1. The problem statement, all variables and given/known data A wedge of mass 4kg and angle 30 degrees is at rest on a smooth horizontal table. a mass of 1kg is placed on the smooth inclined face of the wedge and a horizontal force of 10N is applied to towards the vertical face of the wedge. Show that the acceleration of the wedge is (40 - root(3g) ) / 17 ~= 1.35m/s. Also find the reaction force between the wedge and the mass 2. Relevant equations 3. The attempt at a solution (from diagrams) Equation of wedge horizontally 4A = 10 - Rsin30 (1) Equation of mass horizontally A + acos30 = Rsin30 A = Rsin30 - acos30 (2) Equation of mass vertically asin30 = g - Rcos30 (3) Substitute (2) into (1) 4(Rsin30 - acos30) = 10 - Rsin30 5Rsin30 - 4acos30 = 10 5Rsin30 = 10 + 4acos30 5R* 1/2 = 4a * root(3)/2 + 10 5R/2 = 2 root(3)a + 10 5R = 4root(3)a + 20 R = ((4root(3)a) / 5) + 4 Now I have 'R' in terms of 'a', but I'm not sure where to go. My notes on this trail off...I seem to substitute R back into equation (1), along with equation (2) but that will just give me R in terms of a again.