A mass M is hung with a light inextensible string as shown in figure. Find the tension of the horizontal string.
Show Hint
Always start by drawing a clear Free Body Diagram (FBD) at the point where forces meet. Resolve forces into components along convenient axes (usually horizontal and vertical) and then apply equilibrium conditions.
Step 1: Understanding the Question:
The problem asks for the tension in the horizontal string that supports a mass M, given that another string is inclined at 45\(^{\circ}\) and connected to the same junction. This is a static equilibrium problem. Step 2: Key Formula or Approach:
For a system in equilibrium, the net force in both horizontal and vertical directions is zero.
Let \(T_H\) be the tension in the horizontal string and \(T_I\) be the tension in the inclined string.
The weight of the mass is \(W = Mg\). Step 3: Detailed Explanation:
Draw a Free Body Diagram (FBD) for the junction point O where the three strings meet.
- Downwards force: \( Mg \) (due to the hanging mass).
- Horizontal force (to the left): \( T_H \) (tension in the horizontal string).
- Inclined force: \( T_I \) (tension in the string making 45\(^{\circ}\) with the horizontal).
Resolve the inclined tension \( T_I \) into its horizontal and vertical components:
- Horizontal component of \( T_I \): \( T_I \cos 45^{\circ} \) (to the right).
- Vertical component of \( T_I \): \( T_I \sin 45^{\circ} \) (upwards).
Apply equilibrium conditions:
1. Vertical equilibrium (\( \sum F_y = 0 \)):
Forces acting upwards = Forces acting downwards
\[ T_I \sin 45^{\circ} = Mg \]
\[ T_I \left( \frac{1}{\sqrt{2}} \right) = Mg \Rightarrow T_I = Mg\sqrt{2} \]
2. Horizontal equilibrium (\( \sum F_x = 0 \)):
Forces acting to the right = Forces acting to the left
\[ T_I \cos 45^{\circ} = T_H \]
Substitute \( T_I = Mg\sqrt{2} \) and \( \cos 45^{\circ} = \frac{1}{\sqrt{2}} \):
\[ T_H = (Mg\sqrt{2}) \left( \frac{1}{\sqrt{2}} \right) \]
\[ T_H = Mg \] Step 4: Final Answer:
The tension in the horizontal string is \( Mg \).
