A block is moving on an inclined plane making an angle $45^{\circ}$ with the horizontal and the coefficient of friction is $\mu$. The force required to just push it up the inclined plane is 3 times the force required to just prevent it from sliding down. If we define $\mathrm{N}=10 \mu$, then N is

JEE Advanced · 2011 · 10 Apr 2011 · Paper 1 · Laws of Motion

A block is moving on an inclined plane making an angle

4 5^{\circ}

with the horizontal and the coefficient of friction is

μ

. The force required to just push it up the inclined plane is 3 times the force required to just prevent it from sliding down. If we define

N = 10 μ

, then N is

Answer: 5

Solution

Force to push up:

F_{1} = m g (sin θ + μ cos θ)

Force to prevent sliding down:

F_{2} = m g (sin θ - μ cos θ)

Given

F_{1} = 3 F_{2}

m g (sin θ + μ cos θ) = 3 m g (sin θ - μ cos θ)

sin θ + μ cos θ = 3 sin θ - 3 μ cos θ

4 μ cos θ = 2 sin θ

For

θ = 45°

sin 45° = cos 45° = \frac{1}{2}

4 μ \cdot \frac{1}{2} = 2 \cdot \frac{1}{2}

4 μ = 2 ⟹ μ = 0.5

Therefore,

N = 10 μ = 5

JEE Advanced · 2011 · 10 Apr 2011 · Paper 1 · Integer answer

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