A block of mass 1kg is at rest on a horizontal table. The coefficient of static friction between the block and the table is `0.5` . The magnitude of the force acting upward at an angle of `60^(@)` from the horizontal that will just start the block moving is.

Options

A. 5N
B. `(20)/(2+sqrt3)N`
C. `(20)/(2-sqrt3)N`
D. 10N

Explanation

To solve the problem step by step, we will analyze the forces acting on the block and apply Newton's laws of motion.

### Step 1: Identify the forces acting on the block
The forces acting on the block are:
1. The weight of the block (W = mg) acting downwards.
2. The normal force (R) acting upwards.
3. The applied force (F) acting at an angle of 60 degrees above the horizontal.

### Step 2: Break down the applied force into components
The applied force \( F \) can be broken down into two components:
- Horizontal component: \( F_x = F \cos(60^\circ) = \frac{F}{2} \)
- Vertical component: \( F_y = F \sin(60^\circ) = F \frac{\sqrt{3}}{2} \)

### Step 3: Write the equations for equilibrium
Since the block is at rest initially, we can use the equilibrium conditions:
1. In the vertical direction (y-direction):
\[ R + F_y = W \]
Substituting the values:
\[ R + F \frac{\sqrt{3}}{2} = mg \]
Where \( m = 1 \, \text{kg} \) and \( g = 10 \, \text{m/s}^2 \):
\[ R + F \frac{\sqrt{3}}{2} = 10 \]
Rearranging gives:
\[ R = 10 - F \frac{\sqrt{3}}{2} \quad \text{(Equation 1)} \]

2. In the horizontal direction (x-direction):
The frictional force \( F_f \) that opposes the motion is given by:
\[ F_f = \mu R \]
Where \( \mu = 0.5 \). The equation for the horizontal forces becomes:
\[ F_x = F_f \]
Substituting the values:
\[ F \cos(60^\circ) = \mu R \]
Which simplifies to:
\[ \frac{F}{2} = 0.5 R \]
Rearranging gives:
\[ F = R \quad \text{(Equation 2)} \]

### Step 4: Substitute Equation 1 into Equation 2
Substituting the expression for \( R \) from Equation 1 into Equation 2:
\[ F = 10 - F \frac{\sqrt{3}}{2} \]
Rearranging gives:
\[ F + F \frac{\sqrt{3}}{2} = 10 \]
Factoring out \( F \):
\[ F \left(1 + \frac{\sqrt{3}}{2}\right) = 10 \]
Thus,
\[ F = \frac{10}{1 + \frac{\sqrt{3}}{2}} \]

### Step 5: Simplify the expression for F
To simplify:
\[ F = \frac{10}{\frac{2 + \sqrt{3}}{2}} = \frac{20}{2 + \sqrt{3}} \]

### Final Answer
The magnitude of the force acting upward at an angle of \( 60^\circ \) that will just start the block moving is:
\[ F = \frac{20}{2 + \sqrt{3}} \, \text{N} \]