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butterflyprincess

  • 2 years ago

QUICK QUESTION!!** PIC INCLUDED* Suppose that a box is being towed up an inclined plane as shown in the figure. Find the force w needed to make the component of the force parallel to the inclined plane equal to 2.5 lb. Give answer in component form.

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  1. redshift
    • 2 years ago
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    We will need that figure to know exactly what is being asked.

  2. butterflyprincess
    • 2 years ago
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    Here is a picture!

  3. butterflyprincess
    • 2 years ago
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    Does anyone know how to solve it D=

  4. TuringTest
    • 2 years ago
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    |dw:1352583686628:dw|

  5. TuringTest
    • 2 years ago
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    |dw:1352583791078:dw|

  6. TuringTest
    • 2 years ago
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    the component parallel to the plane is the magnitude of the force W time the cosine of the angle between the force and the plane, which is 30-15=15degrees

  7. butterflyprincess
    • 2 years ago
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    Wait but the angle given says 33 degrees?

  8. TuringTest
    • 2 years ago
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    sorry, good eye

  9. TuringTest
    • 2 years ago
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    |dw:1352584030549:dw|

  10. butterflyprincess
    • 2 years ago
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    ohokkk

  11. butterflyprincess
    • 2 years ago
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    but how would you get the component form..... like what exactly do i solve for? w?

  12. TuringTest
    • 2 years ago
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    yes

  13. butterflyprincess
    • 2 years ago
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    mhm, okay but how do you know what you have to minus 33 with 15?

  14. TuringTest
    • 2 years ago
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    we want the angle between the ramp and the force the ramp is at an angle of 15 deg to the horizontal the force is at an angle of 33 deg to the horizontal therefore the angle between the force and the ramp is 33-15deg

  15. butterflyprincess
    • 2 years ago
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    Oh okay cool, so how can you set it up to get W?cos 18 = 2.5lbs/w..?

  16. TuringTest
    • 2 years ago
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    start with cos(theta)=F/W where F is the component parallel to the plane we set theta=18 and F=2.5 now yuo can solve for W

  17. TuringTest
    • 2 years ago
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    cuz it's a right triangle...|dw:1352584629296:dw|

  18. TuringTest
    • 2 years ago
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    cos(theta)=adj/hyp=F/W

  19. butterflyprincess
    • 2 years ago
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    ohh Okay so W=2.63/

  20. TuringTest
    • 2 years ago
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    I dunno, let me find a calculator...

  21. butterflyprincess
    • 2 years ago
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    lol ok

  22. TuringTest
    • 2 years ago
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    yep, I get the same :)

  23. butterflyprincess
    • 2 years ago
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    oh okay but the answer says in the book <2.20,1.43> so how can i get there?

  24. TuringTest
    • 2 years ago
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    those must be the *components* of W, so it is W*cos(phi) W*sin(phi) where phi is the angle that W makes with the horizontal

  25. TuringTest
    • 2 years ago
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    |dw:1352585011969:dw|here my exaggerated drawing doesn't work so well :P but Wx and Wy are the x and y components respectively

  26. butterflyprincess
    • 2 years ago
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    what so youd get like (2.62 cos (18) , 2.62 sin (18)?

  27. butterflyprincess
    • 2 years ago
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    D:

  28. TuringTest
    • 2 years ago
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    18 is not the angle that W makes with the horizontal...

  29. butterflyprincess
    • 2 years ago
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    wait really then what would it be?O:

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