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Homework Help: Static equilibrium problems

  1. Oct 19, 2006 #1
    hey every1... i'm stuck on this problem and unless i get it and a few others right im gonna fail my engineering statics course :( im desperate! I study aerospace engineering but because i've been ill i missed a few classes and i really dont have time to learn the theory to answer these... thank you for your time.

    Here's the diagram for the 1st Q:

    http://img383.imageshack.us/my.php?image=q1cm6.jpg

    The Q that goes with it is:

    Determine the bending moment at a distance x from the roller support on the left hand side, using the principles of virtual work.

    The values are: F = 2kn, d=2.200m and x=0.600m
     
  2. jcsd
  3. Oct 19, 2006 #2
    2nd Question:

    http://img407.imageshack.us/my.php?image=q2un0.jpg

    Each of the two uniform hinged bars have a mass m and length L. Both bars are connected by hinge S and supported as shown. The structure is loaded by a mass of 4m, applied at the hinge S. Gravitational acceleration is g = 9.81m/s2.

    m= 15kg
    L=5.500m
    angle theta= 130 degrees

    Calculate the torque Ma (in Nm) required for equilibrium
     
  4. Oct 19, 2006 #3

    radou

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    Working on it. :rolleyes:
     
  5. Oct 19, 2006 #4
    your saving my academic life! thankyou!
     
  6. Oct 19, 2006 #5

    radou

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    http://usera.imagecave.com/polkijuhzu322/system1.bmp.jpg"

    Ok, as said, the system is replaced with a mechanism, where a hinge is put in the place where you have to find the moment, and there is a couple of moments M added to that same place. Now, you have to construct an initial relative rotation between the two diskd connected by the hinge, so do it as is done in the displacement sketch. Now all you have to do is apply the principle of virtual work to get the moment M: [tex]F\cdot d_{F}+M(A_{1}+A_{2}) = 0[/tex], where A1 and A2 are the angles of rotation of the two disks.
     
    Last edited by a moderator: Apr 22, 2017
  7. Oct 19, 2006 #6

    radou

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    Regarding the second assignment, replace the system with a system where the force G = mg acts in the center of each bar, and where G' = 4mg acts on the hinge. Try this: [tex]\sum M_{(A)}=0 \Rightarrow R_{B} = \cdots ; \sum F_{yi}=0 \Rightarrow R_{A} = \cdots ; \sum M_{(B)}=0 \Rightarrow M_{A} = \cdots[/tex] I hope this works.
     
  8. Oct 19, 2006 #7
    thanks for the effort, but i dont fully 'get it' :(

    so what i have so far is:

    F . df + M(A1 + A2)=0 which i think is

    2 X df + M(40.36 + 8.53) = 0

    ok and using trigonometry i got that df is 0.606711 m, so

    i get that M = 0.024819 kNm... is that right?
     
    Last edited: Oct 19, 2006
  9. Oct 19, 2006 #8

    radou

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    M should equal -0.54 kNm, I checked on it. You obviously didn't do the trigonometry right. [tex]\frac{0.6}{4}=\frac{d_{F}}{4-2.2}[/tex], [tex]A_{1} = \tan A_{1} = \frac{3.4}{4}[/tex], and [tex]A_{2} = \tan A_{2} = \frac{0.6}{4}[/tex]. I forgot to point this out - A1 and A2 are differential rotations, so in the theory of small displacements you can use the identity [tex]\alpha \approx \tan(\alpha)[/tex].
     
  10. Oct 19, 2006 #9
    double post
     
    Last edited: Oct 19, 2006
  11. Oct 19, 2006 #10
    yea i made a mistake with the trigonometry, am gonna try the 2nd Q now but i dont have much confidence that i'll get the right answer. pleaseee stick around to correct me im certain i'll make a mistake, thx again.

    ok so i drew a free body diagram of the system, with mg acting downwards on the centre of each member, and 4mg acting downwards on hinge A. Then i calculated the vertical reaction force at B from moment equilibrium about point A, and got that Bv = 73.582kN

    then summing forces in the y direction gives that Av = 809.318kN and summing forces in the x direction gives that Ax = 0. But now im not sure what to do! lol.... i know all the forces but am not sure how to calculate the torque at Ma... i mean, i think i can do it without using virtual work... by summing all the moments about point A, and then the reaction moment would be negative of that right.... but have no idea how to do it with virtual work, the idea of displacements confuses me.

    do u have maybe msn or yahoo radou?
     
