Question

Find the contact force on the block having mass 1 kg. (Assume wedge to be fixed​

Answer 1

Answer:

Angle of inclined plane=180-(90+53)=37°

Friction in an inclined plane= µmgcos(Theta)

=0.5(g)Cos37=0.5g(4/5)

Therefore frictional force=0.4g

And normal force=mgcos(theta) in incline plane

Normal force=0.8g