A particle of mass m can just rest on a rough plane inclined at 30 degree to the horizontal without slipping down. show that the least horizontal force needed to maintain its position if the inclination is increased to 45 degree is 0.268mg.

From the first bit of information, the static friction coefficient (Us) is given by

M*g*cos30*Us = M*g*sin30
Us = tan30 = 0.5773

At a 45 degree tilt with that static friction coefficient, the required horizontal force to prevent slipping is given by:

F cos45+ (M*g cos45+Fsin45)*(0.5773) =
= M*g*sin45
F(0.7071 +0.4082) = M*g(.7071 -.4082)
F = .2989/1.115 Mg = 0.268 M g

I have assumed that the applied frorce in horizontal, not up and along the incline. That would lead to a different answer.

To solve this question, we need to analyze the forces acting on the particle when it is on the inclined plane.

When the particle is at rest on the inclined plane, there are two main forces acting on it: gravitational force (mg) and the normal force (N) exerted by the inclined plane.

Let's break down the components of these forces:

1. The gravitational force (mg):
- The gravitational force acts vertically downward.
- We can resolve this force into two components: mg * sin(θ) (force acting parallel to the inclined plane) and mg * cos(θ) (force acting perpendicular to the inclined plane).

2. The normal force (N):
- The normal force acts perpendicular to the surface of the inclined plane.
- Since the particle is at rest, the normal force will be equal in magnitude and opposite in direction to the gravitational force acting perpendicular to the plane.
- Therefore, N = mg * cos(θ).

Now, let's consider the forces acting on the particle when the inclination angle is increased to 45 degrees.

1. The gravitational force (mg):
- As before, the gravitational force can be resolved into two components: mg * sin(θ) (force acting parallel to the inclined plane) and mg * cos(θ) (force acting perpendicular to the inclined plane).
- Since the angle of inclination is now 45 degrees, sin(45) = cos(45) = √2 / 2.
- So, the component of gravitational force parallel to the plane is mg * (√2 / 2) = mg * √2 / 2.

2. The normal force (N):
- The normal force will still be equal in magnitude and opposite in direction to mg * cos(θ), as the gravitational force acting perpendicular to the plane remains the same.
- Therefore, N = mg * cos(θ) = mg * (√2 / 2).

To determine the least horizontal force needed to maintain the particle's position on the inclined plane, we need to find the force acting parallel to the plane.

Considering the forces acting parallel to the plane:

- The frictional force (F) acting against the particle sliding down is given by F = μN, where μ is the coefficient of friction.
- The force acting parallel to the plane is the sum of the gravitational force component parallel to the plane (mg * √2 / 2) and the frictional force (F).
- The minimum horizontal force required to prevent the particle from sliding down is when the frictional force is at its maximum value, which occurs when the particle is at the verge of slipping.
- At this point, the frictional force is given by F = μN = μ * (mg * cos(θ)).

Since the particle is just at the verge of slipping, the frictional force will be equal to the sum of the gravitational force component parallel to the plane and the minimum horizontal force required:

μ * mg * cos(θ) = mg * √2 / 2 + minimum horizontal force

Now, we can solve for the minimum horizontal force:

minimum horizontal force = μ * mg * cos(θ) - mg * √2 / 2

Given that the inclination angle (θ) is 45 degrees, we can simplify further:

minimum horizontal force = μ * mg * cos(45) - mg * √2 / 2
minimum horizontal force = μ * mg * (√2 / 2) - mg * √2 / 2
minimum horizontal force = (μ - 1) * mg * √2 / 2

Given that μ (coefficient of friction) is not provided in the question, we cannot determine the precise value of the minimum horizontal force. However, we can express it in terms of the mass (m) and acceleration due to gravity (g) as follows:

minimum horizontal force = (μ - 1) * mg * √2 / 2
= (μ - 1) * m * g * √2 / 2

Therefore, the least horizontal force needed to maintain the particle's position when the inclination is increased to 45 degrees is equal to (μ - 1) * m * g * √2 / 2, where μ is the coefficient of friction, m is the mass of the particle, and g is the acceleration due to gravity.