Question

In the figure find the minimum value mass m of the rod so that the block of mass M = 20 Kg remains stationary on the inclined plane .
plz help in this fast I will mark brainiest.

Answer 1

Given:

In the figure the block of mass M = 20 Kg remains stationary on the inclined plane .

To find:

Find the minimum value mass m of the rod.

Solution:

The mass of the block = M = 20kg

The coefficient of friction = μ = 0.5

Given, θ = 37°

The angle of response = Ф = tan^{-1} μ = tan^{-1} 0.5 = 26.56°

Ф = 26.56°

as,  θ > Ф

The block will have a tendency to slide down

So, we have,

Mgcosθ + N', N = mgcosθ  

N' = Mgcosθ + mgcosθ ….(1)

For the block to remain stationary, we have,

f = Mgsinθ  ...(2)

If f is static as nature, then,

f < fmax

Mgsinθ < μN'

Mgsinθ < μ(Mgcosθ + mgcosθ)

Mgsinθ − μmgcosθ < μmgcosθn

m > M (tanθ/μ - 1)

m > 20 (2/0.5 - 1)

m > 20 (3)

m > 60

∴ The minimum value mass m of the rod is m > 60