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Jusaquikie

  • 2 years ago

Body A weighs 102 N, and body B weighs 32 N. The coefficients of friction between A and the incline are μs = 0.56 and μk = 0.25. Angle θ is 40°. Let the positive direction of an x axis be up the incline. (b) Find the acceleration of A if A is moving up the incline.

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  1. Algebraic!
    • 2 years ago
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    post the figure or a diagram.

  2. Jusaquikie
    • 2 years ago
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  3. Jusaquikie
    • 2 years ago
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    i have T-102Nsin(40)-19.5=m1a =-6.09

  4. Jusaquikie
    • 2 years ago
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    but i know this is wrong, i think i'm supposed to use the total weight of the system after watching some youtube videos

  5. Algebraic!
    • 2 years ago
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    block on ramp: T -102N*cos(40) - uk*102N*sin(40) = (102N/g)*a suspended block: 32N-T =(32N/g)*a solve for T T=32N -(32N/g)*a plug into first equation and solve for a.

  6. Jusaquikie
    • 2 years ago
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    i think i keep confusing my cos and sin

  7. Algebraic!
    • 2 years ago
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    hold up. I think I may have confused sin and cos. lol.

  8. Algebraic!
    • 2 years ago
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    yeah I switched them, sorry: block on ramp: T -102N*sin(40) - uk*102N*cos(40) = (102N/g)*a

  9. Jusaquikie
    • 2 years ago
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    this isn't different than what i am getting then my numbers are 24N-65.6N-19.5N divided by 10.4

  10. Jusaquikie
    • 2 years ago
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    for x1

  11. Algebraic!
    • 2 years ago
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    |dw:1348867218201:dw|

  12. Algebraic!
    • 2 years ago
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    how'd you find T?

  13. Jusaquikie
    • 2 years ago
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    i thought we drew that angle like this|dw:1348867283133:dw| i got a = 2.45 solving the frictionless problem then solved fro t in x2 as t=32n-(3.27*2.45)

  14. Algebraic!
    • 2 years ago
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    naw, you want components 'along' or 'into' the ramp.

  15. Algebraic!
    • 2 years ago
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    in your sketch mg is a component of some other vector... I don't know what it is

  16. Algebraic!
    • 2 years ago
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    I got -3.8 m/s^2 (g=9.8) that's the answer your book has, no doubt. so try my way:)

  17. Jusaquikie
    • 2 years ago
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    first i did it with no friction o see what would happen and determine my coordnate system. in x2 i solved T=32N-m2a, i put that in x1 and got 32N-m1a-102Nsin(40)=(102N/g)a that turned to 32N-102Nsin(40) = a(10.4+3.37) then through algebra i got a = 2.45

  18. Jusaquikie
    • 2 years ago
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    it's not that i'm not trying your way i think were drawing it different but we have the same 102Nsin(40)

  19. Jusaquikie
    • 2 years ago
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    when i redrew my FBD for the (b) Find the acceleration of A if A is moving up the incline. nothing changed in x2 so i assumed it would be the same. maybe i have to redo the problem and see if i can get the answer you got.

  20. Jusaquikie
    • 2 years ago
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    Thanks Algebraic, i was being lazy and thought the tension would be the same. your right i t was -3.88 m/s^2

  21. Algebraic!
    • 2 years ago
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    cool!

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