1. The problem statement, all variables and given/known data A wedge, all of whose faces are smooth, resting on a smooth horizontal table.Its faces make an angle A and B with the table, where tan A = 4/3 and tan b =3/4 .Particles of each of mass m are placed on these faces.If the mass of the wedge is 2m, show that the wedge will not move 2. Relevant equations F = MA 3. The attempt at a solution I tried to make equations for both particles seperately first particle tan A = 4/3 forces and acceleration of particle resolved in to components parallel = mg(4/5) perpendicular = mg(3/5) parallel = a(3/5) perpendicular = a(4/5) a = acceleration of wedge f= acceleration of particle relative to wedge R1=normal force equations (4/5)mg = m(f - (3a/5) )................. simplify 4g = 5f - 3a (I didn't use this equation) 3/5)mg - R1 = 4ma/5 ......................... simplify 3mg - 5R1 = 4ma Forces and acceleration of the wedge along horizontal R1(sinA)........... (4/5)R1 = 2ma simplify R1 = (10/4)ma putting value for R1 into second equation 3mg - (50/4)ma = 4ma simplify a = (2/11)g Second particle tan B = 3/4 forces and acceleration of particle resolved in to components parallel = (3/5)mg perpendicular = (4/5)mg parallel = (4/5)b perpendicular = (3/5)b b = acceleration of wedge e = acceleration of particle relative to wedge R2 =normal force equations (3/5)mg = m(e - 4b/5 ) (I didnt use this equation) (4/5)mg - R2 = (3/5)b simplify 4mg - 5R2 = 3mb Acceleration of wedge along the horizontal R2(sinB) R2(3/5) = 2mb ......... R2= (10/3)mb put R2 into second equation 4mg - (50/3)mb = 3mb .....multiply by 3.... 12mg = 50mb + 9mb 59b = 12g b = (12/59)g a-b = (2/11)g - (12/59)g is not zero I had no problem with these questions when there was only one particle on the wedge.Now that there is two particles im not sure how to solve them. Should I have tried to make equations including both the particles instead of trying to solve both seperately? Any help would be appreciated.