Solved Problem on Static Equilibrium
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Three pulleys rotate fixed on the same shaft, in the pulleys, there are ropes with negligible mass that
support spheres. Data: pulley 1, r1 = 0.2 m and m1 = 2.7 kg, pulley
2, r2 = 0.4 m, pulley 3, m3 = 1.8 kg. Determine:
a) If m2 = 4.0 kg, what is the radius of the pulley 3 so that the torque of the forces on the system, relative to the axis, be equal to zero;
b) If r3 = 0.8 m, what is the mass attached to pulley 2, so the system rotates clockwise relative to the shaft.
a) If m2 = 4.0 kg, what is the radius of the pulley 3 so that the torque of the forces on the system, relative to the axis, be equal to zero;
b) If r3 = 0.8 m, what is the mass attached to pulley 2, so the system rotates clockwise relative to the shaft.
Problem data:
- Radius of pulley 1: r1=0,2 m;
- Masss of pulley 1: m1=2,7 kg;
- Radius of pulley 2: r2=0,4 m;
- Masss of pulley 3: m3=1,8 kg.
This system is equivalent to a bar, with negligible mass, supported by the center, with the gravitational forces \( {\vec F}_{g1} \), \( {\vec F}_{g2} \), and \( {\vec F}_{g3} \), due to the masses m1, m2, and m3 acting at distances r1, r2, and r3. We choose the counterclockwise direction of rotation of the body as positive (Figure 1-B).
As we want the torque of the system to be zero, we have the condition
\[
\begin{gather}
\bbox[#99CCFF,10px]
{\sum M=0} \tag{I}
\end{gather}
\]
The torque of a force relative to the point is given by
\[
\begin{gather}
\bbox[#99CCFF,10px]
{M=Fd} \tag{II}
\end{gather}
\]
where F is the force that acts on the system, in the problem the gravitational forces, and d
is the distance between the point of application of the force and the point relative to which the torque is
calculated, in the problem the radius of the pulleys.
a) Relative to the support, the gravitational forces \( {\vec F}_{g1} \) and \( {\vec F}_{g3} \) try to make the system rotate in the counterclockwise direction, positive, and gravitational force \( {\vec F}_{g2} \) tries rotate in the clockwise direction, negative (Figure 1-B). Applying the expression (II) in the condition (I)
\[
\begin{gather}
F_{g1}r_{1}-F_{g2}r_{2}+F_{g3}r_{3}=0
\end{gather}
\]
the gravitational force is given by
\[
\begin{gather}
\bbox[#99CCFF,10px]
{F_{g}=mg}
\end{gather}
\]
\[
\begin{gather}
m_{1}gr_{1}-m_{2}gr_{2}+m_{3}gr_{3}=0\\[5pt]
g(m_{1}r_{1}-m_{2}r_{2}+m_{3}r_{3})=0\\[5pt]
m_{1}r_{1}-m_{2}r_{2}+m_{3}r_{3}=\frac{0}{g}\\[5pt]
m_{1}r_{1}-m_{2}r_{2}+m_{3}r_{3}=0 \tag{III}\\[5pt]
2.7\times 0.2-4.0\times 0.4+1.8 r_{3}=0\\[5pt]
0.54-1.6+1.8 r_{3}=0\\[5pt]
-1.06+1.8 r_{3}=0\\[5pt]
1.8 r_{3}=1.06\\[5pt]
r_{3}=\frac{1.06}{1.8}
\end{gather}
\]
\[
\begin{gather}
\bbox[#FFCCCC,10px]
{r_{3}\simeq 0.6\;\text{m}}
\end{gather}
\]
b) Using the expression (III)
\[
\begin{gather}
m_{1}r_{1}-m_{2}r_{2}+m_{3}r_{3}=0\\[5pt]
2.7\times 0.2-m_{2} \times 0.4+1.8\times 0.8=0\\[5pt]
0.54-0.4m_{2}+1.44=0\\[5pt]
-0.4m_{2}+1.98=0\\[5pt]
0.4m_{2}=1.98\\[5pt]
m_{2}=\frac{1.98}{0.4}\\[5pt]
m_{2}=4.95 \;\text{kg}
\end{gather}
\]
If m2 = 4.95 kg, the system will be in equilibrium, to it turns clockwise mass 2 must be
greater than this value
\[
\begin{gather}
\bbox[#FFCCCC,10px]
{m_{2}>4.95\;\text{kg}}
\end{gather}
\]
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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 .