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Static equilibrium question

  1. May 25, 2004 #1
    Hi,

    I was wondering how to analyze the forces in a simple static equilibrium problem I have been contemplating. Suppose you have two pulleys attached to a wall and aligned horizontally with a loop of wire around them, and you hang a weight from the lower strand of wire. I'm stuck trying to figure out how much force the weight exerts on each pulley. I can figure out the case where a single strand of wire is tied to two nails and then a weight is hung from the wire, but the continuous loop of wire with the pulleys has me stumped. I guess too many years have passed since my high school physics lessons. :(

    Thanks in advance.
     
    Last edited: May 25, 2004
  2. jcsd
  3. May 26, 2004 #2
    its hard for me to visualise exactly what ur asking

    the downward force is always going to be due to gravity = m*g

    if there are two pulleys at the same height, they must be taking equal load so
    f=1/2 * m * g
    on each pulley

    thats my guess anyway - the rope around the pulleys may have an effect on components of the force but the net force is always going to be the same
     
  4. May 26, 2004 #3

    Doc Al

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    You would find the tension in the wire exactly the same way in both cases: Set the vertical components of the tension equal to the weight of the hanging mass.
     
  5. May 26, 2004 #4

    Gokul43201

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    Either a figure or a better description would help.

    Anyway, draw the free body force diagrams for each of the pulleys. Resolve all the forces into horizontal and vertical components. Set the resultants to zero, using the fact that the tension is uniform throughout the string and always acts in the direction along the line of the string, away from the pulleys.
     
  6. May 26, 2004 #5
    Thanks for the replies. raymes K, sorry but you're off.

    O------O
    .\......../.
    ...\..../...
    .....\/.....
    .....<>....

    O = frictionless pulley
    <> = weight

    Doc Al and Gokul43201, thanks for the hints. The force diagram still has me stumped. If I can figure it out, then I can do the trig to get the answer.

    The force on a pulley is from the tension in the wire:

    O---->
    ..\
    ....v

    The force exerted by the horizontal wire has no vertical component so it doesn't help counteract the weight, so I think the tension in the wires is the same as in the simple case of two nails with a wire strung between them and a weight hung in the middle. But, the tension in the horizontal wire does have a component along the diagonal wire, so it adds to the force exerted on the pulley along the diagonal.

    Given the weight hanging on the wire is 100lbs, then the sum of the vertical components of the tension in the diagonal wires must equal 100lbs. If the angle at the bottom of the wires is 40 degrees, then given

    left_diagonal=tension in left diagonal wire
    left_vertical= vertical component

    right_diagonal=tension in right diagonal wire
    right_vertical= vertical component

    I get

    cos(40/2)= left_vertical / left_diagonal
    ==>left_vertical = cos(20) * left_diagonal

    and

    cos(40/2)= right_vertical / right_diagonal
    ==> right_vertical = cos(20) * right_diagonal

    Adding the two equations together:

    left_vertical + right_vertical = cos(20) * left_diagonal + cos(20) * right_diagonal

    Since,

    left_vertical + right_vertical = 100 lbs

    I get:

    100 = cos(20) * left_diagonal + cos(20) * right_diagonal

    and since,

    left_diagonal = right_diagonal

    substituting for left_diagonal gives

    100= cos(20)* right diagonal + cos(20) * right_diagonal
    ==> 2 * cos(20) * right_diagonal = 100
    ==> right_diagonal = 53

    So, on either pulley, there is a force of 53 pulling horizontally as well as along the diagonal from the tension in the wires, which I think gives me a force diagram like this:

    O---->53
    ..\
    ....v
    .....53

    The horizontal force has a component along the diagonal, which with a little trig comes out to:

    cos(70) = diagonal_component / 53
    ==>diagonal_component = 18

    So, I get the total force along the diagonal equal to 71. Is that the maximum force in any direction on the pulley? I don't think that's the correct answer.
     
    Last edited: May 26, 2004
  7. May 26, 2004 #6

    arildno

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    What you have found (right_diagonal), is the correct expression for the TENSION in the wire.

    To put it in a slightly different form, let the unit vectors up along the wires from the position of the weight be described like this:
    [tex]\vec{i}_{R}=\sin\frac{\theta}{2}\vec{i}+\cos\frac{\theta}{2}\vec{j}[/tex]
    [tex]\vec{i}_{L}=-\sin\frac{\theta}{2}\vec{i}+\cos\frac{\theta}{2}\vec{j}[/tex]

    ([tex] \theta[/tex] is the opening angle above the weight)

    We then have your expression for the tension in the wire:
    [tex]\mathcal{T}=\frac{mg}{2\cos\frac{\theta}{2}}[/tex]

    In order now to find the force FROM THE WIRE on a PULLEY, we consider a segment of the wire cut somewhere on the skew part of the wire, past around the pulley, and ending somewhere on the horizontal part of wire.

