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

Step 1: Picture the forces on the block.
The hand pushes the block horizontally onto the wall with force $F$. The wall pushes back with an equal normal reaction $N = F$. Gravity pulls the block down with weight $mg$, and friction along the wall acts upward to hold it.
Step 2: Find the weight.
$mg = 0.1 \times 10 = 1$ N.
Step 3: Write the friction available.
The largest friction the wall can supply is $f_{\max} = \mu N = \mu F$.
Step 4: Condition for the block not to slide.
To just hold the block, friction must balance the weight, so at the minimum force $\mu F = mg$.
Step 5: Solve for $F$.
\[ F = \frac{mg}{\mu} = \frac{1}{0.4} \]
Step 6: Compute the value.
\[ F = 2.5\ \text{N} \]
Any smaller push would let the block slip down, so $2.5$ N is the minimum needed.
\[ \boxed{F = 2.5\ \text{N}} \]