Question

A body that weighs 50lbs is placed in contact with an inclined plane surface. The coefficient of friction between the surfaces in contact is 0.25. If the angle between the inclined surface and a horizontal reference plane is 28 degrees, what would be the angle of repose for the body and the material used for the surface?

Answer 1

Normal force, N=m*g*cos(theta)

frictional force, f=u*m*g*cos(theta)

and

force along horizontal line, Fx=m*g*sin(theta)

here,

Fx=f

m*g*sin(theta)=u**m*g*cos(theta)

==>

tan(theta)=u

tan(theta)=0.25

===> theta=14.04 degrees

angle of repose, theta=14.04 degrees