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Leaning ladder

  1. Jan 15, 2005 #1
    I am having problems with the following problem.

    A person is standing on a leaning ladder rested on a wall, where m1 is the mass of the person and m2 is the mass of the ladder. He is up at the distance d on the ladder, L (the length). There is also an angle at the bottom.
    The first part of the question asks to find the minimum coefficient friction, so that the ladder does not slip, which I found was 0.5*L*m_2*g+m_1*g*d)/(L*tan(theta)*(m_2*g+m_1*g))

    Then, part B of the question is to find the magnitude of the force of friction that the floor applies to the ladder. The coefficient of the static friction force, mu_s is equal to (3/2)*mu_min which is given. In this case, I know that f doesn't equal mu_s*N. but it is less than mu_s*F_normal .

    In the end, I got (3/2)*(d/L*m_1*g+0.5*m_2*g)*tan(theta) , which is completely wrong.

    Any help is appreciated.
    Thank you
     
    Last edited: Jan 16, 2005
  2. jcsd
  3. Jan 15, 2005 #2

    Q_Goest

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    These kind of problems use summing of moments around a point and summing forces in various orthogonal directions.

    Start by summing forces in the verticle (Y) direction and horizontal (X) direction. You should find verticle forces don't really matter here, so look at forces in the horizontal direction. In this case, the force against the wall, N, is equal to the frictional force. Note there are no other forces in the X direction, only these two, so the force against the wall is equal to the frictional force.

    Now find the force against the wall, N, by summing moments around a point. If the point you select is the point where the ladder rests on the ground, you should find moments created by the verticle forces due to weight for the ladder and the person are countered by the normal force on the wall, N times the moment arm. This solves for N, and since N = frictional force, you also have the frictional force.
     
  4. Jan 15, 2005 #3

    Gokul43201

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    laminated,

    Rather than just writing down what you got, if you tell us how you got that, it would be possible for us to show you where you made your mistake. Now we really don't know how you got what you got.
     
  5. Jan 16, 2005 #4
    I started summing forces from the x, y direction

    Fx = fs+n1=0
    Fy=n2+(-m1g)+(-m2g)=0

    Then also the torque
    TB=n1Lsin(theta)-0.5(m2g)(Lcos(theta)-m1g(Dcos(theta))+n2(0)+fs(0)=0

    fs=mu_min*n2, plugging in all the values and solving for mu_min...

    This yielded to 0.5*L*m_2*g+m_1*g*d)/(L*tan(theta)*(m_2*g+m_1*g)) which is CORRECT.

    But, how do I find the force of friction that the floor applies to the ladder? Ignoring the slipping...
     
  6. Jan 16, 2005 #5
    So this means f=mu_min*N right? N = (m2g+m1g)

    Fx = fs+n1=0
    Fy=n2+(-m1g)+(-m2g)=0

    Kind of confused...
     
  7. Jan 16, 2005 #6

    Gokul43201

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    Should one assume that there is no friction from the wall ?
     
  8. Jan 16, 2005 #7

    Gokul43201

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    Assuming this, I get, for the first part :

    [tex]\mu _{min} = \frac {m1(d/L) + (m2/2)}{(m1+m2)tan \theta } [/tex]

    You have the same result for this. So far, so good.

    For the second part, the only importance of the additional data is to tell you that the coefficient is large enough for stable equilibrium.

    In this case, you simply redo the calculations that you did for (A), but instead of [itex] \mu _{min} N [/itex], you just call the frictional force F, and find F. Balancing horizontal forces will tell you that F = R, the reaction force from the wall. So, replacing R with F in the torque equation gives :

    [tex]F = \frac {g[m1(d/L) + (m2/2)]}{tan \theta} [/tex]

    Is this what the answer is supposed to be ?
     
  9. Jan 16, 2005 #8
    Yes, that is the correct answer! I have been pondering about that second part for days. I thought that I had to integrate 3/2mu_min into the equation somehow. But, I am still a rookie at this stuff.

    In any case, thank you for your help. I really appreciate your time.
     
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