A rectangular block of mass 2 kg is to be held against a rough vertical wall by applying a force of 98 N perpendicular to the wall. What is the coefficients of friction if the applied force is the minimum required force?

Options

A. 0.1
B. 0.15
C. 0.2
D. 0.3

Explanation

To find the coefficient of friction (μ) for a rectangular block of mass 2 kg held against a rough vertical wall by an applied force of 98 N, we can follow these steps:

### Step 1: Identify the forces acting on the block
- The weight (W) of the block acting downwards:
\[ W = m \cdot g = 2 \, \text{kg} \cdot 9.8 \, \text{m/s}^2 = 19.6 \, \text{N} \]
- The applied force (F) acting perpendicular to the wall:
\[ F = 98 \, \text{N} \]
- The normal force (R) exerted by the wall on the block, which is equal to the applied force when the applied force is the minimum required force to hold the block in place:
\[ R = F = 98 \, \text{N} \]
- The frictional force (f_s) acting upwards to prevent the block from sliding down.

### Step 2: Set up the equilibrium condition
For the block to be in equilibrium vertically, the frictional force must balance the weight of the block:
\[ f_s = W \]
Substituting the weight:
\[ f_s = 19.6 \, \text{N} \]

### Step 3: Relate the frictional force to the normal force
The maximum static frictional force can be expressed in terms of the coefficient of friction (μ) and the normal force (R):
\[ f_s = μ \cdot R \]
Substituting the values we have:
\[ 19.6 \, \text{N} = μ \cdot 98 \, \text{N} \]

### Step 4: Solve for the coefficient of friction (μ)
Rearranging the equation to solve for μ:
\[ μ = \frac{f_s}{R} = \frac{19.6 \, \text{N}}{98 \, \text{N}} = 0.2 \]

### Final Answer
The coefficient of friction (μ) is:
\[ μ = 0.2 \]