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Ladder problem

  1. Apr 15, 2004 #1
    A painter wishes to know whether or not she can safely stand on a ladder. The ladder has a mass M1 = 12 kg which is uniformly distributed throughout its length L = 7.4 m. The ladder is propped up at an angle theta = 53o. The coefficient of static friction between the ground and the ladder is mus = 0.42, and the wall against which the ladder is resting is frictionless. Calculate the maximum mass of the painter for which the ladder will remain stable when she climbs a distance d = 6.6 m up the ladder. (The painter's mass might be so low that only Lilliputian painters can safely ascend the ladder.)

    I think that I've gotten the problem but my answer is wrong. Just curious if anyone could take the time to look over my formulas and work.

    M_L = 12 kg
    L= 7.4m
    Theta = 53 degrees
    Static Friction (mus) = .42

    d = 6.6m

    I drew my force diagrams and such, I have a the Normal force of the wall against the top of the ladder, The normal force of the ground against the bottom of the ladder (straight up), Static friction pointing towards the wall. The weight of the ladder at the center of mass of the ladder. And the weight of the Painter/Man.
    For this to be in equilibrium, Net Torque and Net force must equal Zero.

    I suppose this problem is more tough on the algebraic side?
    Here's some of my work.
    Positive torque - counterclockwise
    Negative torque - Clockwise
    Goal - To find the mass of the maximum mass of the man.

    Torques:
    (L/2)*M_L*g Cos(theta)
    d*M_man*g*Cos(theta)

    X force N_w= mus*Normal Force
    Y force (M_man + M_ladder)g = Normal Force

    -L*mus*(M_man + M_L)g*sin(Theta)

    So the formula i made to find the mass of the man was this.

    (L/2)*M_L*g Cos(theta) + d*M_man*g*Cos(theta) - L*mus*(M_man + M_L)g*sin(Theta) = 0

    After plugging in numbers i got
    M_man(38.92 - 291.9019) = - 261.8617529

    I got 1.035 kg. This is incorrect.. What am i doing wrong?
     
  2. jcsd
  3. Apr 15, 2004 #2

    NateTG

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    Well, perhaps you should list all of the forces that you need to account for. There are five (you're only accounting for four of them).
     
  4. Apr 15, 2004 #3
    I drew my force diagrams and such, I have a the Normal force of the wall against the top of the ladder, The normal force of the ground against the bottom of the ladder (straight up), Static friction pointing towards the wall. The weight of the ladder at the center of mass of the ladder. And the weight of the Painter/Man.

    Normal force of wall against ladder
    Normal foce of ground against ladder
    Static Friction between ladder and ground
    Weight of Man
    Weight of Ladder.

    Friction and Normal for of wall against ladder cancel out because net force is = 0 and they are the only x forces.

    Haven't I accounted for them all?
     
  5. Apr 15, 2004 #4

    NateTG

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    But the torque for friction, and the torque for the normal force on the wall might not cancel.
     
  6. Apr 15, 2004 #5
    Don't they have to to make Tnet= 0?

    How would you find them with the information given?
     
    Last edited: Apr 15, 2004
  7. Apr 15, 2004 #6
    Maybe this matters, I have my pivot point at the bottom of the ladder.
     
  8. Apr 15, 2004 #7

    NateTG

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    Right, so the torque due to friction with the floor will be zero, but the torque due to the normal force of the wall will not.

    You should be able to figure out the mangitude of the (maximum) normal force that the wall can exert since [tex]F_{net x}=0[/tex].
     
  9. Apr 15, 2004 #8
    Didn't I do that?

    -L*mus*(M_man + M_L)g*sin(Theta)

    That's torque.. nm.

    The normal force would just be the .. mus(Mass of man + Mass of Ladder)*g...?
     
    Last edited: Apr 15, 2004
  10. Apr 15, 2004 #9

    NateTG

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    Ugh.. I gues the text just makes things hard to read....

    Let's see:

    [tex]\tau_{ladder}=\frac{L}{2} M_{ladder} g \cos \theta \approx 3.7 \times 12 \times 9.81 \times 0.601 \approx 261[/tex]
    [tex]\tau_{man}=d M_{man} g\cos \theta \approx 6.6 M_{man}\times 9.81 \times 0.601 \approx 39 M_{man}[/tex]
    [tex]\tau_{wall}\geq -\mu_s g (M_{ladder}+M_{man})L \sin\theta [/tex]
    [tex]\approx -.42 \times 9.81 \times (M_{man}+12) 7.4 \times 0.798 \approx - 24.3 M_{man} - 292[/tex]
    now
    [tex]\tau_{ladder}+\tau_{man}\leq -\tau_{wall}[/tex]
    so
    [tex]261 + 39 M_{man} \leq 24.3 M_{man} + 292[/tex]
    [tex]31 \geq 15 M_{man} [/tex]
    [tex]2.0 \geq M_{man}[/tex]

    Not sure if that's the right answer, but:

    Your result doesn't match the second to last equation, but the equation would, except for some grouping problems, indicate that you lost one of the terms in the algebra.
     
  11. Apr 15, 2004 #10
    Well that's actually the right answer, thanks. But where did 24.3M_man come from?
    I was just getting 292M_man, as opposed to 292 +24.3M_man.. Grr I think it was just math, i didn't distribute to the M_man.

    Thanks alot for ALL your help : )
     
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