1. Set up forces along the incline

Let the mass of the block be $m$, so its weight is $mg$.

The plane is inclined at $\theta = 45^\circ$.

Normal reaction: $R = mg\cos\theta$

Friction magnitude at limiting condition: $f = \mu R = \mu mg\cos\theta$

Since the block is on the verge of motion in each case, friction acts opposite to the impending motion.

2. Force required to just push it up the incline

If the block is just about to move upward, friction acts down the plane.

So the applied force upward must balance:

• component of weight down the plane: $mg\sin\theta$
• friction down the plane: $\mu mg\cos\theta$

Hence, $F_1 = mg\sin\theta + \mu mg\cos\theta$

3. Force required to just prevent it from sliding down

To just prevent sliding down, the block is on the verge of moving downward, so friction acts up the plane.

Thus the required upward force balances the remaining tendency downward: $F_2 + \mu mg\cos\theta = mg\sin\theta$

So, $F_2 = mg\sin\theta - \mu mg\cos\theta$

4. Use the given condition

Given: $F_1 = 3F_2$

Substitute: $mg\sin\theta + \mu mg\cos\theta = 3\left(mg\sin\theta - \mu mg\cos\theta\right)$

Cancel $mg$: $\sin\theta + \mu \cos\theta = 3\sin\theta - 3\mu \cos\theta$

Rearrange: $4\mu \cos\theta = 2\sin\theta$

$2\mu \cos\theta = \sin\theta$

$\mu = \frac{\sin\theta}{2\cos\theta} = \frac{\tan\theta}{2}$

Since $\theta = 45^\circ$ and $\tan 45^\circ = 1$, $\mu = \frac{1}{2}$

5. Compute $N = 10\mu$

$N = 10\mu = 10\times \frac{1}{2} = 5$

6. Comparison with stored answer

Derived answer is $5$, which matches the stored correct answer.