Solved Problem on Dynamics
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In the system shown in the figure, p1 is a mobile pulley, p2 is a fixed pulley, the weight of block B is 2000 N, and the angle of the inclined plane is equal to 30°. Determine which should be the weight of block A so that block B has a speed of 20 m/s after a path of 40 m in the upward direction. Neglect the masses of the ropes and the pulleys. Neglect the friction between the ropes and the pulleys and between block B and the plane. Assume g = 10 m/s2.


Problem data:
  • Weight of block B:    FgB = 2 000 N;
  • Angle of the inclined plane:    θ = 30°;
  • Displacement of the block B:    ΔSB = 40 m;
  • Initial speed:    v0 = 0;
  • Final speed of block B:    vB = 20 m/s;
  • Acceleration due to gravity:    g = 10 m/s2.
Solution

In the rope that supports block B, we have the tension force \( \vec{T} \) due to block B. The cord transmits this tension to the mobile pulley p1 through the fixed pulley p2. On the other side of the mobile pulley p1, the same tension \( \vec{T} \) is transmitted to the fixed point on the wall. On the rope, that holds the mobile pulley p1, there is a tension \( 2\vec{T} \) due to the tension \( \vec{T} \) that acts on the rope on the two sides of the pulley, this tension is transmitted by the rope to block A (Figure 1).
Figure 1

Drawing a free-body diagram, we have the forces that act on the block.
  • Block A:
We choose a reference frame with the x-axis parallel to the inclined plane and in the direction downward. In this body acts the gravitational force \( {\vec F}_{gA} \), the tension force \( 2\vec{T} \) on the rope, and the normal reaction force \( \vec{N} \) (Figure 2-A).

Figure 2

The gravitational force can be decomposed into two components, a parallel component to the x-axis, \( {\vec F}_{gAP} \), and o another component normal or perpendicular \( {\vec F}_{gAN} \).
In the triangle on the left, in Figure 2-B, we see that the gravitational force \( {\vec F}_{gA} \) is perpendicular to the horizontal plane, it makes a 90° angle, the angle between the inclined plane and the horizontal plane is given as 30°, as angles of a triangle add 180º the angle α between gravitational force \( {\vec F}_{gA} \) and the parallel component \( {\vec F}_{gAP} \) should be
\[ 30°+90°+\alpha=180°\Rightarrow \alpha=180°-30°-90°\Rightarrow \alpha=60° \]
In the triangle on the right, we have that the normal component \( {\vec F}_{gAN} \) makes with the inclined plane a 90° angle, then the angle β between the gravitational force \( {\vec F}_{gA} \) and the normal component \( {\vec F}_{gAN} \) should be
\[ 60°+\beta=90°\Rightarrow \beta=90°-60°\Rightarrow \beta=30° \]
they are complementary angles.
We draw the forces in a coordinate system (Figure 2-C), we apply Newton's Second Law.
\[ \bbox[#99CCFF,10px] {\vec{F}=m\vec{a}} \tag{I} \]
As there is no movement in the y direction, the normal reaction force and the normal component of gravitational force cancel.
Direction x:
\[ \begin{gather} F_{gAP}-2T=m_{A}a_{A} \tag{II} \end{gather} \]
in the Figure 2-C, we have the gravitational force given by
\[ \begin{gather} F_{gAP}=F_{gA}\sin\theta \tag{III} \end{gather} \]
substituting the expression (III) into expression (II)
\[ \begin{gather} F_{gA}\sin\theta -2T=m_{A}a_{A} \tag{IV} \end{gather} \]
  • Block B:
We choose a reference frame with the x-axis in the upward direction. In this body acts the gravitational force \( {\vec F}_{gB} \) and the tension force \( \vec{T} \) on the rope (Figure 3).
Applying the expression (I), in the horizontal direction, we do not have forces acting, in the vertical direction we have the gravitational force and the tension force
\[ \begin{gather} T-F_{gB}=m_{B}a_{B} \tag{V} \end{gather} \]
Figure 3

To find the acceleration of block B, we use the displacement given in the problem (Figure 4). From the Kinematics we used the equation of velocity as a function of displacement
\[ \bbox[#99CCFF,10px] {v^{2}=v_{0}^{2}+2a\Delta S} \tag{VI} \]
  • Block B:
Final speed:    \( v=v_{B}=20\ \text{m/s} \);
Initial speed:    \( v_{0}=v_{0B}=0 \);
Displacement:    \( \Delta S=\Delta S_{B}=40\ \text{m} \).
Figure 4

