Solved Problem on Static Equilibrium
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FR
PT-BR
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For the system in equilibrium shown in the figure, determine the tension forces on the strings A
and B, knowing that body C has 100 N.
Problem data:
- Weight of body C: W = 100 N.
Problem diagram:
The forces that act on the system are the gravitational force \( {\vec F}_{g} \) on block C that points down and tension forces. The cord holding the block only transmits the gravitational force of the block to the point where it is attached to the other ropes.
Cord A makes a 60° angle with the ceiling, drawing a horizontal line that passes through the point where the body is attached, we have the tension force \( {\vec T}_{A} \) also makes a 60° angle with the horizontal, they are alternate angles.
Cord B makes a 60° angle with the vertical wall, and the angle between the tension force \( {\vec T}_{B} \) and the cord that holds block C is also 60°, they are alternates angles. The angle between the horizontal line and the tension force \( {\vec T}_{B} \) is 30° with the horizontal line, they are complementary angles, adding 90°.
The forces that act on the system are the gravitational force \( {\vec F}_{g} \) on block C that points down and tension forces. The cord holding the block only transmits the gravitational force of the block to the point where it is attached to the other ropes.
Cord A makes a 60° angle with the ceiling, drawing a horizontal line that passes through the point where the body is attached, we have the tension force \( {\vec T}_{A} \) also makes a 60° angle with the horizontal, they are alternate angles.
Cord B makes a 60° angle with the vertical wall, and the angle between the tension force \( {\vec T}_{B} \) and the cord that holds block C is also 60°, they are alternates angles. The angle between the horizontal line and the tension force \( {\vec T}_{B} \) is 30° with the horizontal line, they are complementary angles, adding 90°.
Solution
We draw the forces in a coordinate system and decompose the forces in these directions (Figure 2).
The gravitational force
\( {\vec F}_{g} \)
has only the component in the negative y direction. The tension force
\( {\vec T}_{A} \)
has the component
\( {\vec T}_{Ax} \)
in the positive x direction and the component
\( {\vec T}_{Ay} \)
in the positive y direction. The tension force
\( {\vec T}_{B} \)
has the component
\( {\vec T}_{Bx} \)
in the negative x direction and the component
\( {\vec T}_{By} \)
in the negative y direction.
Since the system is in equilibrium, the resultant of the forces acting on it is equal to zero.
Since the system is in equilibrium, the resultant of the forces acting on it is equal to zero.
\[
\begin{gather}
\bbox[#99CCFF,10px]
{\sum \vec{F}=0}
\end{gather}
\]
\[
\begin{gather}
{\vec T}_A+{\vec T}_B+\vec P=0 \\[5pt]
{\vec T}_{Ax}+{\vec T}_{Ay}-{\vec T}_{Bx}-{\vec T}_{By}-{\vec F}_g=0 \tag{I}
\end{gather}
\]
- x-direction:
\[
\begin{gather}
{\vec T}_{Ax}=T_A\cos 60° \tag{II}
\end{gather}
\]
\[
\begin{gather}
{\vec T}_{Bx}=T_B\cos 30° \tag{III}
\end{gather}
\]
The gravitational force has no component in the x-direction.- y-direção:
\[
\begin{gather}
{\vec T}_{Ay}=T_A\operatorname{sen}60° \tag{IV}
\end{gather}
\]
\[
\begin{gather}
{\vec T}_{Bx}=T_B\operatorname{sen}30° \tag{V}
\end{gather}
\]
Substituting equations (II), (III), (IV), (V), and (VI) into equation (I) and separating the components in
the x and y directions:
- x-direção:
\[
\begin{gather}
T_A\cos 60°-T_B\cos 30°=0
\end{gather}
\]
- y-direção:
\[
\begin{gather}
T_A\operatorname{sen}60°-T_B\operatorname{sen}30°-F_g=0
\end{gather}
\]
Da Trigonometria
\( \cos 30°=\dfrac{\sqrt{3\;}}{2} \),
\( \sin 30°=\dfrac{1}{2} \),
\( \cos 60°=\dfrac{1}{2} \), \( \operatorname{sen}60°\dfrac{\sqrt{3\;}}{2} \).
\( \cos 60°=\dfrac{1}{2} \), \( \operatorname{sen}60°\dfrac{\sqrt{3\;}}{2} \).
\[
\left\{
\begin{array}{l}
\dfrac{1}{2}T_A-\dfrac{\sqrt{3\;}}{2}T_B=0 \\
\dfrac{\sqrt{3\;}}{2}T_A-\dfrac{1}{2}T_B-100\;\mathrm N=0
\end{array}
\right.
\]
That is a system of two equations with two unknowns (T and TB).
Solving the first equation for T and substituting in the second equation
\[
\begin{gather}
\frac{1}{\cancel 2}T_A=\frac{\sqrt{3\;}}{\cancel 2}T_B\\[5pt]
T_A=\sqrt{3\;}\;T_B \tag{VI}
\end{gather}
\]
\[
\begin{gather}
\frac{\sqrt{3\;}}{2}\times\sqrt{3\;}\;T_B-\frac{1}{2}T_B-100\;\mathrm N=0 \\[5pt]
\frac{3}{2}T_B-\frac{1}{2}T_B=100\;\mathrm N \\[5pt]
\frac{2}{2}T_B=100\;\mathrm N
\end{gather}
\]
\[
\begin{gather}
\bbox[#FFCCCC,10px]
{T_B=100\;\mathrm N}
\end{gather}
\]
substituting this value of the normal reaction force into equation (VI), we obtain the value of the
tension force
\[
\begin{gather}
T_A=\sqrt{3\;}\times 100\;\mathrm N
\end{gather}
\]
\[
\begin{gather}
\bbox[#FFCCCC,10px]
{T_A \approx 173\;\mathrm N}
\end{gather}
\]
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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 .