Question

The minimum force required to start pushing a body up a rough (coefficient of) inclined plane is \(\mathrm{F}_{1}\). While the minimum force needed to prevent it from sliding down is \(\mathrm{F}_{2}\). If the inclined plane makes an angle \(\theta\) from the horizontal. such that \(\tan \theta=2 \mu\) than the ratio \(\left(\mathrm{F}_{1} / \mathrm{F}_{2}\right)\) is (A) 4 (B) 1 (C) 2 (D) 3

Short answer

The ratio of the minimum force required to start pushing a body up a rough inclined plane is F₁ / F₂ = 3 (Option D).

Step-by-step solution

Identify the forces acting on the body

There are three forces acting on the body: 1. Gravitational force (W = mg) acting vertically downwards. 2. Normal force (N) acting perpendicular to the inclined plane (upwards) 3. Frictional force (Ff) acting parallel to the inclined plane, opposing the motion of the body We have to analyze these forces to calculate F₁ and F₂.

Resolve the forces into components

We can resolve W into two components: one is parallel to the inclined plane Wp (opposing the upward motion) and the other is perpendicular to the plane Wn (counteracting Normal force). Wp = W * sinθ Wn = W * cosθ

Calculate the frictional force

The frictional force is given by the formula Ff = μN, however, since Wn counteracts N, Ff = μWn = μW * cosθ

The minimum force required to push the body up

To start pushing the body up, the force applied should be just equal to the sum of component of weight acting parallel to inclined plane i.e., Wp and the frictional force i.e., Ff. F₁ = Wp + Ff = W * sinθ + μW * cosθ

The minimum force required to prevent the body from sliding downwards

To prevent the body from sliding down, the force applied should be just equal to the net force acting downward. F₂ = Wp - Ff = W * sinθ - μW * cosθ

Use tanθ = 2μ to find the ratio

To eliminate μ from the expressions, we can use the given expression tanθ = 2μ, which gives: μ = (tanθ) /2 Now, divide F₁ by F₂ to find the required ratio: \( \frac{F_1}{F_2} = \frac{W \times \sin \theta + \frac{\tan \theta}{2} \times W \times \cos \theta}{W \times \sin \theta - \frac{\tan \theta}{2} \times W \times \cos \theta} \)

Simplify and find the ratio

The term W can be canceled out from both the numerator and denominator: \( \frac{F_1}{F_2} = \frac{\sin \theta + \frac{\tan \theta}{2} \times \cos \theta}{\sin \theta - \frac{\tan \theta}{2} \times \cos \theta} \) Now, substitute tanθ = 2μ in the above expression and simplify: \( \frac{F_1}{F_2} = \frac{\sin \theta + \mu \times \cos \theta}{\sin \theta - \mu \times \cos \theta} = \frac{\sin \theta + \frac{\tan \theta}{2} \times \cos \theta}{\sin \theta - \frac{\tan \theta}{2} \times \cos \theta} = \)\( \frac{2 \sin \theta + \sin \theta}{2 \sin \theta - \sin \theta} \) Finally, we get: \( \frac{F_1}{F_2} = \boxed{3} \) Therefore, the ratio (F₁ / F₂) is 3 (Option D).