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Forces, blocks, cords, and springs

  1. Sep 18, 2006 #1
    sorry, couldn't think of a better title

    anyways, i'm somewhat stuck on 2 questions i have in my homework assignment


    first one:


    Three blocks are supported using the cords and two pulleys. If they have weights of WA = WC= W, WB= kW, determine the angle (theta) for equilibrium.

    Given:

    k = 0.25

    [​IMG]








    alright, for the first one, i thought it would be easy and i thought i had a good plan

    Wa = Wc = W Wb = kW and k = 0.25

    i want equilibrium so, sum of the forces in x = 0, sum of forces in y = 0

    so then i break everything into components to add the x and y forces respectively

    in the y the two forces going up should equal the force going down

    Wsin(fi) + 0.25Wsin(theta) = - W

    W cancels and i end up with

    sin(fi) + 0.25sin(theta) = -1

    here is where my math skills fail me, instead of trying to write theta or fi in terms of the other one (i get some ugly things), i see what i can do with the other part


    forces in the x, only two of them, so:

    Wcos(fi) = -0.25Wcos(theta)

    once again, the W cancels, and i dared to solve for one angle in terms of the other

    cos(fi) = -0.25cos(theta)

    fi = arccos(-0.25)*theta

    fi = 104*theta

    theta = fi/104 <---theta will be really really tiny, that's ok, because block b has to fight block c in the x direction and block c has 4 times the mass

    anyways, this is where i pretty much got stuck,

    have i done something screwy, or is it just a lack of trig knowledge?




    and


    Determine the mass of each of the two cylinders if they cause a sag of distance d when suspended from the rings at A and B. Note that s=0 when the cylinders are removed.


    Given:

    d = 0.5m
    l1 = 1.5m
    l2 = 2m
    l3 = 1m
    k = 100N/m
    g = 9.81m/s^2


    [​IMG]





    for the 2nd problem

    i got stuck pretty early on

    i started drawing a FBD at point A, i had t1, t2, and m1

    then i started second-guessing my strategy and came to the conclusion that i had no real idea what i was going to do with that

    i trigged out some angles, wrote some equations of equilibrium, but...when i saw 5 unknowns, i backed down and decided i'd check if i am on the right path before taking on that beastly chore

    i'll just briefly outline where i was headed

    Fxa = t1cos48.59 - t2
    Fya = t1sin48.59 -m1g

    Fxb = t3cos48.59 - t2
    Fyb = t3sin48.59 - m2g

    i must admit, t2 confuses me in the FBD, i need to have it opposing t3 and t1, which means it opposes itself, hmm

    i also tried something like

    m1 + m2 = 2kd/g and i get the combined mass of m1 and m2 is 10.2 kg, but surely, it can't be that simple


    alright...that's where i am :)

    any help would be appreciated
     
  2. jcsd
  3. Sep 19, 2006 #2

    nazzard

    User Avatar
    Gold Member

    Hello mybsaccownt,

    be careful whenever you try to compare forces. As vectors they have both: magnitude and direction. Your approach to look at the components of the acting forces is promising. By comparing the magnitudes, the 2 equations you'll get in your first problem will be:

    [tex]\sin{\phi}+\frac{1}{4}\sin{\theta}=1 \,\,\,(1)\\[/tex] and
    [tex]\cos{\phi}=\frac{1}{4}\cos{\theta} \,\,\,(2)[/tex]

    The following step in your attempt was wrong:

    [tex]\arccos(x*y)\neq\arccos(x)*\arccos(y)[/tex]

    To solve the problem I'd use the following equation:

    [tex]\sin^2{x}+\cos^2{x}=1 \,\,\,(3)[/tex]

    Starting with rearranging and squaring (1):

    [tex]\sin^2{\phi}=(1-\frac{1}{4}\sin{\theta})^2 \,\,\,(1')[/tex]

    squaring (2) and using (3) to get:

    [tex]1-\sin^2{\phi}=\frac{1}{16}\cos^2{\theta} \,\,\,(2')[/tex]

    Using (1') in (2'),...

    Regards,

    nazzard
     
    Last edited: Sep 19, 2006
  4. Sep 20, 2006 #3

    andrevdh

    User Avatar
    Homework Helper

    For the first one:

    Using Newton's second law in component form we have that

    [tex]\Sigma F_Y = 0[/tex]

    so that your equation:

    Wsin(fi) + 0.25Wsin(theta) = - W

    Should actually be

    Wsin(fi) + 0.25Wsin(theta) - W = 0

    But I would rather use the fact that the forces form a closed triangle. Two of the legs are the same length and one is a quarter of the length of the other. So one get an isosceles triangle with the two legs of length 1 and the base of length a quarter. You just need to figure out where the angles should go in such a triangle.
     
  5. Sep 20, 2006 #4
    thanks a lot nazzard! :-)

    i got it, theta = arcsin(1/8)

    my trig skills need some work, hehe



    andrevdh, blocks b and c will lift block a

    that's why their vertical components will be equal to a, but opposite in direction, in order to achieve equilibrium

    so, equal but opposite, hence vertical components of b, c = -Wa

    and Wa is just W

    so

    sum of vertical components of b and c = - W


    also, i'm not sure i understand what you mean by the forces forming a closed triangle

    i'm just breaking the forces into components and applying the equations of equilibrium, it's a good method and it's simple, i'm not saying your method wouldn't work, but it seems like i should just stick to what i know

    i think the only trouble i had was with my trig skills (i didn't remember the identity that was needed to solve for theta)

    the TA mentioned it today in tutorial, but i still would have been stuck for a while without some help


    i appreciate your input, thanks for the help, guys
     
    Last edited: Sep 20, 2006
  6. Sep 21, 2006 #5

    andrevdh

    User Avatar
    Homework Helper

    Yes their magnitudes need to be the same in order to cancel. In Newton's equation direction is taken into account. That is from

    [tex]\Sigma F_y = 0[/tex]

    which is the same condition as your remark that the forces in the y-direction need to balance - their sum need to be zero. So from this we have that

    [tex]W_{BY} + W_{CY} - W_{AY} = 0[/tex]

    this is the scalar component form of N2 = 0 (taking up as positive). Which then comes to

    [tex]W_{BY} + W_{CY} = W_{AY}[/tex]
     
  7. Sep 21, 2006 #6
    oh...I see what you mean

    alright, thanks i got it
     
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