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Equilibrium of rigid bodies

  1. Aug 3, 2010 #1
    Hi everybody!How is it going!
    I'm having kind of a trouble trying to understand how to solve this following simple problem.If somebody is willing to help me i'll be very greatfull!
    I'll try to translate it from portuguese and hope everything written is correct.




    1. The problem statement, all variables and given/known data

    A ladder 10 metters in length and whose gravity center lies in its center is supported on the ground( with friction ) and a wall (without friction).The weight of the ladder is 400 N and a person's weight is 600 N, sin(theta)=0,80 and cos (theta)=0,60,Calculate:

    a)The intensity of the force applied on the ladder by the wall.
    Obs.:The person climbed the ladder 8 metters.
    2. Relevant equations





    3. The attempt at a solution

    What i found out of nothing was this:

    M1 + M2 + M3 = 0
    4,8(600) + 3(400) = 8F
    F = 510N
    But i could not understand it.Can somebody please show me how is this fitted?And why?

    Thanks in advanced!

    I hope it is not too confusing.
     
  2. jcsd
  3. Aug 3, 2010 #2
    What's the angle that the ladder makes with the horizontal??
     
  4. Aug 3, 2010 #3
    The angle with the floor is sin(theta)=0,80 and cos (theta)=0,60.
     
  5. Aug 3, 2010 #4
    I'm studying this right now.

    To answer this question, you have to sum up the torques around a point, then sum up the vertical and horizontal components.
     
  6. Aug 3, 2010 #5
    Yes,that's the way.But have you tried this question.It would really help you put it here;)
     
  7. Aug 3, 2010 #6

    diazona

    User Avatar
    Homework Helper

    We can help guide you, Brunno, but we're not going to just give away the solution to the problem.

    For the first step, draw a free-body diagram that show the forces acting on the ladder. Or at least, make a list of those forces: for each one, you will need to specify its magnitude, its direction, and the location where it acts. Once you do that, it may be easier to see how to use that information to get an equation that will tell you what the force of the wall on the ladder is.
     
  8. Aug 3, 2010 #7
    Yes,i don't really need the answer,i just need to know how to get there,I'll try once more to dry all the forces and everything and i will put here what i'm not quite understanding.
     
  9. Aug 4, 2010 #8
    What your 2nd equation (4.8*600+3*400-8F=0) tells you is that the total torque around the point of contact of the ladder with the ground is zero, and should be so since the ladder is at equilibrium hence the total torque around any point is zero. We choose that point of contact because the friction with the ground is unknown, but the torque of friction is zero because the distance of friction from that point is zero.

    Remember Torque for a force F is equal to F*d*sin(theta) where d is the distance of the force from the point and theta is the angle that the direction of force makes with the direction of distance.

    The torque of the force F from wall is -8F because it is -F*(ladder length)*sin(theta)=-F*10*0.8=-8F.

    Similarly for the other 2 forces (weight of ladder and weight of person) which will have positive signs because they tend to rotate the ladder in an opposite way than the force from the wall. The angle they make with the distance is now 90-theta but sin(90-theta)=cos(theta)
     
  10. Aug 21, 2010 #9
    Thank you everybody for the help!
    Delta² ,can you believe that in my book there's not this equation: F = F*d*sin(theta)?
    But the book still one of my fellows,is not the best one but you know!
    Thank you again everybody!!!
     
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