    We will look at such a segment around the righthand pulley:
    The wire segment experinces tensile forces from the rest of the wire, in addition to a force [tex]\vec{F}_{p}[/tex] from the right pulley.
    Since the wire segment is in equilibrium, we have:
    [tex]\vec{F}_{p}-\mathcal{T}\vec{i}-\mathcal{T}(\sin\frac{\theta}{2}\vec{i}+\cos\frac{\theta}{2}\vec{j})=\vec{0}[/tex]

    By Newton's 3.law then, the force on the righthand pulley from the wire is:
    [tex]\vec{R}=-\vec{F}_{p}=-\frac{mg}{2\cos\frac{\theta}{2}}((1+\sin\frac{\theta}{2})\vec{i}+\cos\frac{\theta}{2}\vec{j})[/tex]
     
  8. May 26, 2004 #7

    arildno

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    You've changed yor reply somewhat since I read it last; if something above doesn't make sense to you, make a new reply..
     
  9. May 26, 2004 #8

    Doc Al

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    So far, so good.

    I'm not sure what you're doing here. I would assume you want the total force that the wires are pulling on each pulley. So you need to add the two 53 lb forces:
    Horizontal components: 53 + 53 sin(20) = 53 (1 + sin(20))
    Vertical components: - 53 cos(20)

    Resultant: 87 lbs (unless I goofed the arithmetic)
    That's the net force that the wires exert on each pulley.
     
  10. May 26, 2004 #9
    Doc Al,

    I'm trying to figure out the maximum force on the pulley in any direction. Does that necessarily have to occur along the vertical or horizontal? My thought was that it would occur along the diagonal(the line from the pulley to the weight)

    O---->53
    ..\
    ....v
    .....53

    With a little observation, it can be determined that the angle between the horizontal and the diagonal is 70 degrees, so I used that angle to compute the component of the horizontal force that lies along the diagonal:

    cos(70) = diagonal_component_of_horizontal / 53

    Ahhhh, someone just hit the light switch. The resultant force is the sum of the horizontal and vertical. I thought you could just pick a direction and determine the components in that direction and then sum them to get the total force and then discard any perpendicular forces--which now seems quite stupid.


    arildno,

    Thanks for directing me to look at a section of the wire around the pulley. I can't quite work out the force diagram to match your Fp equation


    53<---|----O
    .............../
    ..........._ /_
    ............/
    ..........v
    ........53

    It seems to me the pulley should exert a counterbalancing force horizontally to the right as well as a counterbalancing force diagonally up and to the right to offset the tensile force pulling on the ends of the wire segment. I can't figure out how to get the resultant force since it isn't a right triangle and I can't figure the angles out.

    .....53
    ....^
    .../
    ./
    O----->53
     
    Last edited: May 26, 2004
  11. May 27, 2004 #10
    I think another light bulb went off:

    I don't need to work on the the Fp vectors because the resultant Fp vector is opposite and equal to the resultant T vector from the forces pulling on the ends of the wire segment, i.e the tension(T) in the wire. So,

    Fp_horizontal = T + Tsin20
    Fp_vertical = Tcos20

    Fp_resultant = [tex]\sqrt{(T + Tsin20)^2 + (Tcos20)^2} [/tex]

    I'm still not seeing how you got around using the Pythagorean theorem in your Fp equation. The angles change for the resultant T vector, so I'm not sure how you can use theta/2, etc. in solving for the resultant T vector.
     
    Last edited: May 27, 2004
  12. May 27, 2004 #11

    arildno

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    "It seems to me the pulley should exert a counterbalancing force horizontally to the right as well as a counterbalancing force diagonally up and to the right to offset the tensile force pulling on the ends of the wire segment. I can't figure out how to get the resultant force since it isn't a right triangle and I can't figure the angles out."

    Certainly, I wrote:
    [tex]\vec{F}_{p}-\mathcal{T}\vec{i}-\mathcal{T}(\sin\frac{\theta}{2}\vec{i}+\cos\frac{\theta}{2}\vec{j})=\vec{0}[/tex]

    To be precise:
    [tex]-\mathcal{T}\vec{i}[/tex] is the tensile force from the rest of the wire acting upon that part of the segment which lies between the two pulleys, that is, where the wire is horizontal.

    The other tensile force works on the "diagonal" segment.

    Do not worry about the ANGLE here, just add these two vectors together; that gives you the total force.
     
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