Solved Problem on Dynamics
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The bodies A and B have masses of 3m and 2m respectively and slide without friction over the horizontal plane, the body C, hanging on the rope, has mass m. Consider that the rope has a negligible mass, the pulley is without friction, and the gravitational acceleration is g. Calculate the magnitudes of:
a) The acceleration of the body C;
b) The force of reaction of body B on A.


Problem data:
  • Mass of body A:    mA = 3m;
  • Mass of body B:    mB = 2m;
  • Mass of body C:    mC = m;
  • Acceleration due to gravity:    g.
Problem diagram:

We choose acceleration in the direction in which body C is descending (Figure 1).
Figure 1

Solution

Drawing a free-body diagram, we have the forces that act in each body and we apply Newton's Second Law
\[ \begin{gather} \bbox[#99CCFF,10px] {\vec{F}=m\vec{a}} \tag{I} \end{gather} \]
Body A:
  • \( {\vec F}_{gA} \): gravitational force on the body A;
  • \( {\vec N}_{A} \): normal reaction force;
  • \( \vec{T} \): tension force on the rope;
  • \( {\vec F}_{BA} \): force of reaction of body B on A, \( \left|{\vec F}_{BA}\right|=\left|{\vec{F}}_{AB}\right| \).

Figure 2

In the vertical direction there is no movement, the gravitational force \( {\vec F}_{gA} \) and normal reaction force \( {\vec N}_{A} \) cancel out. In the horizontal direction, applying the expression (I)
\[ \begin{gather} T-F_{AB}=m_{A}a \tag{II} \end{gather} \]
Body B:
  • \( {\vec F}_{gB} \): gravitational force on the body B;
  • \( {\vec N}_{B} \): normal reaction force;
  • \( {\vec F}_{AB} \): force of action of body A on B.

Figure 3

In the vertical direction there is no movement, the gravitational force \( {\vec F}_{gB} \) and normal reaction force \( {\vec N}_{B} \) cancel out. In the horizontal direction, applying the expression (I)
\[ \begin{gather} F_{AB}=m_{B}a \tag{III} \end{gather} \]
Body C:
  • \( {\vec F}_{gC} \): gravitational force on the body C;
  • \( \vec{T} \): tension force on the rope.

Figure 4

In the horizontal direction, no forces are acting. In the vertical direction, applying the expression (I)
\[ \begin{gather} F_{gC}-T=m_{C}a \tag{IV} \end{gather} \]
a) The equations (II), (III), and (IV) can be written as a system of three equations with three variables, a, T and FAB
\[ \left\{ \begin{array}{l} T-F_{AB}=m_{A}a\\ F_{AB}=m_{B}a\\ F_{gC}-T=m_{C}a \end{array} \right. \]
adding the three equations
\[ \frac{ \begin{aligned} \cancel{T}-\cancel{F_{AB}}=m_{A}a\\ \cancel{F_{AB}}=m_{B}a\\ \text{(+)}\qquad F_{gC}-\cancel{T}=m_{C}a \end{aligned} } {F_{gC}=m_{A}a+m_{B}a+m_{C}a} \]
the gravitational force is given by
\[ \bbox[#99CCFF,10px] {F_{g}=mg} \]
for the body C
\[ F_{gC}=m_{C}g \]
factoring the acceleration a on the right-hand side
\[ m_{C}g=a\left(m_{A}+m_{B}+m_{C}\right) \]
substituting the data
\[ \begin{gather} mg=a\left(3m+2m+m \right)\\ mg=6ma\\ a=\frac{\cancel{m}g}{6\cancel{m}} \end{gather} \]
\[ \bbox[#FFCCCC,10px] {a=\frac{g}{6}} \]

b) To calculate the force of reaction of bocy B on A, FBA, we use the expression (III), remembering that \( \left|{\vec{F}}_{BA}\right|=\left|{\vec{F}}_{AB}\right| \)
\[ F_{AB}=m_{B}a \]
substituting the mass of body B given in the problem and the acceleration found in the previous item
\[ F_{AB}=\cancel{2}m\frac{g}{\cancelto{3}{6}} \]
\[ \bbox[#FFCCCC,10px] {F_{AB}=\frac{1}{3}mg} \]
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