Hello, i have been asking alot of question lately, i had 60 questions to do for hw, and i'm kinda confused on some of them... A 0.5kg block is placed on top of a 1.0kg block, the coefficient of static friction between the two blocks is 0.35. The cofficient of kinetic friction between the lower block and the lever table is 0.20. What is the maxinimum horizontal force that can be applied to the lower block without the upper block slipping? Some Thoughts: First i found out the kinetic frictional force,between the lower block and the table. (i don't know if you should add the two masses together, since block 2 is on top of block 1, so does it change block 1's normal force? anyways i didn't add 0.5kg.) Block 1: mg(us) = (0.5kg)(9.8N/kg)(0.35) = 1.715N Block 2: mg(uk) = (1.0kg)(9.8N/kg)(0.20) = 1.96N when Force applied = Frictional, no accleration. So, F applied must first overcome block 2's frictional force, then you have a net force, but that force must not be greater than 1.715N So i added the number forces: 1.96N + 1.715N = 3.675N. The force applied must be less than this... but i was wrong, our book says the correct answer is 8.1N... i wonder how they got that. Thanks.