Calculating Rope Tension in Equilibrium Problem

Understanding the Physical Scenario

This problem asks us to find the tension in a specific segment of a rope supporting a mass. The situation involves a mass hanging from a rope, with an additional horizontal force applied at a point along the rope, causing it to be in equilibrium (i.e., stationary).

Given Information:

• Mass ($m$): $10 \text{ kg}$
• Acceleration due to gravity ($g$): $10 \text{ ms}^{-2}$
• Length of the rope segment from the roof to point P ($L_{OP}$): $2 \text{ m}$
• Vertical distance from the roof attachment point (O) to point P ($h$): $1 \text{ m}$
• A horizontal force ($F$) is applied at point P.
• The mass ($m$) is suspended from point P.
• The system is in equilibrium.

Determining the Rope Angle:

Let O represent the attachment point on the roof. Let P be the point on the rope where the mass ($m$) is attached and the horizontal force ($F$) is applied. The rope segment OP has a length of $L_{OP} = 2 \text{ m}$. We are told that point P is $1 \text{ m}$ vertically below O.

We can determine the angle ($\theta$) that the rope segment OP makes with the vertical line passing through O. Consider the right-angled triangle formed by:

• Point O (roof attachment).
• Point P.
• A point directly below O on the same horizontal level as P.

In this triangle, the hypotenuse is the rope segment $L_{OP} = 2 \text{ m}$, and the adjacent side (vertical distance) is $h = 1 \text{ m}$.

Using trigonometry, the cosine of the angle $\theta$ (between the rope OP and the vertical) is:

$ \cos(\theta) = \frac{\text{Adjacent}}{\text{Hypotenuse}} = \frac{h}{L_{OP}} $

Plugging in the values:

$ \cos(\theta) = \frac{1 \text{ m}}{2 \text{ m}} = \frac{1}{2} $

Therefore, the angle $\theta$ is:

$ \theta = \arccos\left(\frac{1}{2}\right) = 60^\circ $

The rope segment OP is at an angle of $60^\circ$ to the vertical.

Analyzing Forces Acting on Point P

For the system to be in equilibrium, the vector sum of all forces acting on point P must be zero. The forces involved are:

1. Tension ($T_{upper}$): The force exerted by the rope segment OP, pulling point P towards the roof attachment O.
2. Applied Horizontal Force ($F$): The external force applied horizontally at P.
3. Weight ($W$): The gravitational force acting on the mass ($m$) attached at P, pulling it vertically downwards.

First, calculate the weight of the mass:

$ W = m \times g = 10 \text{ kg} \times 10 \text{ ms}^{-2} = 100 \text{ N} $

Now, let's resolve the forces into horizontal (X-axis) and vertical (Y-axis) components. We'll consider the upward direction as positive Y and the rightward direction as positive X.

Force Components Summary:

• Tension ($T_{upper}$): Acts along the rope towards O, at $60^\circ$ from the vertical.
  • Vertical Component ($T_{y}$): $T_{upper} \cos(60^\circ) = T_{upper} \times \frac{1}{2}$ (Upwards)
  • Horizontal Component ($T_{x}$): $-T_{upper} \sin(60^\circ) = -T_{upper} \times \frac{\sqrt{3}}{2}$ (Leftwards)
• Horizontal Force ($F$): Acts horizontally.
  • Vertical Component ($F_{y}$): $0$
  • Horizontal Component ($F_{x}$): $F$ (Rightwards)
• Weight ($W$): Acts vertically downwards.
  • Vertical Component ($W_{y}$): $-W = -100 \text{ N}$ (Downwards)
  • Horizontal Component ($W_{x}$): $0$

Step-by-Step Calculation of Tension

To find the tension $T_{upper}$ in the upper part of the rope (segment OP), we use the equilibrium condition that the sum of forces in the vertical direction must be zero.

Vertical Equilibrium Analysis:

Summing the vertical components of all forces acting on P:

$ \sum F_y = T_{y} + F_{y} + W_{y} = 0 $

$ T_{upper} \cos(60^\circ) + 0 + (-100 \text{ N}) = 0 $

Substitute the value of $\cos(60^\circ) = \frac{1}{2}$:

$ T_{upper} \times \frac{1}{2} - 100 \text{ N} = 0 $

Now, solve for $T_{upper}$:

$ T_{upper} \times \frac{1}{2} = 100 \text{ N} $

Multiply both sides by 2:

$ T_{upper} = 2 \times 100 \text{ N} $

$ T_{upper} = 200 \text{ N} $

Horizontal Equilibrium Check (Optional):

We can also check the horizontal equilibrium condition:

$ \sum F_x = T_{x} + F_{x} + W_{x} = 0 $

$ -T_{upper} \sin(60^\circ) + F + 0 = 0 $

$ F = T_{upper} \sin(60^\circ) $

Using the value $T_{upper} = 200 \text{ N}$ and $\sin(60^\circ) = \frac{\sqrt{3}}{2}$:

$ F = 200 \text{ N} \times \frac{\sqrt{3}}{2} = 100\sqrt{3} \text{ N} $

This calculation confirms the required horizontal force, but the primary goal was to find the tension $T_{upper}$.

Final Result

Based on the vertical force equilibrium analysis, the tension in the upper part of the rope (the segment from the roof to point P) is found to be $200 \text{ N}$.