Solved Problem on Static Equilibrium

A body with a mass of 200 kg is maintained in equilibrium on an inclined plane at 30° relative to the horizontal by a cord passing through a fixed pulley and the other end supports a body with a mass M. The rope makes with the inclined plane an angle of 45°. Determine:
a) The mass M;
b) The force exerted by the body on the plan.

Problem data:

Mass of the body on the plane: m=200 kg;
Angle of the inclined plane: 30°;
Angle of the string with the inclined plane: 45°;
Acceleration due to gravity: g=9.8 m/s².

Problem diagram:

Drawing free-bodies diagrams, we have the forces acting on the blocks, and we apply for the equilibrium condition, the sum of all forces is equal to zero

\[ \begin{gather} \bbox[#99CCFF,10px] {\sum \vec{F}=0} \tag{I} \end{gather} \]

Body of mass M:

\( \vec{T} \): tension force on the rope;
\( {\vec F}_{gM} \): gravitational force of the suspended body.

In the horizontal direction, no forces are acting. In the vertical direction, the gravitational force \( {\vec F}_{gM} \) and the tension force \( \vec{T} \) cancel (Figure 1)

\[ \begin{gather} T-P_{{M}}=0 \tag{II} \end{gather} \]

Figure 1

Body with mass 200 kg (Figure 2):

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

We chose a reference frame with an x-axis parallel to the inclined plane and pointing upwards (Figure 3)

Figure 2

We must find the angle that the gravitational force makes with the directions perpendicular (y) and parallel (x) to the inclined plane (Figure 3).

The angle \( Q\hat{A}M \) is given in the problem as 30°, the segment \( \overline{{QM}} \) (the direction of the gravitational force) is perpendicular to the segment \( \overline{{AC}} \), as the sum of the interior angles of a triangle equals 180°, then the angle \( A\hat{Q}M \) should be

\[ \begin{gather} A\hat{Q}M+30°+90°=180°\\[5pt] A\hat{Q}M=180°-30°-90°\\[5pt] A\hat{Q}M=60° \end{gather} \]

Figure 3

To determine the value of the angle α (Figure 4), we enlarge the red region of Figure 3. The angle \( A\hat{Q}M \) is equal to 60° and the segment \( \overline{QN} \) is perpendicular to the segment \( \overline{AB} \), makes an angle of 90°, then the sum of these angles with the angle α should be 180°

\[ \begin{gather} 60°+90°+\alpha=180°\\[5pt] \alpha=180°-60°-90°\\[5pt] \alpha=30° \end{gather} \]

Drawing the forces in a coordinate system xy (Figure 5) we can have the components along the x and y directions.

Figure 4

Components in the x direction

\( N_{x}=0 \)
\( T_{x}=T\cos 45° \)
\( F_{ix}=-F_{{i}}\cos 60° \)

Applying the equilibrium condition (I)

\[ \begin{gather} N_{x}+T\cos 45°-F_{gi}\cos 60°=0\\[5pt] T\cos 45°-F_{gi}\cos 60°=0 \tag{III} \end{gather} \]

Figure 5

Components in the y direction

\( N_{y}=N \)
\( T_{y}=T\sin 45° \)
\( F_{giy}=-F_{gi}\sin 60° \)

Applying the equilibrium condition (I)

\[ \begin{gather} N+T\sin 45°-F_{gi}\sin 60°=0 \tag{IV} \end{gather} \]

Solution

a) The gravitational force is given by

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

From the Trigonometry

\[ \cos 45°=\sin 45°=\frac{\sqrt{2\,}}{2} \]

\[ \cos 60°=\frac{1}{2} \]

\[ \sin 60°=\frac{\sqrt{3\,}}{2} \]

Equations (II), (III), and (IV) can be written as a system of three equations in three unknowns, N, T and M

\[ \begin{gather} \left\{ \begin{array}{l} T-Mg=0\\ \dfrac{\sqrt{2\,}}{2}T-\dfrac{1}{2}mg=0\\ N+\dfrac{\sqrt{2\,}}{2}T-\dfrac{\sqrt{3\,}}{2}mg=0 \tag{V} \end{array} \right. \end{gather} \]

solving the first equation of the system (V) for the tension force

\[ \begin{gather} T=Mg \tag{VI} \end{gather} \]

substituting in the second equation system (V)

\[ \begin{gather} \frac{\sqrt{2\,}}{2}Mg-\frac{1}{2}mg=0\\[5pt] \frac{\sqrt{2\,}}{\cancel{2}}M\cancel{g}=\frac{1}{\cancel{2}}m\cancel{g}\\[5pt] \sqrt{2\,}M=m\\[5pt] M=\frac{m}{\sqrt{2\,}} \end{gather} \]

substituting the mass m given in the problem and \( \sqrt{2\,}\approx 1,4142 \)

\[ \begin{gather} M=\frac{200}{1.4142} \end{gather} \]

\[ \begin{gather} \bbox[#FFCCCC,10px] {M=141.4\ \text{kg}} \end{gather} \]

b) The force exerted on the plane \( F_{p} \) is given by the y component of the gravitational force on the inclined plane

\[ \begin{gather} F_{p}=F_{giy}=-F_{gi}\sin 60°\\[5pt] F_{p}=-mg\sin 60°\\[5pt] F_{p}=-200\times 9.8\times \frac{\sqrt{3\,}}{2} \end{gather} \]

and \( \sqrt{3\,}\approx 1.7321 \)

\[ \begin{gather} \bbox[#FFCCCC,10px] {F_{{p}}=-1697\ \text{N}} \end{gather} \]

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