A mass M kg is suspended by a weightless string. The horizontal force required to hold the mass at `60^(@)` with the vertical is

Options

A. Mg
B. `Mgsqrt3`
C. `Mg(sqrt3+1)`
D. `(Mg)/(sqrt3)`

Answer

To solve the problem of finding the horizontal force required to hold a mass \( M \) kg suspended by a weightless string at an angle of \( 60^\circ \) with the vertical, we can follow these steps:

### Step 1: Understand the Forces Acting on the Mass
The mass \( M \) experiences three forces:
1. The weight \( W = Mg \) acting downwards.
2. The tension \( T \) in the string acting along the string.
3. The horizontal force \( F \) acting horizontally.

### Step 2: Draw a Free Body Diagram
Draw a diagram showing:
- The weight \( W \) acting vertically downward.
- The tension \( T \) acting at an angle of \( 60^\circ \) to the vertical.
- The horizontal force \( F \) acting horizontally.

### Step 3: Resolve the Tension into Components
The tension \( T \) can be resolved into two components:
- Vertical component: \( T \cos(60^\circ) \)
- Horizontal component: \( T \sin(60^\circ) \)

### Step 4: Set Up the Equations
For the mass to be in equilibrium:
1. The vertical forces must balance:
\[ T \cos(60^\circ) = Mg \]
2. The horizontal forces must balance:
\[ F = T \sin(60^\circ) \]

### Step 5: Solve for Tension \( T \)
From the vertical forces equation:
\[ T \cos(60^\circ) = Mg \]
Since \( \cos(60^\circ) = \frac{1}{2} \):
\[ T \cdot \frac{1}{2} = Mg \implies T = 2Mg \]

### Step 6: Substitute \( T \) into the Horizontal Force Equation
Now substitute \( T \) into the horizontal force equation:
\[ F = T \sin(60^\circ) \]
Since \( \sin(60^\circ) = \frac{\sqrt{3}}{2} \):
\[ F = (2Mg) \cdot \frac{\sqrt{3}}{2} = Mg\sqrt{3} \]

### Step 7: Conclusion
The horizontal force required to hold the mass at \( 60^\circ \) with the vertical is:
\[ F = Mg\sqrt{3} \]

### Final Answer
Thus, the required horizontal force is \( F = Mg\sqrt{3} \).

---