This problem involves a block on an inclined plane and the relationship between the forces required to move it up and prevent it from sliding down. Let's break it down step by step.

Step 1: Identify the Forces

For a block on an inclined plane making an angle θ with the horizontal, two main forces act along the plane:

1. Component of Gravitational Force pulling it down the plane: $mg\sin \theta$
2. Force of Friction which opposes motion. Its magnitude is $\mu N$, where $N=mg\cos \theta$ is the normal force.

Step 2: Force to Prevent Sliding Down (F_down)

To just prevent the block from sliding down, an applied force must counteract the net force pulling it down. The gravitational component ($mg\sin \theta$) tries to move it down, while friction ($\mu mg\cos \theta$) acts up the plane to oppose this motion. Therefore, the minimum force required to prevent sliding is:

$F_{down}=mg\sin \theta -\mu mg\cos \theta$

Step 3: Force to Push Up (F_up)

To just push the block up the plane, the applied force must overcome both the gravitational component pulling it down and the friction which now acts down the plane (opposing the upward motion). Therefore, the minimum force required is:

$F_{up}=mg\sin \theta +\mu mg\cos \theta$

Step 4: Apply the Given Condition

The problem states that the force to push up is 3 times the force to prevent sliding down:

$F_{up}=3F_{down}$

Substituting the expressions:

$mg\sin \theta +\mu mg\cos \theta =3(mg\sin \theta -\mu mg\cos \theta )$

Step 5: Simplify the Equation

Divide both sides by $mg$ (which is common and non-zero):

$\sin \theta +\mu \cos \theta =3\sin \theta -3\mu \cos \theta$

Bring like terms together:

$\sin \theta +\mu \cos \theta -3\sin \theta +3\mu \cos \theta =0$

$-2\sin \theta +4\mu \cos \theta =0$

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

Divide both sides by 2:

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

Therefore, $\mu =\frac{\sin \theta }{2\cos \theta }=\frac{1}{2}\tan \theta$

Step 6: Substitute the Given Angle

The angle θ is 45°, and $\tan 45°=1$. Substituting this:

$\mu =\frac{1}{2}\times 1=0.5$

Step 7: Find the Value of N

The problem defines $N=10\mu$. Substituting μ = 0.5:

$N=10\times 0.5=5$

Final Answer

N = 5

Related Topics & Formulae

Friction on an Inclined Plane: The force of friction is given by $f=\mu N$, where $N=mg\cos \theta$ is the normal force. The component of gravity along the plane is $mg\sin \theta$.

Limiting Equilibrium: The minimum force required to move an object is calculated when it is on the verge of moving, and the friction force is at its maximum value, $\mu N$.