A body of mass 'M' starts sliding down on the inclined plane where the critical angle is 30 the coefficient of kinetic friction will be ?answer is square root 3 but how?i am not getting please help.
Why did the body go to see a therapist after sliding down the inclined plane? Because it was feeling a little fricti-onely!
But let me try to explain it to you. When the body starts sliding down the inclined plane, the force of gravity is acting straight downwards. This force can be resolved into two components: one that acts perpendicular to the plane (mg cosθ), and one that acts parallel to the plane (mg sinθ). Here, θ is the angle of the inclined plane, which is 30 degrees in this case.
The kinetic friction force (fk) acts parallel to the plane, in the opposite direction of motion. It can be expressed as fk = μkN, where μk is the coefficient of kinetic friction and N is the normal force exerted by the inclined plane on the body.
In this scenario, the normal force (N) can be calculated as N = mg cosθ. Therefore, the kinetic friction force can be written as fk = μk(mg cosθ).
At the critical angle (θ = 30 degrees), the body is on the verge of sliding and the friction force is at its maximum value, which is just about to overcome the parallel component of the weight mg (mg sinθ).
Now, equating the force of friction with the maximum force it can achieve, we get:
μk(mg cosθ) = (mg sinθ)
Dividing both sides by mg cosθ, we get:
μk = (mg sinθ) / (mg cosθ)
Simplifying, we have:
μk = tanθ
Substituting θ = 30 degrees, we get:
μk = tan(30) = √3
Therefore, the coefficient of kinetic friction is √3.
To find the coefficient of kinetic friction, we need to consider the forces acting on the body as it slides down the inclined plane. These forces include the force of gravity and the force of friction.
1. Force of gravity (mg): The force of gravity acts vertically downward and can be calculated using the equation Fg = mg, where m is the mass of the body and g is the acceleration due to gravity (approximately 9.8 m/s²).
2. Normal force (N): The normal force acts perpendicular to the surface of the inclined plane and is equal in magnitude but opposite in direction to the component of the force of gravity acting perpendicular to the plane. This can be calculated using the equation N = mg cos(θ), where θ is the angle of inclination (30° in this case).
3. Force of friction (Ff): The force of friction opposes the motion of the body and acts parallel to the surface of the inclined plane. In this case, since the body is sliding down, we are concerned with the coefficient of kinetic friction (µk).
The equation to calculate the force of friction is Ff = µkN.
Now, in order to find the coefficient of kinetic friction (µk), we need to find the magnitude of the components of the forces acting on the body.
1. Magnitude of the force of gravity component parallel to the plane (mg sin(θ)).
2. Magnitude of the component of the normal force perpendicular to the plane (N = mg cos(θ)).
As the body slides down the inclined plane, the force of friction balances the component of the force of gravity parallel to the plane. Therefore, we equate the magnitudes of these two forces:
mg sin(θ) = µkN
Substituting the values we have:
mg sin(30°) = µk mg cos(30°)
Simplifying the equation by canceling out mg, we get:
sin(30°) = µk cos(30°)
Now, solving for µk:
µk = sin(30°) / cos(30°)
µk = (√3 / 2) / (1/2)
µk = √3
Thus, the coefficient of kinetic friction (µk) is equal to √3 in this case.