Solved Problem on Dynamics
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What horizontal force should be constantly applied to the mass M = 21 kg so that the mass m1 = 5 kg does not move relative to the mass m2 = 4 kg? Neglect the friction and assume g = 10 m/s2.


Problem data:
  • Mass of car M:    M = 21 kg;
  • Mass of car m1:    m1 = 5 kg;
  • Mass of the sphere m2:    m2 = 4 kg;
  • Acceleration due to gravity:    g = 10 m/s2.
Problem diagram:

We choose a frame of reference, with the x-axis pointing to the right and the y-axis pointed up.
The system has acceleration \( \vec{a} \) in the same direction as the applied force \( \vec{F} \). The acceleration due to gravity \( \vec{g} \) is pointed down, and we consider the rope that connects the m1 and m2 masses, without mass and inextensible, and the pulley without mass and frictionless (Figure 1).
Figure 1

Solution

Drawing free-bodies diagrams, we have the forces that act in each of them, and we apply Newton's Second Law
\[ \begin{gather} \bbox[#99CCFF,10px] {\vec{F}=m\vec{a}} \tag{I} \end{gather} \]
Body m1 (Figura 2):

  • \( \vec{T} \): tension force on the rope;
  • \( {\vec F}_{g1} \): gravitational force on block m1;
  • \( {\vec N}_{1} \): normal reaction force.
In the vertical direction, the gravitational force \( {\vec F}_{g1} \) and the normal reaction force \( {\vec N}_{1} \) cancel.
Figure 2

In the horizontal direction applying the expression (I)
\[ T=m_{1}a \]
substituting the mass of the body
\[ \begin{gather} T=5a \tag{II} \end{gather} \]
As the rope attached to the bodies only transmits the tension of the body m1 to the body m2.

Body m2 (Figura 3-A):

  • \( \vec{T} \): tension force on the rope;
  • \( {\vec F}_{g2} \): gravitational force on the sphere m2.
In the vertical direction, the gravitational force \( {\vec F}_{g2} \) and the component of the tension force in the y direction \( {\vec T}_{y} \) are in equilibrium (Figure 3-B)
\[ \begin{gather} T_{y}=F_{g2} \tag{III} \end{gather} \]
Figure 3

The gravitational force is given by
\[ \begin{gather} \bbox[#99CCFF,10px] {F_{g}=mg} \tag{IV} \end{gather} \]
substituting the expression (IV) into expression (III)
\[ T_{y}=m_{2}g \]
substituting the mass of the sphere and the acceleration due to gravity
\[ \begin{gather} T_{y}=4.10\\ T_{y}=40\;\text{N} \tag{V} \end{gather} \]
In the horizontal direction applying the expression (I)
\[ T_{x}=m_{2}a \]
substituting the mass of the sphere
\[ \begin{gather} T_{x}=4a \tag{VI} \end{gather} \]
The tension force \( \vec{T} \) and its components in x and y directions \( {\vec{T}}_{x} \), \( {\vec{T}}_{y} \), represent the sides of a right triangle (Figure 3-C), using expressions (II), (V), and (VI) applying the Pythagorean Theorem to determine the acceleration of the system
\[ \begin{gather} \vec{T}={\vec{T}}_{x}+{\vec{T}}_{y}\\ T^{2}=T_{x}^{2}+T_{y}^{2}\\ (5a)^{2}=(4a)^{2}+(40)^{2}\\ 25a^{2}=16a^{2}+1600\\ 25a^{2}-16a^{2}=1600\\ 9a^{2}=1600\\ a^{2}=\frac{1600}{9}\\ a=\sqrt{\frac{1600}{9}}\\ a=\frac{40}{3} \end{gather} \]
Applying the expression (I) to the system, the mass will be the total mass given by the sum of the masses of the bodies M, m1, and m2 and acceleration found above
\[ \begin{gather} F=(M+m_{1}+m_{2})a\\ F=(21+5+4)\times \frac{40}{3}\\ F=30\times \frac{40}{3}\\ F=10\times 40 \end{gather} \]
\[ \bbox[#FFCCCC,10px] {F=400\;\text{N}} \]
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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 .