1. The problem statement, all variables and given/known data The question is as follows: A man of mass 100kg is placed onto platform 1. The maximum weight that he can lift using this machine is twice his own weight (200kg*g). The mechanical advantage of the machine is constant. If the mass of the object is 100kg, 1) what is the force upward felt on the engineer and mass respecitvely and 2) What is their accelerations? 2. Relevant equations Fnet=ma MA = Fout/Fin W=F*d 3. The attempt at a solution So I think I'm lacking a fundamental understanding of what Mechanical Advantage is and means. I know that the man has a downward force of 100*g N and the Normal Force is (as always) determined by his net accleration/the net force (since it's not always the exact opposite of his own weight- right?). Lets say Man = m1 Mass = m2 Fn = Normal Force m1g - Fn1 = -m1a1 m2g - Fn2 (mass) = m2a2 So because of the mechanical advantage applied here, if the man weighs 100g N then the output force exerted on the mass upwards is 2*100g N. So what confuses me here is that I would have though that the normal force would then be 2*100*g Newtons for the mass, but it isnt. Is it because the system is moving? and in assuming the normal force on the mass was simply twice the weight of the man was i making the mistake of assuming normal force is anything other than a factor of the net force? If normal force is critically a factor of net acceleration/force, then we would have, as stated above: Fn1- m1g = m1a1 Fn2 - m2g = m2a2 But that is not enough, as we also need a relationship between acceleration, which brings me to my second point of lacking, the notion of mechanical advantage and what it means. So I'm thinking that mechanical advantage does mean that the force out is double, but it also implies that work is conserved. So if work is conseved then: Work man = Work mass Fman*h= 2Fman * (1/2 h) Is the work done by the man equal and opposite to that done by the mass? I assume it is becuse it is similar to a pulley system so they are connected so it should be (this is all intuition though id love some clarification on this). So the man falls twice the distance as the box in the same amount of time so it's acceleration is twice that of the box. Substituting in: Fn1- m1g = m1a1 Fn2- m2g = m2a2 = - m2*(1/2 a1) Further plugging in: Since a1 = (Fn1-m1g)/m1 Fn2- m2*g = -m2 *1/2* (Fn1-m1g)/(m1)) 2*Fn2 - 2m2*g = - m2 *((Fn1-m1g)/m1) 2*Fn2 = 2m2*g - m2((Fn1-m1g)/m1) 2*Fn2 = m2( 2g - Fn1/m1 + g) 2*Fn2 = m2(3g - Fn1/m1) m1=m2=100kg 2*Fn2= 100(3g - Fn1/100) 2*Fn2 = 300g -Fn1 2Fn2 + Fn1= 300g So this is where i get stuck, and the book i'm using says that Fnormal(mass) = 2*Fnormal(man) due to the mechanical advantage. But I thought mechanical advantage was in regards to force applied vs force out (so the weight of the man vs the normal force applied to the mass?). If you do plug accordingly in you get the right answer: 2*(2Fnormal(man)) + Fnormal(man) = 300g 5Fnormal(man) = 300g Fnormal(man) = 60g = 600 N Fnormal(mass = 1200N WOW that was long. So thanks for bearing with me. thats the correct answer, but what confuses me is that how can that ^ be the normal force when the normal force for the mass when its supposed to be double the input force? Which in this case would be 100kg*g (the weight of the man)? The force up is not the force transmitted, i thought the force transmitted was the weight of the man unto the normal force of the mass? Thanks for reading this far and helping out guys.