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Ac3

  • one year ago

In the figure below, a tin of antioxidants (m1 = 1.3 kg) on a frictionless inclined surface is connected to a tin of corned beef (m2 = 2.3 kg). The pulley is massless and frictionless. An upward force of magnitude F = 5.9 N acts on the corned beef tin, which has a downward acceleration of 5.4 m/s2. What are the tension in the connecting cord and angle β?

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  1. Ac3
    • one year ago
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  2. Ac3
    • one year ago
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    Irishboy any idea?

  3. IrishBoy123
    • one year ago
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    of course have you drawn any free body diagrams? do you have your own thoughts on this?

  4. Ac3
    • one year ago
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    yes

  5. Ac3
    • one year ago
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    |dw:1443919104377:dw|

  6. Ac3
    • one year ago
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    Finding the tension like this gives 4.22. However that's wrong and the reason I know it's wrong is because another one with very similar numbers has the Tension as 2.6

  7. Ac3
    • one year ago
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    oh nvm it's right lol

  8. Ac3
    • one year ago
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    Now how do I find the angle?

  9. IrishBoy123
    • one year ago
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    |dw:1443919249971:dw|

  10. Ac3
    • one year ago
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    |dw:1443919309239:dw|

  11. Ac3
    • one year ago
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    n is the normal force from the surface on the can

  12. Ac3
    • one year ago
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    any ideas on the angle?

  13. anonymous
    • one year ago
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    Explain to me how you find tension earlier?

  14. anonymous
    • one year ago
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    Because you need that for angle.

  15. Ac3
    • one year ago
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    |dw:1443919872349:dw|

  16. IrishBoy123
    • one year ago
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    T = 4.243N is good you need \(\beta\), right?

  17. Ac3
    • one year ago
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    yea irishboy

  18. Ac3
    • one year ago
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    tension is found through that equation

  19. IrishBoy123
    • one year ago
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    obviously so you have for the upper block \(T + mg \sin \beta = ma \) ?? do you see that

  20. Ac3
    • one year ago
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    no not exactly

  21. Ac3
    • one year ago
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    is that all in the y?

  22. IrishBoy123
    • one year ago
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    |dw:1443920229507:dw|

  23. IrishBoy123
    • one year ago
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    |dw:1443920344245:dw|

  24. Ac3
    • one year ago
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    OHHHHHHHHHHHHHHHHHHHHHHHHHHHHHHHHHHHHHHHHHHHHHHHHHHHHHHHHHHHHHHHHHHHHHHHHHHHHHHHHHHHHHHHHH

  25. Ac3
    • one year ago
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    that makes sense

  26. Ac3
    • one year ago
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    wow i can't believe I couldn't figure that out

  27. Ac3
    • one year ago
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    I even drew that exact picture out as well

  28. Ac3
    • one year ago
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    thanks Irishboy saving me again

  29. anonymous
    • one year ago
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    You mean for the upper block? \[mgsin \beta-T=ma\]

  30. IrishBoy123
    • one year ago
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    @Ac3 blows me away too, every time. good luck!

  31. Ac3
    • one year ago
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    no it is indeed +T not -T, that's because they're both going down @Shalante

  32. anonymous
    • one year ago
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    m2 is heavier.

  33. anonymous
    • one year ago
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    If its going down, then it would start with mg is first. But you got the answer right?

  34. Ac3
    • one year ago
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    @IrishBoy123 I use M1 for the mass correct?

  35. Ac3
    • one year ago
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    no not yet

  36. anonymous
    • one year ago
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    You need to derive a lot.

  37. Ac3
    • one year ago
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    I got the angle wrong. Must have messed up the calculation. I got 28 for the positive angle and -61.9 for the negative angle

  38. IrishBoy123
    • one year ago
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    @Ac3 run with @Shalante see where you get to. tag me tomorrow otherwise, gtg. good luck!

  39. Ac3
    • one year ago
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    what did you get for the angle @Shalante

  40. anonymous
    • one year ago
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    Tell me how you got tension first

  41. Ac3
    • one year ago
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    T+F-mg=ma(y)

  42. Ac3
    • one year ago
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    plug in all values and solve for T. That's how you get tension.

  43. Ac3
    • one year ago
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    I got the angle guys

  44. Ac3
    • one year ago
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    I guess plugging in all the values at once in the calculator gives you the right answer. lol

  45. Ac3
    • one year ago
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    Thanks a lot @IrishBoy123.

  46. anonymous
    • one year ago
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    Let me guess the angle.

  47. anonymous
    • one year ago
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    Nvm. Someone tagged me.

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