Solved Problem on Static Equilibrium
Português
English
advertisement
A mass block m = 100 kg is suspended by the string system shown in the figure. Determine the
tension forces on all ropes.
Assume: \( \sin 15°=0.259 \), \( \cos 15°=0.966 \), \( \sin 45°=0.707 \), \( \cos 45°=0.707 \), \( \sin 60°=0.866 \), \( \cos 60°=0.5 \).
Assume: \( \sin 15°=0.259 \), \( \cos 15°=0.966 \), \( \sin 45°=0.707 \), \( \cos 45°=0.707 \), \( \sin 60°=0.866 \), \( \cos 60°=0.5 \).
Problem data:
- Mass of the block: m=100 kg;
- Acceleration due to gravity: g=9.8 m/s2.
Drawing the forces that act on the system (Figure 1).
On the block, the gravitational force \( {\vec F}_{g} \) will be balanced by the tension force \( {\vec T}_{1} \), has as the reaction force \( {\vec T}\text{'}_{1} \) on the ceiling, and by the tension force \( {\vec T}_{2} \), has as the reaction force \( {\vec T}\text{'}_{2} \) that is applied at point D.
At point D the tension force \( {\vec T}\text{'}_{2} \) is balanced by tension forces the \( {\vec T}_{3} \) and \( {\vec T}_{4} \), which have as reaction forces \( {\vec T}\text{'}_{3} \) and \( {\vec T}\text{'}_{4} \) on the ceiling.
On the block, the gravitational force \( {\vec F}_{g} \) will be balanced by the tension force \( {\vec T}_{1} \), has as the reaction force \( {\vec T}\text{'}_{1} \) on the ceiling, and by the tension force \( {\vec T}_{2} \), has as the reaction force \( {\vec T}\text{'}_{2} \) that is applied at point D.
At point D the tension force \( {\vec T}\text{'}_{2} \) is balanced by tension forces the \( {\vec T}_{3} \) and \( {\vec T}_{4} \), which have as reaction forces \( {\vec T}\text{'}_{3} \) and \( {\vec T}\text{'}_{4} \) on the ceiling.
Figure 1
Solution
Dividing the problem into two parts, first studying the forces on the block.
By point C, we draw a vertical line perpendicular to the ceiling, the angle between the ceiling and the rope \( \overline{CE} \) is 75°, then the angle between the line and the rope \( \overline{CE} \) is 15°, they are complementary angles, add 90°.
From the block at point E, we draw a vertical line dividing the angle of 30° into two parts as the angle between this line and rope \( \overline{CE} \) is an internal angle with the angle found before it will also measure 15°. This line divides the angle of 30° into two equal parts, is a bisector of the angle of 30° (Figure 2-A).
Drawing the forces in a coordinate system xy and decomposing the forces, the gravitational force \( {\vec F}_{g} \) only has the component \( {\vec F}_{gy} \), the tension forces \( {\vec T}_{1} \) and \( {\vec T}_{2} \) have components \( {\vec T}_{1x} \) and \( {\vec T}_{2x} \) in the x direction, and components \( {\vec T}_{1y} \) and \( {\vec T}_{2y} \) in the y direction. As the system is in equilibrium, the resultant force is equal to zero, and we apply the condition
\[
\begin{gather}
\bbox[#99CCFF,10px]
{\sum F=0} \tag{I}
\end{gather}
\]
Direction x: \( T_{1x}-T_{2x}=0 \)
Direction y: \( T_{1y}+T_{2y}-P=0 \)
\[
\begin{gather}
T_{1}\sin 15°-T_{2}\sin 15°=0\\[5pt]
T_{1}\cos 15°+T_{2}\cos 15°-P=0
\end{gather}
\]
These equations can be written as a system of two equations to two variables, T1 and
T2
\[
\begin{gather}
&\left\{
\begin{array}{l}
T_{1}\sin 15°-T_{2}\sin 15°=0\\
T_{1}\cos 15°+T_{2}\cos 15°-P=0
\end{array}
\right.\\[8pt]
&\left\{
\begin{array}{l}
0.259T_{1}-0.259T_{2}=0\\
0.966T_{1}+0.966T_{2}-mg=0
\end{array}
\right.\\[8pt]
&\left\{
\begin{array}{l}
0.259T_{1}-0.259T_{2}=0\\
0.966T_{1}+0.966T_{2}-100\times 9.8=0
\end{array}
\right.\\[8pt]
\;\\
&\left\{
\begin{array}{l}
0.259T_{1}-0.259T_{2}=0\\
0.966T_{1}+0.966T_{2}-980=0 \tag{II}
\end{array}
\right.
