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Clarification on tension problem

  1. May 8, 2016 #1
    • Thread moved from the technical forums, so no HH Template is shown.
    I'm having trouble solving this problem. I did Google the solution with the steps, which I managed to actually get right in all aspects except for a small part in the trigonometry that throws me off.

    0332207068.png

    I drew my FBD's correctly, hopefully. I'm not sure how I can draw nice looking FBDs online, but here goes.

    b9c027fdfb.jpg

    In the solution it says that the acceleration of mass A is mg * sin (theta). Why is that so? I thought that we're trying to get gravity's force on the object with the formula: sin (theta) = (mA * g) / hypotenuse ----------->
    hypotenuse = (mA * g) / sin (theta). That way we'd get the force in the direction opposite to the force of tension, which is caused by mB * g?

    If I'm not wrong, isn't sin (theta = opp / hypotenuse? How did they get sin (theta) * mAg when that isn't even feasible in the trigonometry equation of sin(theta) = (mAg)/hypotenuse?
     
  2. jcsd
  3. May 8, 2016 #2

    BvU

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    Hi Xari, :welcome:

    It says the acceleration is along the inclined plane, the x-axis therefore. The equations shown are in terms of force, not acceleration.
    You are not wrong. Project mg onto the inclined plane and you'll see ##\theta## sitting there
    upload_2016-5-8_23-26-22.png
     
  4. May 8, 2016 #3
    Omg that makes me so angry. Trig is so slippery!
     
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