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Tourque the ladder question

  1. Jan 29, 2006 #1
    Hello.
    I got the following question.
    a 12 m ladder, whose mass is 20 kg is leaning agains a wall with its base being 6 m from the wall and the angle between the floor and the ladder is 60 degrees. The wall and the floor are frictionless. So we tie a rope 0.5 m from the base of the ladder to the wall to keep the ladder from sliding. A 72 kg person climbs 3 quarters up the ladder and stps. We need to find the tension in the rope and the reactive force in the floor.
    For the force on the floor it is easy, u just get the mass and multiply it by 9.i8 to get the total force up in order for thsi to be static equalibirum.
    Now i was trying to find the tension in teh rope, so i did this:
    i set the place where the ladder is tied to the rope to be the pivot and did this:
    0=-(9-0.5)sin 30 * 72 *9.8 -5.5 sin30 * 20 * 9.8 + sin 30*11.5 Fx
    So i solved for Fx adn i get:
    600. The correct answer is 380. The reason i used 30 is becuase that is the angle between the force applied and the arm as opposed to use 60 which is the angle between the floor and the ladder.
    any ideas where i made the mistake?
    Thx
     
  2. jcsd
  3. Jan 30, 2006 #2
    First what if Fx? If it is the tension in the rope then it won't appear in the equation as it is applied at the pivot point.
    If Fx is the reaction of the wall, then the angle is not 30, it would be 60.

    Second, there are 5 forces on the ladder and you missed 1/2 in the equation. (Vertical up reaction from floor, weight of ladder at center, weight of person, tension of rope and horizontal reaction from the wall)

    As you said, the normal reaction from floor = (72+20)*9.8. To calculate th e tension in rope, take the point at rope as pivot,
    0=-(9-0.5)sin 30 * 72 *9.8 -5.5 sin30 * 20 * 9.8 + sin 60*11.5 Fx - 0.5 * sin30 * 92*9.8
     
  4. Jan 30, 2006 #3
    thx alot, i think i get it now
    i applied what i learned to yet another ladder question:D
    Here it is:
    a ladde length 5 m and whose weight is 400 N is leaning agains frictionless wall. coefficient of static friction between the ladder and the ground is 0.46. I need to determine the greatest distance the ladder can be placed without slipping.
    The prolbme here is that i don;t know the correct answer, but i do have a solution, and it got me:
    Angle between the ladder and the floor being 47.4 degrees, therefor the distance is 3.39 m from the wall.
    Could somone check it out and tell me if my solution gave me the correct answer or not? If not then how do u solve this?
    I did this
    0=Weight * half distance cos theta -Total distance * sin theta*Frictional force.
    from this i got
    tan theta=1000/5*Frictional force.
    And i know
    that Ff= 0.46 *400. tehrefor i got the angle and so on.
    So any ideas if this is right?
     
  5. Jan 31, 2006 #4
    I assume you are considering the pivot where the ladder touches the wall. You have forgotten one more force, the normal reaction of wall. A component of that will be reflected in the equation.
    Also, both frictional force and weight of ladder tend to "rotate" in the same direction; and normal reaction in the opposite.

    Include these and try again.
     
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