\end{gather}
\]
from the first equation of the system (II)
\[
\begin{gather}
0.259T_{1}-0.259T_{2}=0\\[5pt]
0.259T_{1}=0.259T_{2}\\[5pt]
T_{1}=T_{2} \tag{III}
\end{gather}
\]
substituting the value of (III) in the second equation system (II)
\[
\begin{gather}
0.966T_{1}+0.966T_{1}-980=0\\[5pt]
1.932T_{1}=980\\[5pt]
T_{1}=\frac{980}{1.932}\\[5pt]
T_{1}=507.3\ \text{N}
\end{gather}
\]
using expression (III)
\[
\begin{gather}
\bbox[#FFCCCC,10px]
{T_{1}=T_{2}=507.3\ \text{N}}
\end{gather}
\]
Studying the forces acting at point D.Drawing a horizontal line by point D, the angle between this line and rope \( \overline{AD} \) measures 45º is an internal angle with the angle between rope \( \overline{AD} \) and the ceiling.
The angle between the \( \overline{BD} \) rope and the horizontal line measures 60°, which is an alternate angle with the angle between rope \( \overline{BD} \) and the ceiling (Figure 3-A).
Drawing a vertical line by point D, the angle between rope \( \overline{DE} \) and this line is 15º, is an alternate angle with the angle found in the first part of the problem.
Drawing the forces in a coordinate system xy and decomposing the forces, tension force \( {\vec T}_{2} \) has already been determined, it has components \( {\vec T}_{2x} \) and \( {\vec T}_{2y} \), the tension forces \( {\vec T}_{3} \) and \( {\vec T}_{4} \) have components \( {\vec T}_{3x} \) and \( {\vec T}_{4x} \) in the x direction and components \( {\vec T}_{3y} \) and \( {\vec T}_{4y} \) in the y direction. As the system is in equilibrium we can apply the condition (I).
Direction x: \( T_{2x}+T_{3x}-T_{4x}=0 \)
Direction y: \( T_{3y}+T_{4y}-T_{2y}=0 \)
\[
\begin{gather}
T_{2}\sin 15°+T_{3}\cos 60°-T_{4}\cos 45°=0\\[5pt]
T_{3}\sin 60°+T_{4}\sin 45°-T_{2}\cos 15°=0
\end{gather}
\]
these equations can be written as a system of two equations to two variables, T3 and
T4, substituting the data and tension force T2 determined above
\[
\begin{gather}
&\left\{
\begin{matrix}
T_{2}\sin 15°+T_{3}\cos 60°-T_{4}\cos 45°=0\\
T_{3}\sin 60°+T_{4}\sin 45°-T_{2}\cos 15°=0
\end{matrix}
\right.\\[8pt]
&\left\{
\begin{matrix}
0.259\times 507.3+0.5T_{3}-0.707T_{4}=0\\
0.866T_{3}+0.707T_{4}-0.966\times 507.3=0
\end{matrix}
\right.\\[8pt]
&\left\{
\begin{matrix}
131.4+0.5T_{3}-0.707T_{4}=0\\
0.866T_{3}+0.707T_{4}-490.0=0
\end{matrix}
\right.\\[8pt]
&\left\{
\begin{matrix}
0.5T_{3}-0.707T_{4}=-131.4\\
0.866T_{3}+0.707T_{4}=490.0 \tag{IV}
\end{matrix}
\right.
\end{gather}
\]
adding the two equations of system (IV), we eliminate the term in T4
\[
\begin{gather}
\frac{
\begin{matrix}
0.5T_{3}-0.707T_{4}=-131.4\\
0.866T_{3}+0.707T_{4}=490.0
\end{matrix}}
{1.366T_{3}+0=358.60}\\[5pt]
1.366T_{3}=358.60\\[5pt]
T_{3}=\frac{358.60}{1.366}
\end{gather}
\]
\[
\begin{gather}
\bbox[#FFCCCC,10px]
{T_{3}=262.5\ \text{N}}
\end{gather}
\]
substituting this value in the second equation of system (IV)
\[
\begin{gather}
0.5\times 262.5-0.707T_{4}=-131.4\\[5pt]
131.3-0.707T_{4}=-131.4\\[5pt]
-0.707T_{4}=-131.4-131.3\\[5pt]
-0.707T_{4}=-262.7
\end{gather}
\]
multiplying the equation by (−1)
\[
\begin{gather}
\qquad\qquad -0.707T_{4}=-262.7 \qquad \mathrm{(\times -1)}\\[5pt]
0.707T_{4}=262.7\\[5pt]
T_{4}=\frac{262.7}{0.707}
\end{gather}
\]
\[
\begin{gather}
\bbox[#FFCCCC,10px]
{T_{4}=371.6\ \text{N}}
\end{gather}
\]
advertisement
Fisicaexe - Physics Solved Problems by Elcio Brandani Mondadori is licensed under a Creative Commons Attribution-NonCommercial-ShareAlike 4.0 International License .