The Law Posted December 19, 2009 Report Share Posted December 19, 2009 Hi Everyone, I'm a bit stuck on a problem. First, though, here's a problem that I did solve properly... and I'm not sure why it won't work for the second situation I'll present. First one: man in elevator holding box, standing on scale. The man has mass 90kg, box has mass 10kg. The elevator accelerates and decelerates at 1 m/s^2. What is the reading on the scale during a trip from the 5th to 11th floor? A) 900N 1000N C) 1090N D) 1100N So I chose scale as my system. It has the weight of man and box pushing down on it, and the normal force pushing up. (M + m)g = N Then, the elevator accelerates upwards, so the forces upward are stronger than the ones going down so I added "ma" to the weaker side: ma + Mg + mg = N (100)(1) + 90g + 10g = N 1100 = N And that's the answer. Why won't it work here: The box below has a specific gravity less than 1 and is held under water by a string as shown (basically, you have a big container filled with water. there is a string tied to the bottom of the container and it's holding the box as it tries to float back up). If the entire container is accelerated upward, tension in the string will: A) decrease and Fb decrease decrease and Fb increase C) increase and Fb decrease D) increase and Fb increase So I said, forces acting on the box are: Upward forces: Fb Downward: tension and mg (it's weight) Fb = T + mg It accelerates upwards, so upward forces are greater and thus: Fb = T + mg + ma Since we added more to the right side, tension has to go down for the equation to stay the same. Ah, I see what I'm doing wrong here. This is assuming that Fb stays constant, but evidently both change.... Even if I think about it from a common sense way, I got the wrong answer. I thought, well if the box is going upwards, then fine, you can say Fb goes up too. But then I thought that Tension would decrease since it's acting in the opposite direction as the acceleration... Bah this q is stupid. Can someone explain this to me better? The correct answer is D. Link to comment Share on other sites More sharing options...
astrogirl Posted December 19, 2009 Report Share Posted December 19, 2009 Going with the common sense way rather than the mathematical way, I would explain it like this. The box is going upwards, so Fb has to increase. But in order for the box to not fly out of the water bucket, the downward forces have to balance the increased upward forces. The only way to do that is for T to also increase. So Fb and T both increase. Does that help? Link to comment Share on other sites More sharing options...
The Law Posted December 19, 2009 Author Report Share Posted December 19, 2009 Going with the common sense way rather than the mathematical way, I would explain it like this. The box is going upwards, so Fb has to increase. But in order for the box to not fly out of the water bucket, the downward forces have to balance the increased upward forces. The only way to do that is for T to also increase. So Fb and T both increase. Does that help? Yeah that helps. Thanks. Link to comment Share on other sites More sharing options...
The Ace of Spades Posted December 19, 2009 Report Share Posted December 19, 2009 Sarcastic mister know it all. Link to comment Share on other sites More sharing options...
The Law Posted December 19, 2009 Author Report Share Posted December 19, 2009 haha, who me? I was sincere! that actually did help! Link to comment Share on other sites More sharing options...
The Ace of Spades Posted December 20, 2009 Report Share Posted December 20, 2009 Haha, I was finishing off your quote! Link to comment Share on other sites More sharing options...
The Law Posted December 20, 2009 Author Report Share Posted December 20, 2009 Yeah, that helps. Thanks. JKJK Link to comment Share on other sites More sharing options...
wizy32 Posted December 22, 2009 Report Share Posted December 22, 2009 for these situations you must always, always draw a free body diagram , the block is in static equilibrium in the container, however the system has a constant upward acceleration (hence a constant force upwards) the net force on the block tied to the string must be zero(in the system hence Fnet= MA+bouyant force(density(water) *(displaced)volume*gravity)-mg -tension=0 hence Tension+mg=Bouyant foce +MA , now if the container has zero acceleration the MA term is zero , the bouyant force will never change neither does the mass of the block, hence if the container is accelerated upwards the only thing that changes is the tension, (as now the MA term will be positive) Link to comment Share on other sites More sharing options...
The Law Posted December 25, 2009 Author Report Share Posted December 25, 2009 wizy32, Thanks for the post. Can you please explain more about why you chose MA to be on the side of buyant force? That's the step that got me in this problem. Link to comment Share on other sites More sharing options...
wizy32 Posted December 25, 2009 Report Share Posted December 25, 2009 MA is on the side of bouyant force as the acceleration vector is pointing up as well as the bouyant force, Link to comment Share on other sites More sharing options...
Leon Posted December 26, 2009 Report Share Posted December 26, 2009 Oh God, this stuff is on the MCAT? NOooooooooooooooooooooooooooooooooooooooooooooooooooooooooo (add infinite number of zeros) Link to comment Share on other sites More sharing options...
The Law Posted December 26, 2009 Author Report Share Posted December 26, 2009 my sandwich? MY SANDWICH?!!??!?!?!?!?! Link to comment Share on other sites More sharing options...
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