Solved Problem on Dynamics
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In the system of the figure, body A slides on a horizontal surface without friction, dragged by body B which moves downward. Bodies A and B are tied by a rope of negligible mass parallel to the surface and pass through a frictionless pulley of negligible mass. The masses of A and B are, respectively, 32 kg and 8 kg. Find the acceleration of the system and the tension on the rope.


Problem data:
  • Mass of body A:    mA = 32 kg;
  • Mass of body B:    mB = 8 kg;
  • Acceleration due to gravity:    g = 9.8 m/s2.
Problem diagram:

We choose a frame of reference pointing to the right in the same direction as the acceleration.
Figure 1

Drawing a Free-Body Diagram for each block.

  • Block A (Figure 2):
    • Vertical direction:
      • \( \vec W_{\small A} \): weight of body A;
      • \( \vec N_{\small A} \): normal reaction force of the surface on the body.
    • Horizontal direction:
      • \( \vec T \): tension force on the cord.
Figure 2

  • Body B (Figure 3):
    • \( \vec W_{\small B} \): weight of body B;
    • \( \vec T \): tension force on the rope.
Figure 3

Solution

Applying Newton's Second Law
\[ \begin{gather} \bbox[#99CCFF,10px] {\vec F=m\vec a} \end{gather} \]
  • Body A:
In the vertical direction, the weight and the normal force cancel out
In the horizontal direction
\[ \begin{gather} T=m_{\small A}a \tag{I} \end{gather} \]
  • Body B:
In the horizontal direction, no forces are acting.
In the vertical direction
\[ \begin{gather} W_{\small B}-T=m_{\small B}a \tag{II} \end{gather} \]
The weight is given by
\[ \begin{gather} \bbox[#99CCFF,10px] {W=mg} \end{gather} \]
for body B
\[ \begin{gather} W_{\small B}=m_{\small B}g \tag{III} \end{gather} \]
substituting equation (III) into equation (II)
\[ \begin{gather} m_{\small B}g-T=m_{\small B}a \tag{IV} \end{gather} \]
Expressions (I) and (II) can be written as a system of linear equations with two variables (T and a), adding the two equations
\[ \begin{gather} \frac{ \left\{ \begin{array}{rr} \cancel{T}&=m_{\small A}a\\ m_{\small B}g-\cancel{T}&=m_{\small B}a \end{array} \right.} {m_{\small B}g=\left(m_{\small A}+m_{\small B}\right)a}\\[5pt] a=\frac{m_{\small B}g}{m_{\small A}+m_{\small B}}\\[5pt] a=\frac{(8\;\mathrm{kg})\left(9.8\;\mathrm{\frac{m}{s^2}}\right)}{32\;\mathrm{kg}+8\;\mathrm{kg}}\\[5pt] a=\frac{78.4\;\mathrm{\frac{\cancel{kg}.m}{s^2}}}{40\;\mathrm{\cancel{kg}}} \end{gather} \]
\[ \begin{gather} \bbox[#FFCCCC,10px] {a\approx 1.9\;\mathrm{m/s^2}} \end{gather} \]
Substituting the mass of body A and the acceleration found above, in the first equation of the system the tension force on hope will be
\[ \begin{gather} T=(32\;\mathrm{kg})\left(1.9\;\mathrm{\frac{m}{s^2}}\right) \end{gather} \]
\[ \begin{gather} \bbox[#FFCCCC,10px] {T\approx 62.7\;\mathrm N} \end{gather} \]
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Fisicaexe - Physics Solved Problems by Elcio Brandani Mondadori is licensed under a Creative Commons Attribution-NonCommercial-ShareAlike 4.0 International License .