Student Doctor Network Forums > Pre-Medical Forums > MCAT Discussions > TPR Physics Question PDA View Full Version : TPR Physics Question puravida8512-01-2009, 03:46 PMIn the TPR Hyperlearning Science workbook, there is a question that states: A man with a mass of 100kg sits 5m from the center of the sea saw. Two children, each with a mass of 20kg, are seated on the other side of the sea saw. One child sits 10m from the center. How far from the center should the other child sit to balance the sea saw? A) 5 B) 10 C) 15 D) 20 The proposed answer is C = 15m. Their rational is to use the torque equation and set up an equation such as: TorqueCCW = TorqueCW 100kg * 5m = 20kg * 10m + 20kg * x(m) x = 15 How could they set the man at a positive 5 and also put a child in the positive 10 if they are seated "on the other side of the sea saw". The child would therefore be on the same side of him. What I got from the question stem was that if he is 5m from the center, and if the children are on the opposite side, it would mean that one of them is either: 5m + 10m = 15m if we consider the sea saw a continual integer count. or He is at -5m and the child is at +10m because they are on opposite sides of the center which we could mark zero. My visual aide: ___100kg__^______20kg_____20kg -----(-5m)-0------(+10m)-----?m-- If we use 15 m for the child's location as I have proposed, than we can set up the equation as this: 100kg * 5m = 20kg * 15m + 20kg * x (m) 500 kg*m = 300 kg*m + 20kg x (m) 200kg*m = 20kg x 200kg*m / 20kg = x x = 10 m I've been doing 5 hours of Physics now, so my head could be a bit off. But can someone set me straight? Thanks Steelersfan200912-01-2009, 04:06 PMHi, I believe you have to pick a spot as the pivot point and the torque left of that equal the right or (clockwise balancing counter clockwise) You can also solve this via center of mass, but torques makes it easier. This problem tells you that they are certain distance away from the center. So it is easy to pick this center as your pivot which you can set a distance of zero.. The distance of this first child from the pivot is 10 meters, (again this is his distance from the pivot at 0 not his distance from the man, your stated 15 meters) The distance the man from the center is 5 meters, you can think that he's -5 but I mean they are not going to flat out tell you that He's just 5 meters from the center and since his children are on the opposite side of the center they would not be on the same side as him. Just draw the diagram according to the question and you won't go wrong, but read the wordings carefully. I hope that helped and not just confused you Compass12-01-2009, 04:22 PMIn the TPR Hyperlearning Science workbook, there is a question that states: A man with a mass of 100kg sits 5m from the center of the sea saw. Two children, each with a mass of 20kg, are seated on the other side of the sea saw. One child sits 10m from the center. How far from the center should the other child sit to balance the sea saw? A) 5 B) 10 C) 15 D) 20 The proposed answer is C = 15m. Their rational is to use the torque equation and set up an equation such as: TorqueCCW = TorqueCW 100kg * 5m = 20kg * 10m + 20kg * x(m) x = 15 How could they set the man at a positive 5 and also put a child in the positive 10 if they are seated "on the other side of the sea saw". The child would therefore be on the same side of him. What I got from the question stem was that if he is 5m from the center, and if the children are on the opposite side, it would mean that one of them is either Two things: The equation says man is 5m from the center, and then kids are on the OTHER side. It is important to make this distinction. Else you'll never solve the problem. Here is where you got confused, I think. There is no negative or positive distance here for a reason. You are comparing opposing forces. Both sides should be positive. 100 x 5 <--- left side of center right side of center ---> 20 x 10 + 20 x 15 I think you may be confused with this equation: 0 = 100kg * -5m + 20kg * 10m + 20kg * x(m) The total torsional force in this equation should CANCEL OUT (both sides are equal and zero). The original equation shows that the forces should BALANCE OUT (both sides are equal, neither side is necessarily non-zero). Think of it this way. Numbers on the left are torsional for counter-clockwise rotation, numbers on the right are torsional for clockwise rotation. In order for the see-saw to balance, torsion CW and CCW must be equal. mcat4512-01-2009, 04:57 PMfor the stated question, you are correct in bringing up the issue that there should be a sign difference between the two sides. the easiest way to solve the question is to realize that your net torque should be 0 as there will be no rotational motion when the seesaw is balanced. if you think of the problem this way and also arbitrarily set one side of the seesaw as positive and one side of the seesaw as negative, then at some point during the solution process you will end up with the same set up as the answer from the princeton review hyperlearning book. essentially, they just omitted the basic initial step of summing all possible torques on the body in question and setting that equal to 0. hope this helps.... puravida8512-01-2009, 09:26 PMfor the stated question, you are correct in bringing up the issue that there should be a sign difference between the two sides. the easiest way to solve the question is to realize that your net torque should be 0 as there will be no rotational motion when the seesaw is balanced. if you think of the problem this way and also arbitrarily set one side of the seesaw as positive and one side of the seesaw as negative, then at some point during the solution process you will end up with the same set up as the answer from the princeton review hyperlearning book. essentially, they just omitted the basic initial step of summing all possible torques on the body in question and setting that equal to 0. hope this helps.... It's not a matter of the simple algebra that I didn't understand. I know that setting CW = CCW requires understanding that they are in equilibrium which would make the sum of them equal to 0 to begin with. Anyhow, I recently heard on 'audio osmosis' a huge tip that helped me with this problem: If we have a system in equilibrium, setting forces equal to each other like all the forces upward = forces downward allows one to ignore sine conventions because as Compass said: the forces are opposing forces. Ohh.. the fun of physics! Thanks everyone.