Solved Problem on Dynamics
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Two blocks, of masses mA = 0.35 kg and mB = 1.15 kg, are on a frictionless horizontal surface. The rope that connects the blocks has a negligible mass. A horizontal force of constant magnitude equal to 15 N is applied by pulling the two blocks. Find the magnitude acceleration acquired by the system and the tension in the rope connecting the blocks.


Problem data:
  • Mass of body A:    mA = 0.35 kg;
  • Mass of body B:    mB = 1.15 kg;
  • Force applied to the system:   F = 15 N.
Problem diagram:

We choose a frame of reference oriented to the right in the same direction as the applied force \( \vec F \), this produces an acceleration on the system.
Figure 1

Drawing a Free-Body Diagram for each body.

  • Body A:
    • Horizontal direction:
      • \( \vec T \): tension force to the rope.
    • Vertical direction:
      • \( {\vec N}_{\small A} \): normal reaction force of the surface on the body;
      • \( {\vec W}_{\small A} \): weight of body A.
Figure 2

  • Body B
    • Horizontal direction:
      • \( \vec F \): force applied to the system;
      • \( -\vec T \): tension force on the rope.
    • Vertical direction:
      • \( {\vec N}_{\small B} \): normal reaction force of the surface on the body;
      • \( {\vec W}_{\small B} \): weight of body B.
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, there is no motion, the normal force and weight cancel out.
In the horizontal direction,
\[ \begin{gather} T=m_{\small A}a \tag{I} \end{gather} \]
  • Body B:
In the vertical direction, there is no motion, normal force and weight cancel out.
In the horizontal direction
\[ \begin{gather} F-T=m_{\small B}a \tag{II} \end{gather} \]
Equations (I) and (II) can be written as a system of linear equations with two variables (T and a)
\[ \left\{ \begin{array}{rr} T&=m_{\small A}a\\ F-T&=m_{\small B}a \end{array} \right. \]
substituting equation (I) into equation (II), we have the acceleration
\[ \begin{gather} F-m_{\small A}a=m_{\small B}a\\[5pt] a=\frac{F}{m_{\small A}+m_{\small B}}\\[5pt] a=\frac{15\;\mathrm N}{(0.35\;\mathrm kg)+(1.15\;\mathrm kg)}\\[5pt] a=\frac{15\;\mathrm{\frac{\cancel{kg}.m}{s^2}}}{1.5\;\mathrm{\cancel{kg}}} \end{gather} \]
\[ \begin{gather} \bbox[#FFCCCC,10px] {a=10\;\mathrm{m/s^2}} \end{gather} \]

Note: the rope connecting the two blocks is ideal, we can consider it inextensible and of negligible mass, the rope only transmits the force from one block to the other block. The two blocks form a set subjected to the same force, both have the same acceleration, and the system behaves as if it were a single block of total mass given by the sum of the masses of the two blocks, A and B.

Substituting the acceleration found in expression (I), we have the tension on the rope
\[ \begin{gather} T=(0.35\;\mathrm{kg})\left(10\;\mathrm{\frac{m}{s^2}}\right) \end{gather} \]
\[ \begin{gather} \bbox[#FFCCCC,10px] {T=3.5\;\mathrm N} \end{gather} \]

Note: in the same way, we could substitute the acceleration in equation (II) to obtain the tension on the rope, in this case
\[ \begin{gather} (15\;\mathrm N)-T=(1.15\;\mathrm kg)\left(10\;\mathrm{\frac{m}{s^2}}\right)\Rightarrow (15\;\mathrm N)-T=11.5\;\mathrm N\Rightarrow T=3.5\;\mathrm N \end{gather} \]
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