Solved Problem on Dynamics

An Atwood machine is arranged in such a way that the masses M₁ and M₂ slide without friction over two inclined planes with angles of 30° and 60º to the horizontal, rather than moving vertically. Assuming that the ropes that support the masses M₁ and M₂ are parallel to the planes. Determine:
a) The ratio between M₁ and M₂ for the system to remain in equilibrium;
b) Calculate the acceleration and tension force on the rope when the masses are equal, each at 5 kg.
Assume the acceleration due to gravity equal to 9.81 m/s².

Problem data:

Masses of blocks: M₁=M₂=5 kg;
Acceleration due to gravity: g=9,81 m/s².

Problem diagram:

We choose a direction for acceleration arbitrarily (Figure 1)

Figure 1

Solution

Drawing free-bodies diagrams, we have the forces that act in each block.

Body with mass M₁:

We choose a reference frame with the x-axis parallel to the inclined plane and the same direction of the acceleration. In this body act the gravitational force \( {\vec F}_{g1} \), the tension force on the rope \( \vec{T} \), and the normal reaction force \( {\vec N}_{1} \) (Figure 2-A).

Figure 2

The gravitational force can be decomposeed into two directions, one component is parallel to the x-axis \( {\vec F}_{g1P} \) and the other component normal or perpendicular \( {\vec F}_{g1N} \) (Figure 2-B).
In the triangle on the left, the gravitational force \( {\vec F}_{g1} \) is perpendicular to the horizontal plane, it makes a 90° angle, the angle between the inclined plane and the horizontal plane is given as 60°, as the sum of the interior angles of a triangle equals to 180°, the angle α between the gravitational force \( {\vec F}_{g1P} \) and the parallel component \( {\vec F}_{g1P} \) should be

\[ 60°+90°+\alpha=180°\Rightarrow \alpha=180°-60°-90°\Rightarrow \alpha=30° \]

In the triangle on the right, the component of normal force \( {\vec F}_{g1N} \) makes with the inclined plane a 90° angle, then the angle β between the gravitational force \( {\vec F}_{g1} \) and the component of normal force \( {\vec F}_{g1N} \) should be

\[ 30°+\beta=90°\Rightarrow \beta=90°-30°\Rightarrow \beta=60° \]

they are complementary angles.
We draw the forces in a coordinate system xy (Figure 2 -C), and we apply Newton's Second Law

\[ \begin{gather} \bbox[#99CCFF,10px] {\vec{F}=m\vec{a}} \tag{I} \end{gather} \]

As there is no motion in the y direction, the normal reaction force \( {\vec N}_{1} \) and the component of gravitational force \( {\vec F}_{g1N} \) cancel out.
Direction x:

\[ \begin{gather} T-F_{gP}=M_{1}a \tag{II} \end{gather} \]

the component of gravitational force in the parallel direction is given by

\[ \begin{gather} F_{gP}=F_{g1}\sin 60° \tag{III} \end{gather} \]

the gravitational force is given by

\[ \begin{gather} \bbox[#99CCFF,10px] {F_{g}=mg} \tag{IV} \end{gather} \]

for mass M₁

\[ \begin{gather} F_{g1}=M_{1}g \tag{V} \end{gather} \]

substituting the expression (V) into expression (III)

\[ \begin{gather} F_{gP}=M_{1}g\sin60° \tag{VI} \end{gather} \]

substituting the expression (VI) into expression (II)

\[ \begin{gather} T-M_{1}g\sin 60°=M_{1}a \tag{VII} \end{gather} \]

Body with mass M₂:

Likewise, we choose a reference frame with the x-axis parallel to the inclined plane and the same direction of the acceleration. In this body acts the gravitational force \( {\vec{F}}_{g2} \), the tension force \( \vec{T} \), and the normal reaction force \( {\vec{N}}_{2} \) (Figure 3-A).

Figure 3

Likewise, the gravitational force can be decomposed into two directions, a component parallel to the x-axis \( {\vec F}_{g2P} \) and the other component normal or perpendicular \( {\vec F}_{g2N} \) (Figure 3-B).
In this case, the angle between the gravitational force \( {\vec F}_{g2} \) and the component of normal force \( {\vec F}_{g2N} \) will be 30°.
Drawing the forces in a coordinate system xy (Figure 3-C) and applying the expression (I).
In the y direction, there is no motion, the normal reaction force \( {\vec N}_{2} \) and the normal component of gravitational force \( {\vec{F}}_{g2N} \) cancel out.
Direction x:

\[ \begin{gather} F_{g2P}-T=M_{2}a \tag{VIII} \end{gather} \]

the component of gravitational force in the parallel direction is given by

\[ \begin{gather} F_{g2P}=P_{2}\sin 30° \tag{IX} \end{gather} \]

using the express (IV) for the gravitational force of the mass M₂

\[ \begin{gather} F_{2g}=M_{2}g \tag{X} \end{gather} \]

substituting the expression (X) into expression (IX)

\[ \begin{gather} F_{g2P}=M_{2}g\sin30° \tag{XI} \end{gather} \]

substituting the expression (XI) into expression (VIII)

\[ \begin{gather} M_{2}g\sin 30°-T=M_{2}a \tag{XII} \end{gather} \]

a) Equations (VII) and (XII) can be written as a system of two equations

\[ \left\{ \begin{matrix} T-M_{1}g\sin 60°=M_{1}a\\ M_{2}g\sin 30°-T=M_{2}a \end{matrix} \right. \]

