Solved Problem on Dynamics
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In the system of the figure, the body A slides on a horizontal surface without friction, dragged by body B that moves downward. Bodies A and B are attached by a rope of negligible mass parallel to the surface and passing 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 cord. Assume g = 10 m/s2.
Block A on a horizontal surface connected by a rope through a pulley to a suspended block B.

Problem data
  • mass of body A:    mA = 32 kg;
  • mass of body B:    mB = 8 kg;
  • free-fall acceleration:    g = 10 m/s2.
Problem diagram

We choose the acceleration in the direction in which the body B moves downward. Drawing a free-body diagram for each block and using Newton's Second Law
\[ \bbox[#99CCFF,10px] {\vec{F}=m\vec{a}} \]
Acceleration of the system in the direction in which block B is moving downward.
figure 1

Body A

Vertical direction
  • \( \vec P_{\text{A}} \): weight of body A;
  • \( \vec N_{\text{A}} \): normal force of the surface on the body.
Horizontal direction
  • \( \vec T \): tension on the cord.
Block A where the normal force (NA), weight (PA), and tension (T) act on the rope.
figure 2
In the vertical direction, the weight and the normal force cancel out, there is no vertical motion.
In the horizontal direction using Newton's Second Law we have
\[ T=m_{\text{A}}a \tag{I} \]
Body B
  • \( \vec W_{\text{B}} \): weight of body B;
  • \( \vec T \): tension on the rope.
In the horizontal direction, there are no forces acting, in the vertical direction using Newton's Second Law we have
Block B, where the weight (PB) and tension (T) act on the rope.
figure 3
\[ W_{\text{B}}-T=m_{\text{B}}a \tag{II} \]

Expressions (I) and (II) can be written as a system of linear equations with two variables (T and a), adding the two equations we have
\[ \frac{ \left\{ \begin{array}{rr} \cancel{T}&=m_{\text{A}}a\\ W_{\text{B}}-\cancel{T}&=m_{\text{A}}a \end{array} \right.} {W_{\text{B}}=\left(m_{\text{A}}+m_{\text{B}}\right)a} \]
\[ a\;=\;\frac{W_{\text{B}}}{m_{\text{A}}+m_{\text{B}}} \tag{III} \]
The weight of body B is given by
\[ W_{\text{B}}=m_{\text{B}}g \tag{IV} \]
substituting (IV) into (III) and the values given in the problem
\[ a=\frac{m_{\text{B}}g}{m_{\text{A}}+m_{\text{B}}}\\ a=\frac{8 \times 10}{32+8}\\ a=\;\frac{80}{40} \]
\[ \bbox[#FFCCCC,10px] {a=2\;\text{m/s}^{2}} \]
Substituting the mass of the body A and the acceleration, found above, in the first expression the tension is
\[ T=32 \times 2 \]
\[ \bbox[#FFCCCC,10px] {T=64\ \text{N}} \]


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