Question

A mass of 200 kg is placed on a rough inclined plane of angel 30° . If Coefficient of limiting friction is $\sf{\dfrac{1}{\sqrt{3}}}$ , find the greatest and the least forces in Newton , acting parallel to the plane to keep the mass in equilibrium .​

Answer 1

Explanation:

Greatest Force - 1960N

Least Foce - ON

Explanation:

R=mgcose

fmax= μR = μMG Cose

So the greatest force requires to keep the body in equilibrium

Fmax = mgsin+fmax = mgsine

+μmgcose

= 200 x 9.8 (1/2 +1/√3 x √3/2} =

1960N

Fmin = mgsine-fmax

= mgsin0 - μmgcose

mg{sin - μcose} =200 x 9.8(1/2 - 1/√3 x √3/2}

= 200 x 9.8 x 0

Answer 2

Answer:

hope this can help you. thank you