Welcome to PHY 121 Blog Help. Here's how it works. For each homework question and lab report we will make a post, this will probably contain a few tips on what the problems are about and how to solve them. If you are stuck on something then instead of emailing us directly you should post a comment in reply to the relevant post. We will try to guide you through tough points and help you understand the problems and the concepts behind them.
Tuesday, October 12, 2010
Ch8 3 #4
6 comments:
Anonymous
said...
Can't seem to get this right?
Tnet = 0...
Tarm = Tball Tarm = Fr + mgr Tball = mgr
where the radius for Tarm is the Radius(total) - Radius(Elbow Width)...and radius for Tball is simply Radius(total) - Radius(Arm length)?
Never mind it's actually a really easy question. Look at the diagram on pg 42 of the notes for Chp. 8 and pg 43. Notice that the arm is at a 90 degree right angle. That makes the problem considerably simpler (no need for Tx or Fx). Using the diagram on pg 42 figure out which values you can plug into where for the equation Ty*d-m1gl/2-m2gl=0 The tension in the y direction is equivalent to the force you need to hold the ball parallel.
This can be a bit tricky, especially because of the difference between distance and length. The distance is the short one, .04. The length is the long one, .35. Just use the equation Tyd=(m1gL/2) + m2gL. The answer will end up being under 100 N, so check it out.
6 comments:
Can't seem to get this right?
Tnet = 0...
Tarm = Tball
Tarm = Fr + mgr
Tball = mgr
where the radius for Tarm is the Radius(total) - Radius(Elbow Width)...and radius for Tball is simply Radius(total) - Radius(Arm length)?
There's always that one really hard question on every homework. Can anyone help us out?
Never mind it's actually a really easy question. Look at the diagram on pg 42 of the notes for Chp. 8 and pg 43.
Notice that the arm is at a 90 degree right angle. That makes the problem considerably simpler (no need for Tx or Fx).
Using the diagram on pg 42 figure out which values you can plug into where for the equation
Ty*d-m1gl/2-m2gl=0
The tension in the y direction is equivalent to the force you need to hold the ball parallel.
i been working on this problem forever. can somebody help me out this one.
Thanks for the help above ^.
A more direct equation to use would be on the top of page 44 in the notes:
T=(m1g*L/2+m2g*L)/d
This can be a bit tricky, especially because of the difference between distance and length. The distance is the short one, .04. The length is the long one, .35. Just use the equation Tyd=(m1gL/2) + m2gL. The answer will end up being under 100 N, so check it out.
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