Question

If the inclination of an inclined plane is 30degree a block of mass m kept at rest on the surface of the plane will just slide down.if the inclination of block is raised to 45degree what will be the acceleration of the body?

Answer 1

the forces acting on the block are : Normal reaction of the plane perpendicular to the surface,  weight of the block vertically down wards. Force of friction acting along the surface upwards.  Look at the free body diagram showing forces.

If Ф is the angle of plane = 30 deg, the block just starts sliding down. It means that the force of friction upwards the plane is equal to the component of gravitational force downwards the plane.

mg sin 30 = Mu N            and  mg cos 30 = N

Hence     mg sin 30 = Mu mg cos 30

we get    Mu = tan 30  = 0.5773

Now when the angle is increased to 45 degrees, block accelerates with 'a' meters/sec² down the plane.

mg sin 45 - Mu N = m a
Also,  mg cos 45 = N.            Substitute value of N in the above equation.

So we get  mg sin 45 - Mu mg cos 45 = m a

=>  a = 9.8 /√2  [ 1 - Mu ]  = 9.8/√2  * 0.4227  = 2.929 m/sec²

Answer 2

the forces acting on the block are : Normal reaction of the plane perpendicular to the surface,  weight of the block vertically down wards. Force of friction acting along the surface upwards.  Look at the free body diagram showing forces.

If Ф is the angle of plane = 30 deg, the block just starts sliding down. It means that the force of friction upwards the plane is equal to the component of gravitational force downwards the plane.

mg sin 30 = Mu N            and  mg cos 30 = N

Hence     mg sin 30 = Mu mg cos 30

we get    Mu = tan 30  = 0.5773

Now when the angle is increased to 45 degrees, block accelerates with 'a' meters/sec² down the plane.

mg sin 45 - Mu N = m a

Also,  mg cos 45 = N.            Substitute value of N in the above equation.

So we get  mg sin 45 - Mu mg cos 45 = m a

=>  a = 9.8 /√2  [ 1 - Mu ]  = 9.8/√2  * 0.4227  = 2.929 m/sec²