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Static Equilbrium

  1. Apr 28, 2008 #1
    1. The problem statement, all variables and given/known data
    A window cleaner of mass 95 kg places a 22kg ladder against a frictionless wall at a angle 65 degrees with the horizontal. The ladder is 10 m long and rests on a wet cloor with a coefficient of static friction equal to .40. What is the max lenght that the window cleaner can climb before the ladder slips?




    2. Relevant equations
    sum F(x)=F(fr)-F(w)=0
    sum F(y)=F(n)-mg=0
    sum torque=F(w) * l (sin theta) - mg * 1/2 cos theta=0





    3. The attempt at a solution

    F(w)=mg/2tan theta
    =(117)(9.8)/2tan65
    =1146.6/4.28
    =267.89 N

    F(n)=mg
    =1146.6 N

    maxF(fr)=mu* F(n)=mu*mg
    F(w)=mg/2tan theta=mu*mg
    =tan theta=1/2*mu
    =tan theta=1.25
    =theta=51.34 degrees

    Now, I am lost. Can someone please help? THANKS!!
     
  2. jcsd
  3. Apr 28, 2008 #2

    alphysicist

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    Hi Lma12684,

    I don't think this torque equation is correct; the center of mass is halfway up the ladder, but you are trying to find where the window cleaner is. So his position is the unknown you are looking for.
     
  4. Apr 28, 2008 #3
    Was my attempt at solving anywhere close to what I am looking for? Thanks.
     
  5. Apr 28, 2008 #4

    alphysicist

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    It's close enough so that it would be right if the window cleaner were at the middle of the ladder.

    Start with formula you alread have:

    sum torque=F(w) * l (sin theta) - mg * 1/2 cos theta=0

    So this is okay for the torque from the wall force and from the weight of the ladder, if you interpret the mass m as just the ladder, and also change the (1/2) to an (L/2) (since you have kept the L in the first term). Now you need another term on the left hand side for the window cleaner. Assume he is a distance x up the ladder. What would his torque from his weight be on the ladder?
     
  6. Apr 29, 2008 #5
    I used the following formula: sum torque=F(w) * l (sin theta) - mg * L/2 cos theta=0

    m=22kg

    THis is what I got:
    torque=267.89 * 10 * sin 65 - (22)(9.8) * (10/2) (cos 65)
    =4674.19 N

    What am I missing????
     
  7. Apr 29, 2008 #6

    alphysicist

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    The torque formula needs to have three terms, because there are five forces acting on the ladder overall. Two of those are at the pivot point (where the ladder touches the ground), and so their torques will be zero.

    The remaining three forces will all produce torques, and so there will be three terms in your torque equation. You have the torque from the wall:

    [tex]
    \tau_{\rm wall} = F_{\rm wall} L \sin\theta
    [/tex]

    and you have the torque from the weight of the ladder:

    [tex]
    \tau_{\rm ladder} = -(22)(9.8) (L/2) \cos(\theta)
    [/tex]

    Now you need one more term: the torque from the weight of the window cleaner. He is a distance x up the ladder, and he has a mass of 95 kg. So what would the torque from his weight be?

    Once you have that, you add all three torques together (two of them will be negative numbers), and their total sum must be zero.
     
  8. Apr 29, 2008 #7
    Ok, here is what I found:

    T(wall)=2427.90
    T(ladder)=-455.58
    T(painter)=-931x

    0=2427.90-455.58-931x
    x=2.104 meters up ladder

    How does it look?
     
  9. Apr 29, 2008 #8

    alphysicist

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    I don't think that's right. Let me summarize what you have already done:

    In the vertical direction:

    [tex]
    F_N = m_1 g +m_2 g
    [/tex]

    where [itex]F_N[/itex] is the normal force from the ground. You know the right hand side, so you found the ground's normal force to be [itex]F_N = 1146.6[/itex] in newtons.

    Then in the horizontal direction you had:

    [tex]
    F_w = F_{fr}
    [/tex]

    Since it is on the verge of slippling, you can find the force of friction (using the normal force from the ground), and so you can use this to find the force from the wall. What do you get for [itex]F_w[/itex]?

    The third equation has the torque from the wall, the ladder and the painter.

    Now in your torque equation, you calculated three numbers, but I think only the number for the torque from the ladder is correct. (I think you were using the wrong force F_w in the torque from the wall.) Why don't you post the actual numbers you used in calculating those three torques?
     
