Question

An eraser weighing $ 1\,N $ is pressed against a vertical black board with a normal force of $5\, N$. The coefficient of friction $ \mu $ between eraser and Blackboard is approximately $0.4$. The force along the blackboard required to move the eraser is

• A. 2N
• B. 0.4N
• C. 2.4N
• D. 9.8N

Answer 1

The Correct Option is A

The eraser is pressed against the black board. Taking the vertical and horizontal components of forces,
we have \(F=\mu R\)
where, \(R\) is reaction of the board on the rubber.
Given,
\(R=5N,\mu =0.4\) 
\(\therefore\) \(F=0.4\times 5=2\,N\)

Answer 2

To determine the force required to move the eraser along the blackboard, we need to calculate the maximum static friction force. The maximum static friction force \(f_{\text{max}}\) is given by the product of the coefficient of friction \(\mu\) and the normal force \(N\):
\(f_{\text{max}} = \mu \cdot N\)
Given:
\(N\) = 5N
\(\mu\) = 0.4
Substitute these values into the equation:
\(f_{\text{max}} = 0.4 \times 5 \, \text{N} = 2 \, \text{N}\)

So, the correct option is (A): \(2N\)