substituting these values ​​in the expression (VI)
\[ \begin{gather} v_{B}^{2}=v_{0B}^{2}+2a_{B}\Delta S_{B}\\ 20^{2}=0^{2}+2 a_{B}.40\\ 400=0+80a_{B}\\ 80a_{B}=400\\ a_{B}=\frac{400}{80}\\ a_{B}=5\ \text{m/s}^{2} \tag{VII} \end{gather} \]
The interval of time that blocks B will take to move 40 m will be given by the expression of velocity as a function of time.
\[ \begin{gather} \bbox[#99CCFF,10px] {v=v_{0}+at} \tag{VIII} \end{gather} \]
substituting the data and acceleration found in (VII) in the expression (VIII)
\[ \begin{gather} v_{B}=v_{0B}+a_{B}t\\ 40=0+5t\\ t=\frac{40}{5}\\ t=8\ \text{s} \tag{IX} \end{gather} \]
When block B rises 40 m, the point c1 of the rope will also move the same 40 m from this total displacement, 20 m will be used in the shift of the pulley and block A, and the other 20 m will take back in the pulley and point c1 It will end its displacement in point c2 (Figure 5). Therefore, the displacement of block A will be half the displacement of block B. To find the acceleration of the block A we use the expression
\[ \begin{gather} \bbox[#99CCFF,10px] {S=S_{0}+v_{0}t+\frac{at^{2}}{2}} \tag{X} \end{gather} \]
Figure 5
  • Block A:
Initial speed:    \( v_{0}=v_{0A}=0 \);
Displacement:    \( \Delta S=\Delta S_{A}=\dfrac{\Delta S_{B}}{2}=20\ \text{m} \);
Time interval for displacement:    t = 8 s.
Substituting these values ​​in expression (X), the acceleration of block A will be
\[ \begin{gather} S_{A}=S_{0A}+v_{0A}t+\frac{a_{A}t^{2}}{2}\\ S_{A}-S_{0A}=v_{0A}t+\frac{a_{A}t^{2}}{2} \end{gather} \]
as \( S_{A}-S_{0A}=\Delta S_{A} \)
\[ \begin{gather} \Delta S_{A}=v_{0A}t+\frac{a_{A}t^{2}}{2}\\ 20=0.8+\frac{a_{A}8^{2}}{2}\\ 20=0+\frac{16a_{A}}{2}\\ 20=8a_{A}\\ a_{A}=\frac{20}{8}\\ a_{A}=2.5\ \text{m/s}^{2} \tag{XI} \end{gather} \]
Expressions (IV) and (V) can be written as a system of two equations
\[ \left\{ \begin{array}{r} \;F_{gA}\sin\theta -2T=m_{A}a_{A}\\ \qquad T-F_{gB}=m_{B}a_{B} \end{array} \right. \]
multiplying and dividing the right-hand side of the equation of by g
\[ \begin{gather} \left\{ \begin{array}{r} \;F_{gA}\sin\theta-2T=m_{A}a_{A}\dfrac{g}{g}\\ T-F_{gB}=m_{B}a_{B}\dfrac{g}{g} \end{array} \right.\\[10pt] \left\{ \begin{array}{r} \;F_{gA}\sin\theta-2T=\dfrac{m_{A}ga_{A}}{g}\\ T-F_{gB}=\dfrac{m_{B}ga_{B}}{g} \end{array} \right. \end{gather} \]
in both expressions, mAg and mBg, represent FgA and FgB weights of blocks A and B, respectively, substituting
\[ \left\{ \begin{array}{r} \;F_{gA}\sin\theta-2T=\dfrac{F_{gA}a_{A}}{g}\\ T-F_{gB}=\dfrac{F_{gB}a_{B}}{g} \end{array} \right. \]
substituting the problem data and accelerations found in (VII) and (XI)
\[ \left\{ \begin{array}{r} \;F_{gA}\sin 30°-2T=\dfrac{F_{gA}2.5}{10}\\ T-2000=\dfrac{2000\times 5}{10} \end{array} \right. \]
From the Trigonometry
\[ \sin 30°=\frac{1}{2} \]
\[ \left\{ \begin{matrix} \;\dfrac{F_{gA}}{2}-2T=\dfrac{F_{gA}}{4}\\ \;T-2000=1000 \end{matrix} \right. \]
solving the second equation for tension force T
\[ \begin{gather} T-2000=1000\\ T=1000+2000\\ T=3000\ \text{N} \end{gather} \]
substituting this value in the first expression
\[ \begin{gather} \frac{F_{gA}}{2}-2\times 3000=\frac{F_{gA}}{4}\\ \frac{F_{gA}}{2}-\frac{F_{gA}}{4}=6000 \end{gather} \]
multiplying and dividing by 2 the first term on the left-hand side of the equation
\[ \begin{gather} \frac{2}{2}\times\frac{F_{gA}}{2}-\frac{F_{gA}}{4}=6000\\ \frac{2F_{gA}-F_{gA}}{4}=6000\\ F_{gA}=4\times 6000 \end{gather} \]
\[ \bbox[#FFCCCC,10px] {F_{gA}=24000\ \text{N}} \]
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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 .