Solved Problem on Static Equilibrium
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Two identical spheres, A and B, are placed in a box. The reaction force exerted by the bottom of the box on the sphere is 25 N.
a) Determine the mass of the spheres;
b) Find the ratio between the reaction forces of the box on the spheres.


Problem data:
  • Reaction force of the bottom of the box on sphere B:    FR = 25 N;
  • Acceleration due to gravity:    g = 9.8 m/s2.
Problem diagram:

Figure 1

Drawing a free-body diagram, we have the forces that act in each body

Box:
  • \( -{\vec F}_{R} \): force that sphere B exerts at the bottom of the box;
  • \( {\vec F}_{A} \): force the sphere A exerts on the sidewall of the box;
  • \( -{\vec F}_{B} \): force that sphere B exerts on the sidewall of the box;
the weight of the box was neglected.

Sphere A:
  • \( {\vec F}_{g} \): gravitational force on sphere A;
  • \( {\vec F}_{AB} \): contact force on sphere A due to sphere B;
  • \( -{\vec F}_{A} \): reaction force of the box on sphere A.
Sphere B:
  • \( {\vec F}_{g} \): gravitational force on sphere B;
  • \( {\vec F}_{BA} \): contact force on sphere B due to sphere A, \( |\;{\vec{F}}_{BA}\;|=|\;{\vec{F}}_{AB}\;| \);
  • \( {\vec F}_{R} \): reaction force of the bottom of the box on sphere B;
  • \( {\vec F}_{B} \): reaction force of the box on the sphere B.
Solution

We draw the forces in a coordinate system xy (Figures 2 and 3), we obtain its components in the x and y directions, and we apply the equilibrium condition
\[ \begin{gather} \bbox[#99CCFF,10px] {\sum _{i}F_{i}=0} \tag{I} \end{gather} \]
  • Sphere A:
Direction x:
\[ \begin{gather} F_{ABx}-F_{A}=0 \end{gather} \]
we have \( F_{ABx}=F_{AB}\cos \theta \)
\[ \begin{gather} F_{AB}\cos \theta -F_{A}=0 \tag{II} \end{gather} \]
Direction y:
\[ \begin{gather} F_{ABy}-F_{g}=0 \end{gather} \]
we have \( F_{ABy}=F_{AB}\sin \theta \)
\[ \begin{gather} F_{AB}\sin \theta -F_{g}=0 \tag{III} \end{gather} \]
Figure 2

  • Sphere B:
Direction x:
\[ \begin{gather} F_{B}-F_{BAx}=0 \end{gather} \]
we have \( F_{BAx}=F_{BA}\cos \theta \)
\[ \begin{gather} F_{B}-F_{BA}\cos \theta =0 \tag{IV} \end{gather} \]
Direction y:
\[ \begin{gather} F_{R}-F_{BAy}-F_{g}=0 \end{gather} \]
we have \( F_{BAy}=F_{BA}\sin \theta \)
\[ \begin{gather} F_{R}-F_{BA}\sin \theta -F_{g}=0 \tag{V} \end{gather} \]
Figure 3

a) The gravitational force is given by
\[ \begin{gather} \bbox[#99CCFF,10px] {F_{g}=mg} \tag{VI} \end{gather} \]
substituting the expression (VI) into expressions (III) and (V)
\[ \begin{gather} F_{AB}\sin \theta -9.8m=0 \tag{VII} \end{gather} \]
\[ \begin{gather} F_{R}-F_{BA}\sin \theta -9.8m=0 \tag{VIII} \end{gather} \]
From the expression (VII)
\[ \begin{gather} F_{AB}\sin \theta =9.8m \tag{IX} \end{gather} \]
substituting the expression (IX) into expression (VIII)
\[ \begin{gather} 25-9.8m-9.8m=0\\[5pt] 19.6m=25\\[5pt] m=\frac{25}{19.6} \end{gather} \]
\[ \begin{gather} \bbox[#FFCCCC,10px] {m=1.28\;\text{kg}} \end{gather} \]

b) Using expressions (II) and (IV) in the x direction, we obtain the ratio between the reaction forces of the box on the spheres
\[ \begin{gather} F_{A}=F_{AB}\cos \theta\\[5pt] F_{B}=F_{BA}\cos \theta \end{gather} \]
as FAB=FBA, dividing the equations
\[ \begin{gather} \frac{F_{A}}{F_{B}}=\frac{F_{AB}\cos \theta}{F_{AB}\cos \theta } \end{gather} \]
\[ \begin{gather} \bbox[#FFCCCC,10px] {\frac{F_{A}}{F_{B}}=1} \end{gather} \]
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