Solved Problem on Dynamics
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On an inclined plane of 30° to the horizontal, it slides without friction a mass m1 attached to another mass m2. The system is released from the rest, the mass m2 rises 250 m in 20 s. Calculate the ratio m1/m2. Assume g = 10 m/s2.


Problem data:
  • Mass of block 1:    m1;
  • Angle of the inclined plane:    θ = 30°;
  • Mass of block 2:    m2;
  • Displacement of block 2:    ΔS = 250 m;
  • Initial peed:    v0 = 0;
  • Rise time interval:    Δt = 20 s
  • Acceleration due to gravity:    g = 10 m/s2.
Problem diagram:

We choose the acceleration of the system in the direction of mass m1 descending the plane and the mass m2 rising (Figure 1).
Figure 1

Solution

We draw a free-body diagram, and we have the forces that act in each of them.
Block 1 (Figure 2-A):
We choose a frame of reference with the x-axis in the direction of the inclined plane and pointing downward. In this body act the gravitational force of block 1 \( {\vec F}_{g1} \), the normal reaction force of the plane on block 1 \( {\vec N}_{1} \), and the tension force on rope \( \vec{T} \).

Figure 2

The gravitational force \( {\vec F}_{g1} \) can be decomposed into two components, a component parallel to the x-axis \( {\vec F}_{g1P} \) and the other component normal or perpendicular \( {\vec F}_{g1N} \).
In the triangle on the left in Figure 2-B, the gravitational force is perpendicular to the horizontal plane, makes a 90° angle, the angle between the inclined plane and the horizontal plane is given as 30°, as the sum of the interior angles of a triangle equals to 180°, the angle α between the gravitational force and the parallel component should be
\[ \alpha +30°+90°=180°\Rightarrow \alpha=180°-30°-90°\Rightarrow \alpha=60° \]
In the triangle on the right, the normal component makes with the inclined plane a 90° angle, then the angle β between the gravitational force and the normal component is
\[ \alpha+\beta=90° \Rightarrow \beta =90°-\alpha \Rightarrow 90°-60°\Rightarrow 30° \]

We draw the forces in a coordinate system (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 movement in the y direction the normal reaction force and the normal component of gravitational force cancel.
Direction x:
\[ \begin{gather} F_{g1P}-T=m_{1}a \tag{II} \end{gather} \]
the component of gravitational force in the parallel direction is given by (Figure 2-C)
\[ \begin{gather} F_{g1P}=F_{g1}\cos 60° \tag{III} \end{gather} \]
the gravitational force is given by
\[ \begin{gather} \bbox[#99CCFF,10px] {F_{g}=mg} \tag{IV} \end{gather} \]
for block m1 the gravitational force is given by
\[ \begin{gather} F_{g1}=m_{1}g \tag{V} \end{gather} \]
substituting the expression (V) into expression (III)
\[ \begin{gather} F_{g1P}=m_{1}g\cos 60° \tag{VI} \end{gather} \]
substituting the expression (VI) into expression (II)
\[ \begin{gather} m_{1}g\cos 60°-T=m_{1}a \tag{VII} \end{gather} \]
Block 2 (Figure 3):
  • \( {\vec F}_{g2} \): gravitational force of block 2;
  • \( \vec{T} \): tension force on the rope.
Figure 3

We choose the positive direction upward, in the same direction of acceleration. In the horizontal direction, no forces are acting on the block, in the vertical direction applying the expression (I)
\[ \begin{gather} T-F_{g2}=m_{2}a \tag{VIII} \end{gather} \]
For block m2 the gravitational force is given by
\[ \begin{gather} F_{g2}=m_{2}g \tag{IX} \end{gather} \]
substituting the expression (IX) into expression (VIII)
\[ \begin{gather} T-m_{2}g=m_{2}a \tag{X} \end{gather} \]
Adding the expressions (VII) and (X),, we have the acceleration of the system
\[ \frac{ \begin{matrix} m_{1}g\cos 60°-\cancel{T}=m_{1}a\\ \text{(+)}\qquad\quad\;\cancel{T}-m_{2}g=m_{2}a \quad\; \end{matrix}} {m_{1}g\cos 60°-m_{2}g=m_{1}a+m_{2}a} \]
factoring the acceleration due to gravity g on the left-hand side of the equation and acceleration a on the right-hand side
\[ \begin{gather} g(m_{1}\cos 60°-m_{2})=a(m_{1}+m_{2})\\ a=g\left(\frac{m_{1}\cos 60°-m_{2}}{m_{1}+m_{2}}\right) \tag{XI} \end{gather} \]
From the Kinematics, we use the equation of displacement as a function of time with constant acceleration
\[ \bbox[#99CCFF,10px] {S=S_{0}+V_{0}t+\frac{a}{2}t^{2}} \]
\[ S-S_{0}=V_{0}t+\frac{a}{2}t^{2} \]
Figure 4

with \( \Delta S=S-S_{0} \) and using the value of the acceleration of the expression (XI)
\[ \Delta S=V_{0}t+\frac{1}{2}g\left(\frac{m_{1}\cos 60°-m_{2}}{m_{1}+m_{2}}\right)t^{2} \]
From the Trigonometria
\[ \cos 60°=\dfrac{1}{2} \]

Substituting the values ​​given in the problem and with \( \Delta t=t-t_{0}\Rightarrow 20=t-0\Rightarrow t=20\;\text{s, } \)
\( \Delta t=t-t_{0}\Rightarrow 20=t-0\Rightarrow t=20\;\text{s, } \)
\[ \begin{gather} 250=0\times 20+\frac{1}{2}\times 10\times \left(\frac{m_{1}\dfrac{1}{2}-m_{2}}{m_{1}+m_{2}}\right)\times 20^{2}\\ 250=0+5\times \left(\frac{\dfrac{m_{1}-2m_{2}}{2}}{m_{1}+m_{2}}\right)\times 400\\ 250=\left[\frac{m_{1}-2m_{2}}{2\left(m_{1}+m_{2}\right)}\right]\times 2000\\ \frac{250}{2000}=\frac{m_{1}-2m_{2}}{2m_{1}+2m_{2}} \end{gather} \]
on the left-hand side of the equation, we divide the numerator and the denominator by 250
\[ \begin{gather} \frac{250:250}{2000:250}=\frac{m_{1}-2m_{2}}{2m_{1}+2m_{2}}\\ \frac{1}{8}=\frac{m_{1}-2m_{2}}{2m_{1}+2m_{2}} \end{gather} \]
applying a cross-multiplication
\[ \begin{gather} 2m_{1}+2m_{2}=8\times \left(m_{1}-2m_{2}\right)\\ 2m_{1}+2m_{2}=8m_{1}-16m_{2}\\ 2m_{2}+16m_{2}=8m_{1}-2m_{1}\\ 18m_{2}=6m_{1}\\ \frac{m_{1}}{m_{2}}=\frac{18}{6} \end{gather} \]
\[ \bbox[#FFCCCC,10px] {\frac{m_{1}}{m_{2}}=3} \]
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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 .