A friend and I were imagining a physics problem for fun: A 3kg rectangular cart has a pulley system on its front edge. 1kg weight suspended is down the front, and it's attached by a rope through a mass-less pulley to a 1kg mini-cart on top of the cart. There is no friction between the ground and the cart, or the cart and the mini cart, the cart and the front block, or on the pulley. http://img98.imageshack.us/img98/7416/cartproblem.png [Broken] The question is, how hard do you have to push the cart so the little carts don't roll off? In other words, with some level of F_p, the acceleration of the top mini-cart relative to the ground is equivalent to the acceleration of the big cart relative to the ground. I said the answer was simple: 5kg * g, or 49N. My argument: Part 1: The acceleration of the system must be g, or 9.8m/s/s. That way, the yellow front box is held up by the pull of the blue box backwards. Or in other words, if the cart had dice in its rearview mirror, they would make a 45deg angle when the cart on the windshield is no longer falling, and the cart on the roof is no longer slipping forwards. To make these forces equal, you'd have to put the pedal to the metal and travel at 9.8m/s/s (which is a 2.7s "zero to sixty" time if you want to imagine that). Part 2: The mass under acceleration is 5kg, the same as it would be if the mini-carts were bolted down. Part 3: F = m * a = (1+3+1)kg * g = 49N My friend (call him Bob) said the answer is 4kg * g/2, or 2kg * g, or 19.8N His argument: Part 1: The equations of motion for block 1 on top: T = 1kg*a For block 3 (small one on the side's) vertical direction: 1kg*g - T = 1kg*a -> 1kg*g - T + T = (1kg + 1kg)*a -> a = 1kg*g/ (1kg + 1kg) = g/2 Part 2: "Without friction there is no way for the Blue cart to "know" that the Red cart is being pushed. It is absolutely impossible. There is nothing about Blue's horizontal motion that has anything to do with Red. There is simply no coupling between the horizontal motion of Red and the horizontal motion of Yellow." Therefor he only uses the mass of the Red cart and the Yellow cart, which is being pushed by the normal force of "Red,Yellow." Part 3: F = m*a = (1kg + 3kg)*(g/2) = 2g = 19.6N I replied that (1) T = 1kg*a only works if "a" encompasses the relative acceleration of Blue over Red AND that of Red over the ground. And (2) the horizontal force between Blue and Red comes by virtue of the normal force on the pulley. Who is right?