Remenbering the Trigonometry

\[ \begin{gather} \sin 60°=\dfrac{\sqrt{3\;}}{2}\\ \sin 30°=\dfrac{1}{2} \end{gather} \]

\[ \left\{ \begin{matrix} T-\dfrac{\sqrt{3\;}}{2}M_{1}g=M_{1}a\\ \dfrac{1}{2}M_{2}g-T=M_{2}a \end{matrix} \right. \tag{XIII} \]

For the system to remain in equilibrium, we must have the sum of the forces equal to zero

\[ \bbox[#99CCFF,10px] {\sum F=0} \]

\[ \begin{gather} \left\{ \begin{matrix} T-\dfrac{\sqrt{3\;}}{2}M_{1}g=0\\ \dfrac{1}{2}M_{2}g-T=0 \end{matrix} \right. \\{\,}\\ \left\{ \begin{matrix} \dfrac{\sqrt{3\;}}{2}M_{1}g=T\\ \dfrac{1}{2}M_{2}g=T \end{matrix} \right. \end{gather} \]

dividing the second equation by the first in the system

\[ \begin{gather} \frac{\dfrac{1}{2}M_{2}g}{\dfrac{\sqrt{3\;}}{2}M_{1}g}=\frac{T}{T}\\[5pt] \frac{\dfrac{1}{2}M_{2}g}{\dfrac{\sqrt{3\;}}{2}M_{1}g}=1\\[5pt] \frac{M_{2}\cancel{g}}{M_{1}\cancel{g}}=\frac{\dfrac{\sqrt{3\;}}{2}}{\dfrac{1}{2}}\\[5pt] \frac{M_{2}}{M_{1}}=\frac{\sqrt{3\;}}{\cancel{2}}\times \frac{\cancel{2}}{1} \end{gather} \]

\[ \bbox[#FFCCCC,10px] {\frac{M_{2}}{M_{1}}=\sqrt{3\;}} \]

b) To find the acceleration and tension force on the rope, as the masses are equal we do M₁=M₂=M, and we substitute in the system of equations (XIII)

\[ \left\{ \begin{matrix} T-\dfrac{\sqrt{3\;}}{2}Mg=Ma\\ \dfrac{1}{2}Mg-T=Ma \end{matrix} \right. \]

This is a system of two equations with two variables, a and T, adding these two equations

\[ \begin{gather} \frac{ \begin{aligned} \cancel{T}-\dfrac{\sqrt{3\;}}{2}Mg=Ma\\ \text{(+)}\qquad \dfrac{1}{2}Mg-\cancel{T}=Ma \end{aligned} } {-{\dfrac{\sqrt{3\;}}{2}\cancel{M}g}+\dfrac{1}{2}\cancel{M}g=\cancel{M}a+\cancel{M}a}\\ \frac{1}{2}g(\;-\sqrt{3\;}+1\;)=2a\\ a=\frac{g}{2\times 2}\times (-\sqrt{3\;}+1)\\ a=\frac{g}{4}\times (-\sqrt{3\;}+1) \end{gather} \]

substituting the given value for acceleration due to gravity and \( \sqrt{3\;}\simeq 1.73 \)

\[ \begin{gather} a=\frac{9.81}{4}\times (-1.73+1)\\ a=2.45\times (-0.73)\\ a=-1.79 \end{gather} \]

the negative sign in acceleration indicates that the direction of motion will be opposite to that shown in Figure 1

\[ \bbox[#FFCCCC,10px] {a=1.79\ \text{m/s}^{2}} \]

the actual direction of acceleration is shown in Figure 4.

Figure 4

Subtracting the system equations

\[ \begin{gather} \frac{ \begin{aligned} T-\dfrac{\sqrt{3\;}}{2}Mg=Ma\\ \text{(-)}\qquad \dfrac{1}{2}Mg-T=Ma \end{aligned} } {2T-\dfrac{\sqrt{3\;}}{2}Mg-\dfrac{1}{2}Mg=Ma-Ma}\\ 2T-\frac{\sqrt{3\;}}{2}Mg-\frac{1}{2}Mg=0\\ 2T=\frac{\sqrt{3\;}}{2}Mg+\frac{1}{2}Mg\\ T=\frac{1}{2}\times \left(\frac{\sqrt{3\;}}{2}Mg+\frac{1}{2}Mg\right) \end{gather} \]

substituting the values of the problem

\[ \begin{gather} T=\frac{1}{2}\times \left(\frac{1.73}{2}\times 5\times 9.81+\frac{1}{2}\times 5\times 9.81\right)\\ T=\frac{1}{2}\times \left(42.43+24.53\right) \end{gather} \]

\[ \bbox[#FFCCCC,10px] {T=33.51\ \text{N}} \]

Fisicaexe - Physics Solved Problems by Elcio Brandani Mondadori is licensed under a Creative Commons Attribution-NonCommercial-ShareAlike 4.0 International License .