  10. Apr 29, 2008 #9
    Here we go:

    T(wall)=Fw* L *sin theta
    =267.89*10*sin 65
    =2427.90

    T(ladder)=-22 (9.8) (10/2) (cos 65)
    =-455.58

    T (painter)= mg (h)
    =(95)(-9.8)(h)
    =-931h

    0=2427.90-455.58-931h
    x=2.104 meters

    Where did I go wrong?
     
  11. Apr 29, 2008 #10

    alphysicist

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    The force from the wall is not 267.89 N. You can find the force from the wall using the horizontal force equation, which says in this problem that the force from the wall equals the force of friction.

    Remember how you got the torque from the ladder: it was the force (mg), times the distance from the pivot point (10/2), times an appropriate trigonometric factor. Then the sign of the torque comes from the direction of the torque (clockwise or counterclockwise).

    Sometimes the trig factor is actually equal to one (because it is sin(90 degrees) or cos(0 degrees), but it isn't here; here the torque from the painter needs a trig factor just like the torque from the weight of the ladder.



    If deciding on the trig function itself is causing problems, there are three main ways of writing down the magnitude of the torque:

    1. [itex]\tau=r F \sin\theta[/itex]: here [itex]\theta[/itex] is the angle between the directions of the moment arm and the force. To use this for the weight of the painter, the force is vertically downwards, and the r is 65 degrees above the horizontal. What is the angle between these two directions?

    2. [itex]\tau =r F_{\perp}[/itex]: Multiply the distance from the pivot point to where the force acts, times the component of F that is perpendicular to r. So here you would need to find the component of F that is perpendicular to the ladder.

    3.[itex]\tau=r_{\perp} F[/itex]: Multiply the total force times the component of r that is perpendicular to that force. I think this is the easiest for ladder problems. Since your weight force is vertical, [itex]r_{\perp}[/itex] is just the horizontal component of r for the torque from that weight.

    They all give the same answer, of course, but sometimes one way is more natural for a certain problem. (There are other ways of writing torque but I think these are probably the most useful for introductory physics problems.)
     
  12. Apr 29, 2008 #11
    Ok here we go again:

    I really appreciate your help.

    F(fr)=F(w)
    F(fr)=mu F(n)
    =.4(117)(9.8)
    =458.64 N

    Thus,
    T(wall)=F(w)(L)(sin 65)
    =(458.64)(10)(sin65)
    =4156.69

    Now, I am still lost on T(painter).

    1. t=rFsintheta: here is the angle between the directions of the moment arm and the force. To use this for the weight of the painter, the force is vertically downwards, and the r is 65 degrees above the horizontal. What is the angle between these two directions?

    *This is confusing to me? I would appreciate any help you can offer. Thank you.
     
  13. Apr 30, 2008 #12

    alphysicist

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    Lma12684,

    Here are the three methods.

    1: Use [tex]\tau=r F \sin\theta[/tex] where [itex]\theta[/itex] is the angle between the directions of r and F. In this image, the angle you want to find is the angle in red:

    http://img91.imageshack.us/my.php?image=method1ju9.jpg

    2: Use [tex]\tau=r F_{\rm perp}[/tex] where [itex]F_{\rm perp}[/itex] is the component of F that is perpendicular to the moment arm, it is red in the image:

    http://img91.imageshack.us/my.php?image=method2sg7.jpg

    By extending the force line down to the ground, you can make a right triangle and find the angle needed to get than component.

    3: Use [tex]\tau=r_{\rm perp} F[/tex] where [itex]r_{\rm perp}[/itex] is the component of r that is perpendicular to F; it is red in the image:

    http://img91.imageshack.us/my.php?image=method3pt0.jpg


    All three will give you the same numerical result; but sometimes a particular problem is much easier to solve using one method than the other, so I think it's important to be able to use all three.

    Once you find the torque from the painter, it will be negative just like the torque from the weight of the ladder, since it is a clockwise torque about the pivot.
     
  14. Apr 30, 2008 #13
    Thanks for all your help, but I couldn't figure the last part out, and the assignment was due today. I REALLY APPRECIATE ALL YOU HAVE DONE. Thank you.
     
  15. Apr 30, 2008 #14

    alphysicist

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    Did you figure it out afterwards? Since the weight of the painter is in the same orientation to the ladder as the weight of the ladder, they'll have the same form; they just act at a different distance. So the torque equation would be:

    [tex]
    F_W (10) \sin(65) - 22 (9.8) (5) \cos(65) - 95 (9.8) x \cos(65) = 0
    [/tex]

    with the torques from both weights being negative since they are clockwise.
     
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