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brinethery

  • 2 years ago

Work problem. Please see attached. I'm having trouble with part (b)

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  1. brinethery
    • 2 years ago
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  2. brinethery
    • 2 years ago
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    I struggled with this type of problem when I was in physics last quarter :-(

  3. Outkast3r09
    • 2 years ago
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    did you find the acceleration... since they're roped together you assume that the acceleration of each is the same

  4. Outkast3r09
    • 2 years ago
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    if ou have acceleration you'll have two unknowns that can be related to each other

  5. brinethery
    • 2 years ago
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    Yes, I tried finding it. I got a=4.70 m/s^2

  6. brinethery
    • 2 years ago
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    So for A, I did sum of the forces in x (x-axis being parallel to the inclined plane) and then for B, I did sum of the forces in y. I made sure to make the direction of motion positive. For A, there's a +T in the equation and B, there's a -T in the equation. So I added both equations so that T would cancel out and I could solve for a.

  7. Outkast3r09
    • 2 years ago
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    |dw:1359083523070:dw|

  8. Outkast3r09
    • 2 years ago
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    |dw:1359083623616:dw|

  9. Outkast3r09
    • 2 years ago
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    |dw:1359083742313:dw|

  10. brinethery
    • 2 years ago
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    I just made my x-axis parallel to the plane. So my equation was m1(a) = (-mew_k)(m1)(g)(cos(40)) +T And then for block B, I did m2(a) = (m2)(g) - T And then I added both equations. a(m1+m2) = (-mew_k)(m1)(g)(cos(40)) + (m2)(g) a = (-mew_k)(m1)(g)(cos(40)) + (m2)(g)/(m1+m2)

  11. Outkast3r09
    • 2 years ago
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    \[sin(40)=\frac{W_x}{W}\] \[W_x=mgsin(40)\] \[W_y=mgcos(40)\] \[\sum F_y=N-W\] \[N=W\] \[\sum F_{x1}=m_1a_{1x}=T-W_x-f_k=T-mgsin(40)-(\mu_s)mgcos(40)\] \

  12. Outkast3r09
    • 2 years ago
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    where is your weight x component?

  13. brinethery
    • 2 years ago
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    ha, maybe that's the problem!

  14. brinethery
    • 2 years ago
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    that must've been where I goofed :-p. Just silly mistakes I guess.

  15. Outkast3r09
    • 2 years ago
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    \[m_1a_{1x}=T-m_1gsin(40)-\mu_s(m_1gcos(40))\]

  16. brinethery
    • 2 years ago
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    I didn't bother to label the acceleration since it will be the same. Would that be correct?

  17. Outkast3r09
    • 2 years ago
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    \[m_2a_{2y}=T-m_2g \] \[a_{2y}=-a_{1x}\]

  18. Outkast3r09
    • 2 years ago
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    is the pulley frictionless?

  19. brinethery
    • 2 years ago
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    yes it is

  20. Outkast3r09
    • 2 years ago
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    \[T=-m_2a_{1x}-m_2g=-m_2(a_{1x}+g)\] \[T=m_1a_{1x}+m_1gsin(40)+\mu_km_1gcos(40)\]

  21. brinethery
    • 2 years ago
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    Now that I added in the x-component of weight for Block A, my acceleration is 1.754 m/s^2

  22. Outkast3r09
    • 2 years ago
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    \[-m_2a_{1x}-m_2g=m_1a_{1x}+m_1gsin(40)+\mu_km_1gcos(40)\]

  23. Outkast3r09
    • 2 years ago
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    \[-m_2a_{1x}-m_1a_{1x}=m_2g+m_1gsin(40)+\mu_km_1gcos(40)\] \[a_{1x}(-m_2-m_1)=m_2g+m_1gsin(40)+\mu_km_1gcos(40)\] \[a_{1x}=

  24. Outkast3r09
    • 2 years ago
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    i think i dropped a sign somewhere let me get a paper and pen

  25. brinethery
    • 2 years ago
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    What was your acceleration? m1(a) = (-mew_k)(m1)(g)(cos(40)) -(m1)gsin(40) +T m2(a) = m2(a) = (m2)(g) - T a(m1+m2) = (-mew_k)(m1)(g)(cos(40)) -(m1)gsin(40) + (m2)(g) a = ((-mew_k)(m1)(g)(cos(40)) -(m1)gsin(40) + (m2)(g))/(m1+m2)

  26. brinethery
    • 2 years ago
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    I have a pretty good idea on the setup of the equations, so you can just tell me if I completely missed something. And also, what your acceleration is.

  27. Outkast3r09
    • 2 years ago
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    how'd you define your axis for the y part?

  28. brinethery
    • 2 years ago
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    my axis for the y-part is regular. Vertical for y, horizontal for x.

  29. brinethery
    • 2 years ago
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    I mean for block B, that is.

  30. Outkast3r09
    • 2 years ago
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    isn't downwards negative for the mg?

  31. brinethery
    • 2 years ago
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    When I took physics, my instructor told us to make the positive direction for each object the same as the direction of motion.

  32. brinethery
    • 2 years ago
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    so for block A, up and to the right would be positive. And for block B, down would be positive.

  33. brinethery
    • 2 years ago
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    so you're accounting for the system as a whole.

  34. Outkast3r09
    • 2 years ago
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    i got -1.67

  35. Outkast3r09
    • 2 years ago
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    that's for B it could be rounding error though

  36. Outkast3r09
    • 2 years ago
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    so for mine \[a_{2x}=-1.67\] \[a_{1x}=-a_{2x}=-(-1.67)=1.67\]

  37. Outkast3r09
    • 2 years ago
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    if you didn't round use your answer... min was rounded.... a lot . my graphing calc is out in the car and it's freezing lol. Funny thing, i just got back from physics 2

  38. brinethery
    • 2 years ago
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    lol

  39. brinethery
    • 2 years ago
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    We're working out of Hibbeler's book for dynamics and the chapter on work is confusing. I mean, I was already confused with these types of problems in physics I but now I just feel even more lost. This book doesn't explain things very well. You really have to read and re-read stuff in there.

  40. Outkast3r09
    • 2 years ago
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    so anyways now you have \[S=\frac{1}{2}(1.754)t^2+vt+S_0\] and \[V=at+v_0=1.754t+0\]

  41. Outkast3r09
    • 2 years ago
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    so you can just sub in =]. Never heard of hibbeler. however this book from about 10 years ago im using is confusing. It uses greek lettering for everything and my teacher uses easy things like \[R=\frac{l}{a}\]

  42. brinethery
    • 2 years ago
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    my friend sent me her stuff, so I'll try it out and tell you if it worked.

  43. brinethery
    • 2 years ago
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    Thank you for taking the time to help me.

  44. Outkast3r09
    • 2 years ago
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    no problem if you need anymore help let me know

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