A duster weighs 0.5N. It is pressed against a vertical board with a horizontal force of 11N, If the co-efficient of friction is 0.5 the minimum force that must be applied on the duster parallel to the board to move it upwards is

Options

A. 0.4 N
B. 0.7 N
C. 6 N
D. 7 N

Answer

To solve the problem, we need to determine the minimum force that must be applied on the duster parallel to the board to move it upwards. Here are the steps to find the solution:

### Step 1: Identify the Forces Acting on the Duster
- The weight of the duster (W) acting downwards = 0.5 N.
- The horizontal force (F_horizontal) pressing the duster against the board = 11 N.
- The coefficient of friction (μ) = 0.5.

### Step 2: Calculate the Normal Force (N)
The normal force (N) exerted by the vertical board on the duster is equal to the horizontal force applied:
\[ N = F_{\text{horizontal}} = 11 \, \text{N} \]

### Step 3: Calculate the Frictional Force (F_friction)
The frictional force (F_friction) can be calculated using the formula:
\[ F_{\text{friction}} = \mu \cdot N \]
Substituting the values:
\[ F_{\text{friction}} = 0.5 \cdot 11 \, \text{N} = 5.5 \, \text{N} \]

### Step 4: Determine the Total Force Required to Move the Duster Upwards
To move the duster upwards, we need to overcome both the weight of the duster and the frictional force. Therefore, the total force (F_total) required is:
\[ F_{\text{total}} = W + F_{\text{friction}} \]
Substituting the values:
\[ F_{\text{total}} = 0.5 \, \text{N} + 5.5 \, \text{N} = 6 \, \text{N} \]

### Conclusion
The minimum force that must be applied on the duster parallel to the board to move it upwards is **6 N**.

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