will be negative, applying the expression (I)

\[ \begin{gather} T+T-F_{gB}=m_{B}a\\[5pt] 2T-F_{gB}=m_{B}a \tag{III} \end{gather} \]

Note: It is not necessary to analyze body C, for this, by symmetry, has the same behavior as body A.

Expressions (II) and (III) can be written as a system of two equations

\[ \begin{gather} \left\{ \begin{matrix} F_{gA}-T=m_{A}a\\ 2T-F_{gB}=m_{B}a \end{matrix} \right. \tag{IV} \end{gather} \]

the gravitational force is given by

\[ \begin{gather} \bbox[#99CCFF,10px] {F_{g}=mg} \end{gather} \]

the gravitational forces of bodies A and B are given by

\[ \begin{gather} F_{gA}=m_{A}g \tag{V-a} \end{gather} \]

\[ \begin{gather} F_{gB}=m_{B}g \tag{V-b} \end{gather} \]

substituting expressions (V-A) and (V-B) in the system (IV)

\[ \begin{gather} \left\{ \begin{matrix} \;m_{A}g-T=m_{A}a\\ \;2T-m_{B}g=m_{B}a \end{matrix} \right. \tag{VI} \end{gather} \]

a) Substituting m_A=m_B=m in system (VI)

\[ \begin{gather} \left\{ \begin{matrix} mg-T=ma\\ 2T-mg=ma \end{matrix} \right. \end{gather} \]

solving the first equation for T and substituting on the second equation

\[ \begin{gather} T=mg-ma\\[5pt] 2(mg-ma)-mg=ma\\[5pt] 2mg-2ma-mg=ma\\[5pt] mg=ma+2ma\\[5pt] \cancel{m}g=3\cancel{m}a\\[5pt] 3a=g \end{gather} \]

\[ \begin{gather} \bbox[#FFCCCC,10px] {a=\frac{g}{3}} \end{gather} \]

Direction: Body B rises, bodies A and C descends.

b) Substituting m_A=m and m_B=3m in system (VI)

\[ \begin{gather} \left\{ \begin{array}{l} mg-T=ma\\ 2T-3mg=3ma \end{array} \right. \end{gather} \]

solving the first equation for T and substituting on the second equation

\[ \begin{gather} T=mg-ma\\[5pt] 2(mg-ma)-3mg=3ma\\[5pt] 2mg-2ma-3mg=3ma\\[5pt] -mg=3ma+2ma\\[5pt] -\cancel{m}g=5\cancel{m}a\\[5pt] 5a=-g\\[5pt] a=-{\frac{g}{5}} \end{gather} \]

The negative signal in acceleration indicates that the motion will be in the opposite direction to the direction chosen in Figure 1, the acceleration will be

\[ \begin{gather} \bbox[#FFCCCC,10px] {a=\frac{g}{5}} \end{gather} \]

Direction: Body B descents, bodies A and C rise.

c) We have m_A=m, and we want to determine m_B when a = 0.5g, substituting these values in the system (VI)

\[ \begin{gather} \left\{ \begin{matrix} mg-T=m0.5g\\ 2T-m_{B}g=m_{B}0.5g \end{matrix} \right.\\[5pt] \left\{ \begin{matrix} mg-T=0.5mg\\ 2T-m_{B}g=0.5m_{B}g \end{matrix} \right. \end{gather} \]

solving the first equation for T and substituting on the second equation

\[ \begin{gather} T=mg-0.5mg\\[5pt] T=0.5mg\\[5pt] 2\times (0.5mg)-m_{B}g=0.5m_{B}g\\[5pt] mg=0.5m_{B}g+m_{B}g\\[5pt] m\cancel{g}=1.5m_{B}\cancel{g}\\[5pt] m=\frac{3}{2}m_{B}\\[5pt] m_{B}=\frac{m}{\dfrac{3}{2}} \end{gather} \]

\[ \begin{gather} \bbox[#FFCCCC,10px] {m_{B}=\frac{2}{3}m} \end{gather} \]

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