1) Two particles, A and B, of masses m and 2m, respectively,are placed on a line of greatest slope, l, of a rough inclined plane which makes an angle of 30 degrees with the horizontal. The coefficient of friction between A and the plane is (√3)/6 and the coefficient of frictuon between B and the plane is (√3)/3. The particles are at rest with B higher up l than A and are connected by a light inextensible string which is taut. A force P is applied to B. a) Show that the least magnitude of P for which the two particles move upwards along l is (11√3)mg/8 and give, in this case, the direction in which P acts. B) Find the least magnitude of P for which the particles do not slip downwards along l. I think it's all in the diagram except the angle of inclination of 30. The angle between P and the plane is θ. 1: Resolving parallel to the slope at B gives: Pcosθ - FB - T - 2mgsin30 = 2ma. 2: Resolving perpendicular to the plane at B gives: RB = Psinθ + 2mg cos30. 3: Resolving parallel to the slope at A gives: T - FA - mg sin30 = ma. 4: Resolving perpendicular to the plane at A gives: RA = mg cos30. FB = (√3)/3*RB = (√3)/3*(Psinθ + 2mg cos30). FA = (√3)/6*RA = mg/4. From eq. 3: T = ma + FA + mg/2 = ma + 3mg/4. Now I'm a bit confused - I have 8 unknowns - a, P, θ, FA, FB, RA, RB, T - and only 6 equations. For part b) I can put a=0 and F=uR, but in part a), a =/ 0. I don't know what to do next. Any help please? 2) The points A and B are 180 metres apart and lie on horizontal ground. A missile is launched from A at speed 100m/s and at an acute angle of elevation to the line AB of arcsin (3/5). A time T seconds later, an anti-missile missile is launched from B, at speed 200m/s and at an acute angle of elevation to the line BA of arcsin (4/5). The motion of both missiles takes place in the vertical plane containing A and B, and the missiles collide. Taking g=10m/s and ignoring air resistance, find T. I'm really not too sure on how to solve this, so I've just calculated the heights of each and said that these must be equal. Vertically at A: a=-g, u=100*(3/5), s=yA Using s=ut + 1/2at^2: yA = 100*(3/5)t -5t^2. Vertically at B: a=-g, u=200*(4/5), s=yB Using s=ut + 1/2at^2: yB = 200*(4/5)(t-T) - 5(t-T)^2 100*(3/5)t -5t^2 = 200*(4/5)(t+T) - 5(t+T)^2 => 60t - 5t^2 = 160(t+T) - 5(t+T)^2 => 60t - 5t^2 = 160t + 160T - 5t^2 -10tT -5T^2 => 5T^2 + (10t-160)T -100t = 0. This equation doesn't seem to get me anywhere. Can someone explain what I should be trying to do? Thanks.