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← 05-47 Forces and Trig

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Showing Revision 1 created 07/09/2012 by Amara Bot.

  1. Let's make things interesting. Let's introduce some trigonometry into the equation.
  2. So here's Robin and you can tell by her determined face that she really wants to
  3. push this cart through this brick wall and it seems like she is not having very much luck.
  4. In this example, something is different from last time.
  5. Robin isn't pushing exactly to the right or exactly to the left.
  6. She is pushing at some angle and let's call that angle here α.
  7. I'm sure you can already see the triangles forming in your head. Let's take a look at what's going on.
  8. Over here I'm going to draw a diagram for the cart, the force diagram.
  9. Of course as always we have the carts weight playing downwards
  10. and noticed that I didn't even tell you what m was, you don't even care.
  11. We have seen normal force pointing upwards. We know because the car isn't falling to the ground.
  12. We have Robin who is pushing, well looks like she is pushing down into the right.
  13. I'm actually going to put that force over here and let's label it FR for Robin,
  14. and we can't forget that the wall is pushing to the left.
  15. Of course if the wall weren't there the cart would be free to roll but since it is
  16. and it's stopping the motion, the wall itself must be providing a force.
  17. Let's call that FW for wall. Now when I look at this force diagram something jumps out to me.
  18. I've got these forces which are both vertical.
  19. The wall force which is exactly to the left and then the force of Robin pushing
  20. which is causing me some pain.
  21. It's not to the right. It's not straight down. It's some combination of right and down.
  22. Wait a second though, I know how to handle this.
  23. This is just me breaking down the vector into it's components.
  24. Let's make a right triangle with this would be the hypotenuse.
  25. Well here's a right triangle and now we see we get our angle α back again and let's label these sides.
  26. Well this is the horizontal component of Robin pushing and this is the vertical component.
  27. I'm going to label them FRx and FRy.
  28. One thing that's important to remember is that by writing FR
  29. as these two together, we can now get rid of FR.
  30. This is the exact same trick we did last unit
  31. when we had some initial velocity and we have some angle alpha.
  32. We broke that velocity into x and y components.
  33. You are doing exact same thing but with forces and just like with velocity
  34. once we have expressed it in terms of components x and y, we didn't care about this guy anymore.
  35. Likewise, we're now using the x and y component of Robin's pushing
  36. and we're not going to think about this too much anymore.
  37. But we still have to answer the question, "What are these values
  38. for the x and y components of Robin's pushing?"
  39. Let me give you some numbers so that this is less abstract.
  40. Let's say that the force Robin is pushing with is a 100 N total that's FR
  41. and this angle she's pushing at is 30°.
  42. Can you use your knowledge of trig to tell me what is FRx and what is FRy?
  43. Enter your answers here and here to one decimal place.
  44. I'll be very impress if you can get this right on your first try.