A block of mass m attached to a fixed position O on a smooth inclined wedge of mass M oscillates with amplitude And linear frequency f the wedge is located on a rough horizontal surface if the angle of the wedge is 60 degree then the force of friction acting on the wedge is given by ( coefficient of static friction = u
Answers
The force of friction acting on wedge is 1/2mω2 A sin ( ωt + Ф)
Mass of the block = m (Given)
Angle of wedge = 60° (Given)
Force exerted by the block along an Incline plane = F = ma
= mω2x
= mω² A sin ( ωt + Ф)
Horizontal surface of the wedge = F cos 60°
= f = 1/2mω2 A sin ( ωt + Ф)
The force should be equal to the frictional force.
Therefore, the force of friction acting on wedge is 1/2mω2 A sin ( ωt + Ф)
The force of friction acting on wedge is 1/2mω2 A sin ( ωt + Ф)
Mass of the block = m (Given)
Angle of wedge = 60° (Given)
Force exerted by the block along an Incline plane = F = ma
= mω2x
= mω² A sin ( ωt + Ф)
Horizontal surface of the wedge = F cos 60°
= f = 1/2mω2 A sin ( ωt + Ф)
The force should be equal to the frictional force.
Therefore, the force of friction acting on wedge is 1/2mω2 A sin ( ωt + Ф