Solved Problem on Dynamics

The masses m and M are connected, by two ropes A and B, on an inclined plane elevated, as in the figure. Neglecting the masses of the ropes, and the friction in the pulleys, it is given the inclination angle of the plane equal to θ and the acceleration due to gravity g, determine:
a) The acceleration of the system, knowing that the mass M is descending the plane;
b) The difference between tensions forces T_A and T_B.

Problem data:

Mass of block 1: M;
Mass of block 2: m;
Angle of inclination of the plane: θ;
Acceleration due to gravity: g.

Problem diagram:

The acceleration of the system is in the direction of the block with mass M descending the plane and the block with mass m rising (Figure 1).

Figure 1

Solution

Drawing a free-body diagram, we have the forces that act in each of them and we apply Newton's Second Law.

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

Block 1:

\( {\vec F}_{gM} \): gravitational force of block 1;
\( \vec{N} \): normal reaction force of the plane on block 1;
\( {\vec T}_{A} \): tension force on rope A;
\( {\vec T}_{B} \): tension force on rope B.

We choose a reference frame, with the x-axis parallel to the inclined plane and in a direction downward (Figure 2-A). The gravitational force \( {\vec F}_{gM} \) can be decomposed into two components, a component parallel to the x-axis, \( {\vec F}_{gP} \), and another component normal or perpendicular \( {\vec F}_{gN} \). In the triangle to the right, in Figure 2-B, we see that the gravitational force \( {\vec F}_{gM} \) is perpendicular to the horizontal plane, it makes a 90° angle, the angle between the inclined plane and the horizontal plane is equal to θ, 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

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

The components of the gravitational force in the x and y directions are perpendicular to each other. In the triangle to the left, we have the angle between the gravitational force \( {\vec F}_{gM} \) and the component of the gravitational force in the y direction, \( {\vec F}_{gN} \), is

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

Figure 2

Drawing the forces in a system of coordinates, in y direction the gravitational force \( {\vec F}_{gM} \) and the normal reaction force \( \vec{N} \) cancel, there is no movement in this direction. In the x direction applying the expression (I)

\[ \begin{gather} F_{gP}+T_{B}-T_{A}=Ma \tag{II} \end{gather} \]

The angle &theeta; is measured with the y-axis, unlike it is usually done, when the angle is used with the x-axis. The component of gravitational force in the x direction is given by (Figure 2-C)

\[ \begin{gather} F_{gP}=F_{gM}\sin \theta \tag{III} \end{gather} \]

The gravitational force is given by

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

substituting the expression (IV) into expression (III) to the mass M

\[ \begin{gather} F_{gP}=Mg\sin \theta \tag{V} \end{gather} \]

substituting the expression (V) into expression (II)

\[ \begin{gather} Mg\sin \theta +T_{B}-T_{A}=Ma \tag{VI} \end{gather} \]

Block 2:

\( {\vec F}_{gm} \): gravitational force of block 2;
\( {\vec T}_{A} \): tension force on rope A;
\( {\vec T}_{B} \): tension force on rope B.

Figure 3

We choose the positive direction upward in the same direction as the acceleration of the system. In the horizontal direction, no forces are acting in the block, in the vertical direction applying the expression (I)

\[ \begin{gather} T_{A}-T_{B}-F_{gm}=ma \tag{VII} \end{gather} \]

substituting the expression (IV) into expression (VII) to the mass m

\[ \begin{gather} F_{gm}=mg \tag{VIII} \end{gather} \]

substituting the expression (VIII) into expression (VII)

\[ \begin{gather} T_{A}-T_{B}-mg=ma \tag{IX} \end{gather} \]

a) Adding the expressions (VI) and (IX), we have the acceleration of the system

\[ \frac{\left. \begin{align} \;Mg\sin \theta+\cancel{T_{B}}-\cancel{T_{A}}=Ma\\ \;\cancel{T_{A}}-\cancel{T_{B}}-mg=ma \end{align} \right.} {Mg\sin \theta -mg=Ma+ma} \]

factoring the acceleration a on the right-hand side of the equation, and the acceleration due to gravity g on the left-hand side

\[ g(M\sin \theta -m)=a(M+m) \]

\[ \bbox[#FFCCCC,10px] {a=g\left(\frac{M\sin \theta -m}{M+m}\right)} \]

b) Substituting the value found in the previous item in the expression (IX)

\[ \begin{gather} T_{A}-T_{B}-mg=mg\left(\frac{M\sin \theta-m}{M+m}\right)\\ T_{A}-T_{B}=mg\left(\frac{M\sin \theta-m}{M+m}\right)+mg \end{gather} \]

factoring the term mg on the right-hand side of the equation

\[ T_{A}-T_{B}=mg\left(\frac{M\sin \theta -m}{M+m}+1\right) \]

in the parentheses, the common factor between 1 and (M+m) will be (M+m), placing over the same denominator

\[ \begin{gather} T_{A}-T_{B}=mg\left(\frac{M\sin \theta-m+M+m}{M+m}\right)\\ T_{A}-T_{B}=mg\left(\frac{M\sin \theta+M}{M+m}\right) \end{gather} \]

factoring the term \( \frac{M}{M+m} \)

\[ \bbox[#FFCCCC,10px] {T_{A}-T_{B}=\frac{Mmg}{M+m}\;\left(\sin \theta +1\right)} \]

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