Identify all forces on the block

The block of mass $m=2\,\text{kg}$ is pressed against a vertical wall by a horizontal force of $50\,\text{N}$.

• Horizontal push on block = $50\,\text{N}$
• Normal reaction by wall = $N=50\,\text{N}$
• Weight of block downward = $mg = 2\times 10 = 20\,\text{N}$
• Applied vertical upward force = $F$
• Static friction acts along the wall and opposes the tendency of motion.

Maximum static friction

Given coefficient of static friction: $\mu_s = 0.5$

Hence maximum static friction is $f_{s,\max} = \mu_s N = 0.5\times 50 = 25\,\text{N}$

Condition for maximum upward force without moving upward

If $F$ is increased upward, the block tends to move upward.

Therefore static friction will act downward to oppose this tendency.

For the block to be just on the verge of moving upward, downward forces = upward forces.

Upward force: $F$

Downward forces:

• weight $=20\,\text{N}$
• maximum friction $=25\,\text{N}$

So at limiting equilibrium, $F = mg + f_{s,\max}$ $F = 20 + 25 = 45\,\text{N}$

Answer

The maximum value of $F$ so that the block does not move upward is $\boxed{45\,\text{N}}$

Option check

• A: $10\,\text{N}$ — incorrect
• B: $20\,\text{N}$ — incorrect
• C: $25\,\text{N}$ — incorrect
• D: $45\,\text{N}$ — correct

Comparison with stored answer

Stored correct answer: D

Our derived answer: D ($45\,\text{N}$)

Hence, the derived answer agrees with the stored answer.