Solved Problem on Dynamics

In the system of the figure, body B slides on a horizontal surface without friction. It is connected through ropes and pulleys, lightweight and frictionless, with two bodies, A and C, that move vertically. The masses of A, B, and C are, respectively, 5 kg, 2 kg and 3 kg.Find the acceleration of the system and the magnitude of tension force in the cordss.

Problem data:

Mass of body A: m_A = 5 kg;
Mass of body B: m_B = 2 kg;
Mass of body C: m_C = 3 kg;
Acceleration due to gravity: g = 9.8 m/s².

Problem diagram:

We choose the acceleration in the direction in which body A moves downward.

Figure 1

Drawing a Free-Body Diagram for each block.

Body A (Figure 2):
- Vertical direction:
  - \( \vec W_{\small A} \): weight of body A;
  - \( \vec T_{\small AB} \): tension force in the cord between blocks A and B.

Figure 2

> Body B (Figure 3):
- Vertical direction:
  - \( \vec W_{\small B} \): weight of body B;
  - \( \vec N_{\small B} \): normal reaction force of the surface on the body.
- Horizontal direction:
  - \( \vec T_{\small AB} \): tension force in the rope between blocks A and B;
  - \( \vec T_{\small BC} \): tension force in the rope between blocks B and C.

Figure 3

Corpo C (Figure 4):
- Vertical direction:
  - \( \vec W_{\small C} \): weight of body C;
  - \( \vec T_{\small BC} \): tension force in the rope between blocks B and C.

Figure 4

Solution

Applying Newton's Second Law

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

Body A:

In the horizontal direction, no forces are acting.
In the vertical direction

\[ \begin{gather} W_{\small A}-T_{\small {AB}}=m_{\small A}a \tag{I} \end{gather} \]

Body B:

In the vertical direction, the weight and the normal force cancel out, there is no vertical motion.
In the horizontal direction

\[ \begin{gather} T_{\small {AB}}-T_{\small {BC}}=m_{\small B}a \tag{II} \end{gather} \]

Corpo C:

In the horizontal direction, there are no forces acting.
in the vertical direction

\[ \begin{gather} T_{\small {BC}}-W_{\small C}=m_{\small C}a \tag{III} \end{gather} \]

The weight is given by

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

For the bodies A and C

\[ \begin{gather} W_{\small A}=m_{\small A}g \tag{IV-a} \end{gather} \]

\[ \begin{gather} W_{\small C}=m_{\small C}g \tag{IV-b} \end{gather} \]

substituting equation (IV-a) into equation (I)

\[ \begin{gather} m_{\small A}g-T_{\small {AB}}=m_{\small A}a \tag{V-a} \end{gather} \]

substituting equation (IV-b) into equation (III)

\[ \begin{gather} T_{\small {BC}}-m_{\small C}g=m_{\small C}a \tag{V-b} \end{gather} \]

The equations (II), (V-a) e (V-b) can be written as a system of linear equations with three variables (T_AB, T_BC, and a), adding the three equations

\[ \begin{gather} \frac{ \left\{ \begin{array}{rr} m_{\small A}g-\cancel{T_{\small {AB}}} &=m_{\small A}a\\ \cancel{T_{\small {AB}}}-\cancel{T_{\small {BC}}}&=m_{\small B}a\\ \cancel{T_{\small {BC}}}-m_{\small C}g &=m_{\small C}a \end{array} \right.} {m_{\small A}g-m_{\small C}g=\left(m_{\small A}+m_{\small B}+m_{\small C}\right)a}\\[5pt] a=\frac{m_{\small A}g-m_{\small C}g}{m_{\small A}+m_{\small B}+m_{\small C}}\\[5pt] a=\frac{\left(5\;\mathrm{kg}\right)\left(9.8\;\frac{\mathrm m}{\mathrm{s^2}}\right)-\left(3\;\mathrm{kg}\right)\left(9.8\;\mathrm{\frac{m}{s^2}}\right)}{(5\;\mathrm{kg})+(2\;\mathrm{kg})+(3\;\mathrm{kg})}\\[5pt] a=\frac{19.6\;\mathrm{\frac{\cancel{kg}.m}{s^2}}}{10\;\mathrm{\cancel{kg}}} \end{gather} \]

\[ \begin{gather} \bbox[#FFCCCC,10px] {a\approx 2\;\mathrm{m/s^2}} \end{gather} \]

Substituting the mass of body A and the acceleration, found above, in the first expression of the system, the tension in the cord will be

\[ \begin{gather} m_{\small A}g-T_{\small {AB}}=m_{\small A}a\\[5pt] T_{\small {AB}}=m_{\small A}g-m_{\small A}a\\[5pt] T_{AB}=(5\;\mathrm{kg})\left(9.8\;\mathrm{\frac{m}{s^2}}\right)-(5\;\mathrm{kg})\left(2\;\mathrm{\frac{m}{s^2}}\right) \end{gather} \]

\[ \begin{gather} \bbox[#FFCCCC,10px] {T_{\small {AB}}=39\; \mathrm N} \end{gather} \]

Substituting the mass of body C and the acceleration, found above, in the third expression of the system, the tension in the cord will be

\[ \begin{gather} T_{\small {BC}}-m_{\small C}g=m_{\small C}a\\[5pt] T_{\small {BC}}=m_{\small C}a+m_{\small C}g\\[5pt] T_{BC}=(3\;\mathrm{kg})\left(2\;\mathrm{\frac{m}{s^2}}\right)+(3\;\mathrm{kg})\left(9.8\;\mathrm{\frac{m}{s^2}}\right) \end{gather} \]

\[ \begin{gather} \bbox[#FFCCCC,10px] {T_{\small {BC}}\approx 35.4\; \mathrm N} \end{gather} \]

Fisicaexe - Physics Solved Problems by Elcio Brandani Mondadori is licensed under a Creative Commons Attribution-NonCommercial-ShareAlike 4.0 International License .