Solved Problem on Dynamics

In the system of the figure, the masses of A, B, and C are 10 kg, 20 kg, and 5 kg, and sin θ = 0.8. Neglecting the friction forces, calculate the acceleration of the system and the magnitude of the tension forces on the ropes. Assume g = 10 m/s².

Problem data:

Mass of body A: m_A = 10 kg;
Mass of body B: m_B = 20 kg;
Mass of body C: m_C = 5 kg;
Acceleration due to gravity: g = 10 m/s².

Problem diagram:

We choose the acceleration of the system in the direction that body A is descending and body C is rising (Figure 1).

Figure 1

Solution

We draw a free-body diagram, and we have the forces that act in each of them.
Body A (Figure 2-A):

\( {\vec F}_{gA} \): gravitational force on body A;
\( {\vec N}_{A} \): normal reaction force of the plane on the body A;
\( {\vec T}_{AB} \): tension force between blocks A and B.

We choose a frame of reference with the x-axis parallel to the inclined plane and in the direction of acceleration. The gravitational force \( {\vec F}_{gA} \) can be decomposed into two components, a parallel component to the x-axis \( {\vec F}_{gAP} \), and other component normal or perpendicular \( {\vec F}_{gA} \).

Figure 2

In the triangle on the left in Figure 2-B, we see that the gravitational force is perpendicular to the horizontal plane, making a 90° angle, the angle between the inclined plane and the horizontal plane is given as θ. As the sum of the interior angles of a triangle equals 180°, the angle between the gravitational force \( {\vec F}_{gA} \) and the parallel component \( {\vec F}_{gAP} \) will be

\[ \alpha +\theta +90°=180°\Rightarrow \alpha=180°-\theta -90°\Rightarrow \alpha=90°-\theta \]

The components of the gravitational force in the x and y directions are perpendicular to each other, in the triangle to the right, we have the angle between the gravitational force \( {\vec F}_{gA} \) and the normal component of the gravitational force \( {\vec F}_{gAN} \) will be

\[ 90°-\alpha \Rightarrow 90°-(90°-\theta)\Rightarrow 90°-90°+\theta \Rightarrow \theta \]

Drawing the forces in a coordinate system (Figure 2 -C), applying Newton's Second Law

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

In the y direction, there is no movement, the normal reaction force \( {\vec N}_{A} \), and normal component of gravitational force \( {\vec F}_{gAN} \) cancel.
In the x direction, considering the angle θ measured from the y-axis unlike what is usually done when the angle is measured with the x-axis

\[ \begin{gather} F_{gAP}-T_{AB}=m_{A}a \tag{II} \end{gather} \]

the parallel component to the x-axis is given by

\[ \begin{gather} F_{gAP}=F_{gA}\sin \theta \tag{III} \end{gather} \]

substituting the expression (III) into expression (II)

\[ \begin{gather} F_{gA}\sin \theta -T_{AB}=m_{A}a \tag{IV} \end{gather} \]

The gravitational force is given by

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

substituting the expression (V) into expression (IV) for the body A

\[ \begin{gather} m_{A}g\sin \theta -T_{AB}=m_{A}a \tag{VI} \end{gather} \]

Body B:

\( {\vec F}_{gB} \): gravitational force on body B;
\( {\vec N}_{B} \): normal reaction force.
\( {\vec T}_{AB} \): tension force on the rope between blocks A and B;
\( {\vec T}_{BC} \): tension force on the rope between blocks B and C.

Figure 3

In the y direction, there is no vertical motion, the normal reaction force \( {\vec N}_{B} \) and gravitational force \( {\vec F}_{gB} \) cancel.
In the x direction applying the expression (I)

\[ \begin{gather} T_{AB}-T_{BC}=m_{B}a \tag{VII} \end{gather} \]

Body C:

\( {\vec F}_{gC} \): gravitational force on body C;
\( {\vec T}_{BC} \): tension force on the rope between blocks B and C.

Figure 4

In the horizontal direction, no forces are acting.
In the vertical direction, applying the expression (I)

\[ \begin{gather} T_{BC}-F_{gC}=m_{C}a \tag{VIII} \end{gather} \]

substituting the expression (V) into expression (VIII) for the body C

\[ \begin{gather} T_{BC}-m_{C}g=m_{C}a \tag{IX} \end{gather} \]

The equations (VI), (VII), and (IX) can be written as a system of three equations to three unknowns (T_AB, T_BC e a)

\[ \left\{ \begin{array}{l} m_{A}g\sin \theta-T_{AB}=m_{A}a\\ T_{AB}-T_{BC}=m_{B}a\\ T_{BC}-m_{C}g=m_{C}a \end{array} \tag{X} \right. \]

adding the three equations

\[ \begin{gather} \frac{ \begin{align} m_{A}g\sin \theta-\cancel{T_{AB}}=m_{A}a\\ \cancel{T_{AB}}-\cancel{T_{BC}}=m_{B}a\\ \text{(+)}\qquad\cancel{T_{BC}}-m_{C}g=m_{C}a \end{align} } {m_{A}g\sin \theta-m_{C}g=\left(m_{A}+m_{B}+m_{C}\right)a}\\ a=\frac{m_{A}g\sin \theta-m_{C}g}{m_{A}+m_{B}+m_{C}} \tag{XI} \end{gather} \]

substituting the values given in the problem into expression (XI)

\[ \begin{gather} a=\frac{10\times 10\times 0.8-5\times 10}{10+20+5}\\ a=\frac{80-50}{35}\\ a=\frac{30}{35} \end{gather} \]

\[ \bbox[#FFCCCC,10px] {a=0.86\;\text{m/s}^{2}} \]

Substituting the mass of the body A, the value of sin θ and the acceleration found above, in the first equation of the system (X), the tension force will be

\[ \begin{gather} m_{A}g\sin \theta-T_{AB}=m_{A}a\\ 10\times 10\times 0.8-T_{AB}=10\times 0.86\\ 80-T_{AB}=8,6\\ T_{AB}=80-8,6 \end{gather} \]

\[ \bbox[#FFCCCC,10px] {T_{AB}=71.4\;\text{N}} \]

Substituting the mass of the body C, and the acceleration found above, in the third expression of the system (X), the tension force will be

\[ \begin{gather} T_{BC}-m_{C}g=m_{C}a\\ T_{BC}-5\times 10=3\times 0.86\\ T_{BC}-50=2.58\\ T_{BC}=50+2.58 \end{gather} \]

\[ \bbox[#FFCCCC,10px] {T_{BC}=52.6\;\text{N}} \]

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