With static equillibrium TEE exam in a few days

  1. plz help with static equillibrium TEE exam in a few days

    hi ive got a physics question that is static equillibrium, if anyone can help me sort out the prob i would be very happy :).

    ok the prob i have with it is i get the cos or sin the angle wrong like i get everythign else right exept i get the cos or sin mixed up, if anyone can tell me how i determine which one to use it woul be very nice.

    FIND THE TENSION in cable

    [​IMG]
     
  2. jcsd
  3. arildno

    arildno 12,015
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    Before starting, could you tell me what would your solution strategy consist of?
    That is, how, in general terms, would you try to determine the tension?
     
  4. sure ok what i would do is

    take moments of point B

    sum (ACW) = sum (CW)

    ill just say let z be the cos or sin im not sure of how to work out

    2.25 T = (5x200x z 32 deg) + (2.5x54x z 32 deg)
     
  5. verty

    verty 1,934
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    I would draw a vector diagram with each force following the next. It will look a certain way because the forces are in equilibrium.
     
  6. the only bit i have trouble with is determining if i use sin or cos of the angle
     
  7. can someone please show me how to determine if i use sin or cos of the angle 32?
     
  8. arildno

    arildno 12,015
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    Always remember that you are to find the length of the relevant "lever arm".
    Now, for the weights, they represent forces in the vertical direction.
    What direction should therefore the "lever arm" in question have?
     
  9. we already know the length of the arm (5m). Yeah the weights represent forces vertically down one at the end of the arm (the mass) and the other half way through the arm (weight of the arm). but i still down understand how to work out the weather its cos or sin
     
  10. arildno

    arildno 12,015
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    No, the pole is NOT the "lever arm"!
    The lever arm is the component of the position vecor (i.e the pole) PERPENDICULAR to the force.

    So, what is the direction of the lever arm in this case?
    And what is its length?
     
  11. i think i know what you mean, the length is

    cos32 = x/5
    x = 5cos32 = 4.24 (2dp)

    and the direction is just 32 deg isnt it?
     
  12. arildno

    arildno 12,015
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    You are right in x being 5cos(32) when it comes to the hanging weight.
    Hence, the hanging weight's MOMENT about B is simply the product between the weight and the lever arm (that is, x).
    x has HORIZONTAL direction, by the way.

    Now, compute the other moments about B!
     
  13. so

    2.25T = (54x2.5cos32)+(200x5cos32)
    T= 422.70 N (2dp)

    so for moments questions when they are angular i need to find the length of the postion vector perpendicular to the force?
     
  14. arildno

    arildno 12,015
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    That is one way to do it, yes.

    You can also compute the product of force, the length of the position angle and the sine between them.

    The answers will be equal, since (the length of the position TIMES the sine of the angle between the force vector and the position vector) equal
    the length of the lever arm.
    To see this, remember that:
    [tex]\sin(90-x)=\cos(x),cos(90-x)=\sin(x)[/tex]
    where angles are measured in degrees.
     
  15. thanks for your help!
     
  16. 1. draw diagram
    2. sum forces in x and y direction
    3. sum torques around a chosen point, in this case i would choose the point to be at where the beam and the wall connect because you actually have a reaction force there, and its good to cancel them out if your chosen point is there

    sum of forces in x = Rx (horizontal reaction force of wall) - T = 0
    sum of forces in y = Ry - weight - mg (the gravational pull on pole) = 0

    sum of torques = - mg cos 32 (2.6m) - weight(cos 32 ) 5 m + 4.24 m (Tsin32) = 0

    if a force makes the pole go clockwise around the chosen Point, then that force is negative; if its counterclockwise then its positive in the torque equation

    all forces in the torque equation MUST BE PERPENDICULAR TO THE POLE; mg points straight down, but to get the mg that is perpendicular to the pole, u multiply mg by cos32!! its the same with Tension, if u look more closely, you know you have to get the Tension force that is perpendicular with the pole, so u'd have to multiply the normal Tension by sin32 to get the tension that is PERPENDICULAR to the pole

    and of course you multiply the force by the length from the chosen point, where mg is half the poles length, so its 2.5 m away, and the weight is 5 metres away. tension is 4.24 m away because 2.25 m / sin32 is 4.24 (trig)

    so now you only have one unknown, T, in your torque equation

    it becomes:

    -115N - 848N + 2.25 T = 0
    T = 428 N
     
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