    Last edited: Oct 19, 2006
  12. Oct 19, 2006 #11

    radou

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    P.S. If you're certain you'll make a mistake, then you will make a mistake. Conslusion: don't be certain you'll make a mistake. :wink:
     
  13. Oct 19, 2006 #12
    is there anybody, out there? (pink floyd)
     
  14. Oct 19, 2006 #13

    radou

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    Unfortunately, no msn neither yahoo. :biggrin:

    Btw, it doesn't say anywhere that you have to use virtual work in assignment 2. Your way of thinking about 2 seems correct.
     
  15. Oct 19, 2006 #14
    hey are u still out there radou? could u possibily help me on another couple questions? :!!)
     
  16. Oct 19, 2006 #15

    radou

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    As long as I'm online, I'll help. :smile:
     
  17. Oct 19, 2006 #16
    thankyou! btw, i know this will sound rude as some1 can sound... but i have 8 more Q's and if i dont get them all right i really am gonna fail this course :( could i maybe just have the answers? believe me when i say that, i will learn the theory behind all of this stuff but i reallly am in need of nothing less than a miracle just now...
     
  18. Oct 19, 2006 #17

    radou

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    Depends on how big the questions are. Btw, answers won't help you if you don't understand anything. But nevermind, let's give it a try. :rolleyes:
     
  19. Oct 19, 2006 #18
    ok this is actually Q11

    http://img120.imageshack.us/my.php?image=qmc9.jpg

    a= 0.200m
    b=2m
    theta= 35 degrees

    Determind the torque M (in kNm) on the activating lever of the dump truck necessary to balance the load at the given dump angle theta, g = 9.81m/s2.
     
    Last edited: Oct 19, 2006
  20. Oct 19, 2006 #19
  21. Oct 19, 2006 #20

    radou

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    These are all very similar (and typical) statics assignments, so, if you solved the one at the beginning of the thread (the one without virtual work), then you should know how to solve these one. As said before, use the fact that a system is in equilibrium if the sum of all moments of the forces acting on it (including couples, as in Q3) with respect to any point must vanish, as must the sum of all horizontal and vertical forces (i.e. compoments of forces). Pick the 'moment equation point' wisely - it's a way to elliminate unknowns.
     
    Last edited: Oct 19, 2006
  22. Oct 19, 2006 #21
    ok here is Q5:

    http://img138.imageshack.us/my.php?image=q5ayq9.jpg

    q = 2kN/m
    a= 0.350

    Calculate the moment reaction in A (in kNm) counter clock-wise positive.

    PS i really didn't mange to work out the 1st one either... please please please! do it for me, the thing is... if i dont get these Q's all right then i can't sit the exam... which is a week away. And i would definetely be revising and doing all these things on my own in the exam, i just need the opportunity to sit the exam...
     
    Last edited: Oct 19, 2006
  23. Oct 19, 2006 #22
  24. Oct 19, 2006 #23

    radou

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    Q5.

    You can separate the two disks and look only at the part from the hinge S to the end. Use the sum of moments in the point of the hinge to obtain the reaction in B (which is vertical, of course). Then you can use the sum of all vertical forces to obtain the vertical force in the hinge C (the horizontal force equals zero, since there is no horizontal load applied to the disk). Further on, you can look at the part A-S now separately; the force acting on it is the vertical force from the hinge C (but in the opposite direction of that one which you got while looking at the right disk). Use the sum of moments at point A to obtain the value of the reaction moment in A.
     
  25. Oct 19, 2006 #24
  26. Oct 19, 2006 #25
    q8

    http://img172.imageshack.us/my.php?image=q8qp8.jpg

    F = 25kn
    q = 30kN
    a = 5m

    calculate the reaction force at A in kn, upwards positive.

    PS. if u wont give me the answers then could u possibily make sure i have the right answers if i post what i think is right at the end of this thread? :)
     
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