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1018

  • one year ago

physics problem (question's in the comments)

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  1. 1018
    • one year ago
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  2. Michele_Laino
    • one year ago
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    please see my answer to a similar question http://openstudy.com/users/michele_laino#/updates/55e295d1e4b03567e110576d

  3. 1018
    • one year ago
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    oh hi ms michele! thank you for your help again. haha, ok i will :)

  4. 1018
    • one year ago
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    @Michele_Laino hi are you still online? im having trouble with the problem. if you're not too busy maybe you can help me? thanks

  5. Michele_Laino
    • one year ago
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    please "ms" stands for?

  6. 1018
    • one year ago
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    oh, that's just short for miss. haha. sorry

  7. anonymous
    • one year ago
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    Let's look at block m2 first forces acting on it are |dw:1441042822826:dw|

  8. anonymous
    • one year ago
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    for equilibrium resultant force on block m2=0 this implies for vertical components \[N=m_{2}g+F.\sin(34)\] We know that friction is \[f=\mu.N\] \[\therefore f=\mu(m_{2}g+F.\sin(34))\] For horizontal components we have \[T+f=F.\cos(34)\] From this you can find T \[T=F.\cos(34)-f=F.\cos(34)-\mu(m_{2}g+F.\sin(34))\]

  9. anonymous
    • one year ago
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    But then why is the block 1 given for?Maybe I'm missing out something :p

  10. 1018
    • one year ago
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    i was just about to ask if that is the working equation already. haha

  11. anonymous
    • one year ago
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    Do you have an answer so we can at least verify by plugging in values?

  12. 1018
    • one year ago
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    no i dont sorry. actually the problem changed already, i mean same problem but with different values because i tried to answer it and it was wrong haha so im just trying to see the analysis of it

  13. anonymous
    • one year ago
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    oh

  14. 1018
    • one year ago
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    but do you want to see the new? although it's literally the same problem only with different values

  15. 1018
    • one year ago
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    or i could just type the new values. but really the same question.

  16. 1018
    • one year ago
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    i have an idea but i dont know. haha. there 2 tensions present in the string, so maybe it's tension (right) - tension (left) lol

  17. anonymous
    • one year ago
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    One thing that I think is not given is whether the applied force is keeping the system in equilibrium or causing acceleration, which is a key deciding factor Let's assume it causes acceleration a, cuz that seems to be the case here, Therefore for block 1 we have Force acting towards the right on it is the tension T and towards left have friction equal to normal reaction \[T-\mu.m_{1}g=m_{1}a\] Thus to find T we need to eliminate a \[T=m_{1}(\mu.g+a)\] Now consider the system of box 1 and 2 combined We have \[a=\frac{F_{res}}{m_{1}+m_{2}}\] Where F is the resultant force on the system of 2 blocks I think F is \[F_{res}=T-\mu.m_{1}g+F.\cos(34)-\mu(m_{2}g+F.\sin(34))-T\] The first 2 terms correspond to block 1 and other terms to block 2 cancelling out the Tension we get \[F_{res}=F.\cos(34)-\mu(m_{1}g+m_{2}g+F.\sin(34))\] \[\implies a=\frac{F.\cos(34)-\mu(m_{1}g+m_{2}g+F.\sin(34))}{m_{1}+m_{2}}\] plugging this into equation for Tension we have \[T=m_{1}(\mu.g+\frac{F.\cos(34)-\mu(m_{1}g+m_{2}g+F.\sin(34))}{m_{1}+m_{2}})\] I'm not sure about it, but maybe worth a shot

  18. 1018
    • one year ago
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    yeah that's what i thought the first time, but i just assumed that it was not moving since there's a string involved.

  19. anonymous
    • one year ago
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    Can you try the expression I got for Tension?see if it works

  20. 1018
    • one year ago
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    ill try this equation. thanks! i'll let you know if its correct. haha

  21. 1018
    • one year ago
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    the dot is multiplication right? just to be clear, with the mew(?) and g

  22. anonymous
    • one year ago
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    yeep, I had to do it otherwise it was writing it as mug lol

  23. anonymous
    • one year ago
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    \[mug\]

  24. 1018
    • one year ago
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    on calcu its negative, is that ok?

  25. 1018
    • one year ago
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    oh hold on i think i made a mistake ill try again

  26. 1018
    • one year ago
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    hey! it's correct! thank you so much!

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