1. im stuck on question 8. the equation i got for the wedge is .6R -.5S=ma r stands for relative force of the particle s stands for relative force of the wedge . i got the equation 4mg-R=5m(.6a) for the particle where a stands for acceleration of the wedge. does anyone know where to go from here. are my equations wrong? i can post a picture of my diagram if anyone needs it

2. The forces acting on the wedge are mg, friction, S the normal reaction between the wedge and the surface, and R the normal reaction between the wedge and the particle.
The forces actin on the particle are 5mg and R the normal reaction between the wedge and the particle
I have called Î± the angle with tan(Î±) = 3/4
If we call the acceleration of the wedge a then the equations of motion for the wedge are
1) vertically S = mg + R cos(Î±)
2) horizontally ma = R sin(Î±) - 0.5S

The equations of motion for the particle are
3) Horizontally 5m(bcos(Î±) - a) = Rsin(Î±) since bcos(Î±) - a is the total acceleration in this direction
4) Vertically 5mbsin(Î±) = 5mg - Rcos(Î±)

Eliminating S from 1) and 2) gives R = 5ma +2.5mg

Eliminating b from 3) and 4) gives 4mg - 3ma = R

Equating the two above expressions for R gives a = 3g/16

Substituting in 4 then gives b = 3g/4

can equation 3 also be 5m(b-.8a) = 3mg
and also i dont understand equation 4 5mbsin(Î±) = 5mg - Rcos(Î±) should it not be 4mg-R=5m(.6a) are you not supposed to look at the forces horizontal and vertical to the plane

4. Originally Posted by markosheehan
can equation 3 also be 5m(b-.8a) = 3mg
and also i dont understand equation 4 5mbsin(Î±) = 5mg - Rcos(Î±) should it not be 4mg-R=5m(.6a) are you not supposed to look at the forces horizontal and vertical to the plane
To obtain my equations 3) and 4) I looked at the motion vertically and horizontally but you can certainly also look at motion along the plane and perpendicular to the plane. I believe your equations are correct for motion resolved the second way and these should also lead to the same solutions.