2007-10-31

Today's nonsense
- Slang physics ("convo")

Bryan and Dakota went mountain climbing, they tied each other with ropes, and one slipped and fell, and so one person is hanging from the rope, and now you have to get the tension, it's "just another typical problem," and that's how you should look at it. It's the same problem all over again. Need to include free body diagrams.

Here's the story: this is example six. Bryan, tied by a long rope to Dakota, slips off an icy edge. Before setting his climbing ax, Dakota slides along the frictionless ice. What is the acceleration of each and the tension in the rope? Dakota will be m_{_1} and a_{_1}.

Bryan has no i and Dakota has no j.

Sum of F_{_x} = T_{_1} = m_{_1}a_{_1}
Sum of F_{_y} = T - m_{_2}g = m_{_2}a_{_2}

The physics problems become more difficult when you try to rationalize what's going on, instead of just following straight the laws of forces and balancing the forces. Overthinking. Just realize. Well. What actually makes a difference? If both of you are attached, then both must accelerate at the same rate. Unless the rope was an elastic band. And then that's just like a hypothetical related rate with a ladder that's on fire etc.

So: a_{_1} = -a_{_-2} = a (instead of a_{_1} or a_{_2}-- so Dakota has a positive a and Bryan has a negative a).
The tensions: those tensions are the exact same. T_{_1} = T_{_2} = T because it's all the same rope.

m_{_1}a = m_{_2}(-a) + m_{_2}g -- just solving for T in both cases. T_{_1} = m_{_1}a etc. (be careful)
Our goal is to solve for "a." So you want to move the (-a) to the other side. And then factor out the a.
m_{_1}a + m_{_2}a = m_{_2}g
a = m_{_2}g / (m_{_1} + m_{_2})

While working in space, you push on a box of mass m_{_1} with force F. The box is in direct contact with a second box of mass m_{_2}