A block of mass 20 kg is placed on a rough horizontal surface. When a force of 80 N is applied at an angle of 30°with the horizontal, the block just begins to slide. What is the coefficient of static friction ? (g=`10m//s^2`)

Options

A. 0.253
B. 0.433
C. 0.63
D. 0.75

Explanation

To find the coefficient of static friction (μs) for the block, we will follow these steps:

### Step 1: Identify the forces acting on the block
- The weight of the block (W) acts downwards and is given by \( W = mg \), where \( m = 20 \, \text{kg} \) and \( g = 10 \, \text{m/s}^2 \).
- The applied force (F) is 80 N at an angle of 30° with the horizontal.
- The normal reaction force (R) acts upwards.
- The frictional force (f) acts opposite to the direction of motion.

### Step 2: Calculate the weight of the block
\[ W = mg = 20 \, \text{kg} \times 10 \, \text{m/s}^2 = 200 \, \text{N} \]

### Step 3: Resolve the applied force into components
- The horizontal component of the applied force (F_x):
\[ F_x = F \cos(30°) = 80 \cos(30°) = 80 \times \frac{\sqrt{3}}{2} = 40\sqrt{3} \, \text{N} \]
- The vertical component of the applied force (F_y):
\[ F_y = F \sin(30°) = 80 \sin(30°) = 80 \times 0.5 = 40 \, \text{N} \]

### Step 4: Apply Newton's second law in the vertical direction
Since the block is just about to slide, we can set up the equation for vertical forces:
\[ R + F_y = W \]
Substituting the known values:
\[ R + 40 = 200 \]
Solving for R:
\[ R = 200 - 40 = 160 \, \text{N} \]

### Step 5: Apply Newton's second law in the horizontal direction
Since the block is just about to slide, the frictional force (f) equals the horizontal component of the applied force:
\[ f = F_x \]
The frictional force can also be expressed as:
\[ f = \mu_s R \]
Setting these equal gives:
\[ \mu_s R = F_x \]
Substituting the known values:
\[ \mu_s \cdot 160 = 40\sqrt{3} \]

### Step 6: Solve for the coefficient of static friction (μs)
\[ \mu_s = \frac{40\sqrt{3}}{160} = \frac{\sqrt{3}}{4} \]
Calculating the numerical value:
\[ \sqrt{3} \approx 1.732 \implies \mu_s \approx \frac{1.732}{4} \approx 0.433 \]

### Final Answer
The coefficient of static friction (μs) is approximately **0.433**.

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