Question

A block of mass \(0.1\) kg is held against a vertical wall by applying a horizontal force \(F\) on the block. If the coefficient of friction between the wall and the block is \(0.4\), then what is the magnitude of the minimum force \(F\) needed to keep the block at rest? \[ g=10\ \text{m s}^{-2} \]

• A. \(4\) N
• B. \(0.4\) N
• C. \(2.5\) N
• D. \(25\) N

Hint:

For a block pressed against a vertical wall: \[ N=F \] and \[ f_{\max}=\mu F. \] For limiting equilibrium, \[ \mu F=mg. \] This directly gives the minimum force required.

Answer 1

The Correct Option is C

Concept: The horizontal force produces a normal reaction on the wall. \[ N=F \] The maximum static friction is \[ f_{\max}=\mu N \] For the block to remain at rest, \[ f_{\max}\ge mg. \]

Step 1: Calculate the weight of the block. \[ m=0.1\ \text{kg} \] \[ g=10\ \text{m s}^{-2} \] \[ mg=0.1\times10=1\ \text{N} \]

Step 2: Apply the equilibrium condition. For minimum force, \[ \mu F=mg \] \[ 0.4F=1 \] \[ F=\frac{1}{0.4} \] \[ F=2.5\ \text{N} \]

Step 3: Write the final answer. \[\begin{aligned} \boxed{2.5\ \text{N}} \end{aligned}\] Hence, option \(\mathbf{(C)}\) is correct.