1. The problem statement, all variables and given/known data A 3.0-kg block sits on top of a 5.0-kg block which is on a horizontal surface. The 5.0-kg block is pulled to the right with a force F. The coefficient of static friction between all surfaces is 0.60 and the kinetic coefficient is 0.36. a)What is the minimum value of F needed to move the two blocks? b)If the force is 10% greater than your answer for (a), what is the acceleration of each block? 2. Relevant equations Sum of forces = ma 3. The attempt at a solution Forces acting on the bottom box are normal force up the y axis, normal force from the top box and mg going down the y axis. On the x axis, the F we need to find is positive and the friction of 1 on 2, the kinetic friction on the ground and the tension of the rope is going the opposite way. For the box on top, there is normal force up y axis, mg down, friction of 2 on 1 going the positive x direction and tension going the other way. I know for the top box that n=mg so n =3(9.8) regarding the y axis for the x axis: F=ma Friction of 2 on 1 = u*n = .6(3*9.8) = 17.64 so 17.64 - T = -3a Then for the bottom box I know that : n2-mg-n1= 0 n2 = 5(9.8)+3(9.8) so n2=78.4 for the x direction: F-T-kinetic friction-Friction1on2 = ma so F-T-28.224-17.64 = 5a And now I'm stuck!