Solved Problem on Static Equilibrium

A man with a mass equal to 70 kg crosses a rope bridge over a river, as shown in the figure. The angles that the rope makes with a horizontal line, at the position where the man is, are equal to 15° and 25°. What are the tension forces acting on the rope?
Data: cos 15° = 0.9659, sin 15° = 0.2419, cos 25° = 0.9063, sin 25° = 0.4226.

Problem Data:

Mass of the man: m = 70 kg;
Angle 1 between the rope and the horizontal: α₁ = 25°;
Angle 2 between the rope and the horizontal: α₂ = 15°;
Acceleration due to gravity: g = 9,8 m/s².

Problem Diagram:

For simplicity, we consider that all the weight force of the man is applied at a single point of the rope (Figure 1-A).
The forces acting on the system are the weight force of the man \( \vec P \), which points downward, and the tensions in the ropes. The rope on the left side makes an angle of 25° with the horizontal; this is the same angle formed between the tension force 1, \( {\vec T}_1 \), and the x-axis. The rope on the right side makes an angle of 15° with the horizontal; this is the same angle formed between the tension force 2, \( {\vec T}_2 \), and the x-axis.

Figure 1

We draw the forces in an xy coordinate system and decompose the forces in these directions (Figure 1-B). The weight force has only a component in the negative y direction. The tension force 1 has a component in the positive x direction and a component in the negative y direction. The tension force 2 has a component in the positive x direction and a component in the positive y direction.

Solution:

The gravitational force is given by

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

the gravitational force of the man will be

\[ \begin{gather} F_g=70\times 9,8 \\[5pt] F_g=686\;\mathrm N \end{gather} \]

Since the system is in equilibrium, the resultant of the forces acting on it is equal to zero.

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

x direction: \( -{\vec T}_{1x}+{\vec T}_{2x}=0 \)

y direction: \( -\vec P+{\vec T}_{1y}+{\vec T}_{2y}=0 \)

in magnitude we have

\[ \begin{gather} -T_1\cos 25°+T_2\cos 15°= 0 \\[5pt] -F_g-T_1\sin 25°+T_2\sin 15°= 0 \end{gather} \]

substituting the given values for the sine and cosine and the weight calculated above, these equations form a system of two equations with two unknowns (T₁ and T₂).

\[ \begin{gather} \left\{ \begin{array}{l} -0,9063 T_1+0,9659 T_2=0 \tag{I} \\ -686-0,4226 T_1+0,2588 T_2=0 \end{array} \right. \end{gather} \]

from the first equation of the system (I), we isolate the value of T₁.

\[ \begin{gather} 0,9063 T_1=0,9659 T_2 \\[5pt] T_1=\frac{0,9659}{0,9063}T_2 \\[5pt] T_1=1,0669 T_2 \tag{II} \end{gather} \]

substituting equation (II) into the second equation of the system (I), we obtain the value of T₂.

\[ \begin{gather} -686-0,4226\times 1,0669 T_2+0,2588T_2=0 \\[5pt] 0,4509 T_2+0,2588 T_2=686 \\[5pt] 0,7097 T_2=686 \\[5pt] T_2=\frac{686}{0,7097} \end{gather} \]

\[ \begin{gather} \bbox[#FFCCCC,10px] {T_2=966,6\;\mathrm N} \end{gather} \]

Substituting the value found above into equation (III), we obtain the value of T₁.

\[ \begin{gather} T_1=1,0669\times 966,6 \end{gather} \]

\[ \begin{gather} \bbox[#FFCCCC,10px] {T_1=1031,3\;\mathrm N} \end{gather} \]

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