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Math Help - Forces in Equilibrium (Vectors) II

  1. #1
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    Forces in Equilibrium (Vectors) II

    Here is another question I am suck on, it seems simple but I cant get it.



    Maybe i am drawing the free body diagram wrong. Does the tension in the rope still need to be include since its the same on both sides? (since there is one rope). Thanks again!
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    Rhymes with Orange Chris L T521's Avatar
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    Quote Originally Posted by Oblivionwarrior View Post
    Here is another question I am suck on, it seems simple but I cant get it.



    Maybe i am drawing the free body diagram wrong. Does the tension in the rope still need to be include since its the same on both sides? (since there is one rope). Thanks again!
    The tension in the rope must be the same at the end of both boxes.

    Let the blue box have a mass of m_1 and let the orange box have a mass of m_2

    Doing a FB analysis on the blue box [the x direction is what matters] , we see that

    \sum F_x=0:

    -m_1g\cos\alpha+T=0\implies T=m_1g\cos\alpha

    Doing a FB analysis on the orange box [the x direction is what matters] , we see that

    \sum F_x=0:

    m_2g\cos(30^{\circ})-T=0\implies T=m_2g\cos(30)

    Therefore, since the tensions are the same, we see that m_2g\cos(30)=m_1g\cos(66)

    Can you take it from here?

    --Chris
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    Ooh that makes sense now, I was adding the normal force into the mix before. It was just making it complicated. So you don't need the normal force?
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    Sorry, shouldn't the force due to gravity along the incline be mg\sin \theta? Otherwise it'd be the rest of the analysis follows: m_{1} \sin 66^{\circ} = m_{2} \sin 30^{\circ}

    unless I've thoroughly confused myself...
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    Rhymes with Orange Chris L T521's Avatar
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    Quote Originally Posted by o_O View Post
    Sorry, shouldn't the force due to gravity along the incline be mg\sin \theta? Otherwise it'd be the rest of the analysis follows: m_{1} \sin 66^{\circ} = m_{2} \sin 30^{\circ}

    unless I've thoroughly confused myself...
    I'm the one that confused myself!

    --Chris
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    Rhymes with Orange Chris L T521's Avatar
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    Quote Originally Posted by Oblivionwarrior View Post
    Ooh that makes sense now, I was adding the normal force into the mix before. It was just making it complicated. So you don't need the normal force?
    No. It just tells us that N_1=m_1g and N_2=m_2g. It wouldn't tell us much, because the masses aren't known. Since their is tension in the string and they are inclined at different angles, it would have been best to look at the forces along the incline, rather than perpendicular to the incline.

    --Chris
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    Are you sure its ? I got the same thing but with Cos.
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  8. #8
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    sine, not cosine.
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    Would the ratio be 1.83:1?
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    Rhymes with Orange Chris L T521's Avatar
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    Quote Originally Posted by Oblivionwarrior View Post
    Would the ratio be 1.83:1?
    Yes!

    1.83:1 would signify that m_2 is 1.83 times larger than m_1.

    --Chris
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    Quote Originally Posted by Chris L T521 View Post
    Yes!

    1.83:1 would signify that m_2 is 1.83 times larger than m_1.

    --Chris
    I put 1.83:1 as the answer and it was wrong, did I screw up the ratio? Should It be 0.55:1 or 1:0.55 or 1:1.83
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    an "online" response, no doubt. the ratio for \frac{M}{m} is correct ... how many decimal places do you need?

    \frac{\sin(66)}{\sin(30)} \approx 1.82709091529
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    Quote Originally Posted by skeeter View Post
    an "online" response, no doubt. the ratio for \frac{M}{m} is correct ... how many decimal places do you need?

    \frac{\sin(66)}{\sin(30)} \approx 1.82709091529
    I got it to the right amount of decimals, but after reading the questions again " Determine the ratio of the mass of the right box to the mass of the left box" would the ratio be reversed? 0.55:1?
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  14. #14
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    left side ... right side

    mg\sin(66) = Mg\sin(30)

    \frac{\sin(66)}{\sin(30)} = \frac{M}{m}

    what else can I say?
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  15. #15
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    Quote Originally Posted by skeeter View Post
    left side ... right side

    mg\sin(66) = Mg\sin(30)

    \frac{\sin(66)}{\sin(30)} = \frac{M}{m}

    what else can I say?
    Hmmm, maybe I should just enter just the 1.827 instead of 1.827:1?
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