A block weighing `10kg` is at rest on a horizontal table. The coefficient of static friction between the block and the table is `0.5`. If a force acts downward at `60^(@)` with the horizontal, how large can it be without causing the block to move ? .`(g =100ms^(-2))` .

Options

A. `346N`
B. `446N`
C. `746N`
D. `846N`

Explanation

To solve the problem step by step, we will analyze the forces acting on the block and apply the principles of static friction.

### Step 1: Identify the forces acting on the block
1. **Weight of the block (W)**: The weight of the block is given by \( W = mg \), where \( m = 10 \, \text{kg} \) and \( g = 100 \, \text{m/s}^2 \).
\[ W = 10 \, \text{kg} \times 100 \, \text{m/s}^2 = 1000 \, \text{N} \]

2. **Applied force (F)**: A force \( F \) acts downward at an angle of \( 60^\circ \) with the horizontal. This force can be resolved 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 2: Calculate the normal force (N)
The normal force \( N \) acting on the block is affected by the weight of the block and the vertical component of the applied force. The normal force can be expressed as:
\[ N = W + F_y = W + F \sin(60^\circ) \]
Substituting the values we have:
\[ N = 1000 \, \text{N} + F \frac{\sqrt{3}}{2} \]

### Step 3: Calculate the maximum static friction (f_s)
The maximum static friction force is given by:
\[ f_s = \mu_s N \]
Where \( \mu_s = 0.5 \). Thus:
\[ f_s = 0.5 \left( 1000 \, \text{N} + F \frac{\sqrt{3}}{2} \right) \]

### Step 4: Set up the equilibrium condition
For the block to remain at rest, the horizontal component of the applied force must equal the maximum static friction force:
\[ F_x = f_s \]
Substituting the expressions we have:
\[ \frac{F}{2} = 0.5 \left( 1000 + F \frac{\sqrt{3}}{2} \right) \]

### Step 5: Solve for F
Expanding the right side:
\[ \frac{F}{2} = 500 + 0.25 F \sqrt{3} \]
Multiplying through by 2 to eliminate the fraction:
\[ F = 1000 + 0.5 F \sqrt{3} \]
Rearranging gives:
\[ F - 0.5 F \sqrt{3} = 1000 \]
Factoring out \( F \):
\[ F (1 - 0.5 \sqrt{3}) = 1000 \]
Thus:
\[ F = \frac{1000}{1 - 0.5 \sqrt{3}} \]

### Step 6: Calculate the value of F
To find the numerical value, we need to calculate \( 1 - 0.5 \sqrt{3} \):
\[ \sqrt{3} \approx 1.732 \implies 0.5 \sqrt{3} \approx 0.866 \]
So:
\[ 1 - 0.5 \sqrt{3} \approx 1 - 0.866 = 0.134 \]
Now substituting back:
\[ F \approx \frac{1000}{0.134} \approx 7460 \, \text{N} \]

### Final Answer
The maximum force \( F \) that can act downward at \( 60^\circ \) without causing the block to move is approximately \( 746 \, \text{N} \).

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