consider the friction between A,B. F=20g*.5=98.2 N max, so the forces P are less than this, so no slipping occurs.
Therefore, force P acts on the total mass (30kg). I don't understand, if you claim your acceleration of B is correct.
Block A has a mass of 20kg and is placed on top of cart B which has a mass of 100 kg. Blosk A has a pulley attatched to it with a force of P acting towards the right. Cart B is on rollers. The coefficient of kinetic and static friction are both the same value of 0.5. Find the acceleration of each part for (a) P= 60N and (b) P= 40N
Ok so here are my equations I came up with. For block A sum of the forces in the x = P-F = m(a1) {where F is the friction force and a1 is acceleration of block A} and sum of the forces in the y= N1-W1 = 0 {where N1 is the normal force between the blocks and W1 is the weight of block A}
For block B sum of the forces in the x = F = m(a2) {where F is the friction force and a2 is the accleration of block B} and sum of the forces in the y = N2-N1-W2 = 0 {where N2 is the normal force between the ground and the cart and W2 is the weight of block B}
I solved part (a) for block B correctly (0.981m/s^2) but none of the other ones come out right
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