Determine Constant N for Friction on Inclined Plane

To determine the value of $N$, we analyze the forces acting on the block on an inclined plane with an angle of inclination $\theta = 45^\circ$ and coefficient of friction $\mu$.

1. Forces acting on the block:

• Component of gravitational force down the incline:
  $F_g = mg \sin\theta$
• Normal force perpendicular to the incline:
  $N_{\text{normal}} = mg \cos\theta$
• Maximum static/limiting friction force:
  $f_{\text{max}} = \mu N_{\text{normal}} = \mu mg \cos\theta$

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2. Force required to push the block up the incline ($F_1$):

When the block is pushed up the incline, friction acts down the incline (opposing motion). For the block to just move up, the applied force must balance both gravity and friction:
$F_1 = mg \sin\theta + \mu mg \cos\theta = mg (\sin\theta + \mu \cos\theta)$

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3. Force required to just prevent the block from sliding down ($F_2$):

When the block tends to slide down, friction acts up the incline to oppose the motion. To just prevent it from sliding, the required upward force is:
$F_2 = mg \sin\theta - \mu mg \cos\theta = mg (\sin\theta - \mu \cos\theta)$

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4. Applying the given condition:

We are given that $F_1 = 3 F_2$:
$mg (\sin\theta + \mu \cos\theta) = 3 mg (\sin\theta - \mu \cos\theta)$

Dividing both sides by $mg \cos\theta$:
$\tan\theta + \mu = 3(\tan\theta - \mu)$

Substitute $\theta = 45^\circ$ so that $\tan 45^\circ = 1$:
$1 + \mu = 3(1 - \mu)$
$1 + \mu = 3 - 3\mu$
$4\mu = 2 \implies \mu = 0.5$

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5. Calculating $N$:

Given $N = 10\mu$:
$N = 10 \times 0.5 = 5$