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Homework Help: Coefficient of static friction between a ladder and a friction-less wall

  1. Oct 28, 2011 #1
    1. The problem statement, all variables and given/known data

    A uniform 7.7 m tall aluminum ladder is leaning against a friction-less vertical wall.
    The ladder has a weight of 291 N.
    The ladder slips when it makes a 59.0◦ angle with the horizontal floor.
    Determine the coefficient of static friction between the ladder and the floor.

    2. Relevant equations

    τ = (r) (F) (sin(Θ))

    r = (1/2)(h) (because it is a solid figure and force applies from the center of it)

    τx = τ cos(Θ)

    F(friction max) = µs * N
    µ(static) = F(friction max) / N

    3. The attempt at a solution

    h = 7.7
    r = 3.85


    τ = (3.85) (291) (sin(59◦))
    τ = 960.3274

    τx = 960.3247 * cos(59◦)
    τx = 494.6052

    so if τ(x) is the amount of force moving in the x direction, then shouldn't τx = Force(friction max) ?

    µs = F(friction max) / N

    µs = 494.6052 / 291 which is more than 1 so it is obviously wrong.

    I am clearly going in the wrong direction and I would really appreciate any help that you could offer, even if you tell me i need to scrap this! If you could steer me in the right direction I would greatly appreciate it!
  2. jcsd
  3. Oct 28, 2011 #2


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    Hello hanaelise23,

    Welcome to physics forums!

    You might want to start over, with first making a free body diagram (FBD). Label all the forces involved. The forces are:
    • The normal force from the ground to the ladder.
    • The frictional force from the ground on the ladder.
    • The normal force from the frictionless wall on the ladder.
    • The gravitational force (and you are correct that this should be placed in the middle of the ladder, since the ladder is uniform).

    Note that there are only two vertical forces. Since the ladder is in static equilibrium and not accelerating vertically, you know that the sum of these two forces must be zero (taking direction into account, of course). This gives you an important relationship.

    Also that there are only two horizontal forces. Since the ladder is not accelerating horizontally, you know that the sum of these two forces must be zero (taking direction into account, of course). This gives you another important relationship.

    Now you need to sum "the moments" (torques) around some point. It doesn't really matter what point that is*. But a logical choice is the point at which the ladder touches the ground or the point at which the ladder touches the wall. Since the ladder is not twisting/spinning, you know that the sum of all the moments (torques) about this point must be zero.

    *Hypothetically though, you could pick any point in the universe if you wanted to. Picking one of the suggested points simply makes the math easier.

    I assume here that "τ" here represents some moment (torque). But at what point is this moment taken? I can't determine a easy choice of of point that would involve a sinΘ multiplied by the gravitational force and half the radius. (maybe a cosΘ, but not a sinΘ).
    Again, see above regarding the sinΘ, and choice of reference point for the moment. Also, you will need sum at least one other moment (torque), such that the sum of all torques (about the point you choose) equals zero. The number of moments you need to sum depends upon which point you choose.
    You lost me here. I thought τ was a torque, not a force. Torques don't have an 'x' component.
    I think you're mixing torques and forces. They are related, however. But you can't just set them equal to each other.
    Why is that? There's nothing that says that a friction coefficient can't be greater than 1. There are many real-world examples where the coefficient of friction is much greater than 1.

    However, according to my calculations, the coefficient of friction is less than one for this particular problem. :wink:
    Last edited: Oct 28, 2011
  4. Oct 31, 2011 #3
    Thank you!
    so the normal force of the ground on the ladder(Ng) is equal to the gravitational force (Fg).

    so Fg= Ng

    and the friction force(Ff) is equal to the normal force of the wall on the ladder Nw.

    so Ff = Nw

    to get the torque, knowing that τ= 1/2 r F and solving about the point the ladder touches the ground would get

    τ=1120.35 N-m

    is this right? i am confused about Torques. and i know i'm supposed to solve for another one, but i dont know how that works. if you could clarify what i need to do with the torques, i would greatly appreciate it.
  5. Oct 31, 2011 #4


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    Very nice! :approve: The above is correct, at least in terms of the magnitudes (directions are opposite, but your relationships are correct in terms of the magnitudes). Keep those relationships in your back pocket. You'll need to come back to them later.
    Something is not quite right there. Remember, the angle θ is defined as the angle of the ladder with respect to the horizontal floor, not the vertical wall.

    On your FBD, there should be a gravitational force vector starting at the middle of the ladder and pointing down. At the tip of this vector, draw a dotted line such that it intersects the ladder at a 90o angle. You now have a right triangle with Fg being the hypotenuse. This new triangle is a similar triangle formed by the ladder, floor and wall. One of the remaining two angles (not the 90o angle) is equal to θ. Determine which angle this is.

    Is the Fg component perpendicular to the ladder opposite [STRIKE]Fg[/STRIKE] [Edit: Ooops I meant "opposite θ"] (thus related to sinθ) or adjacent to [STRIKE]Fg[/STRIKE] [Edit: adjacent to θ] (thus related to cosθ)?
    I assume the '8' is a typo. But you still need to bring in the correct trigonometric function.
    You need to bring in the correct trigonometric function to that particular torque (the one caused by the gravitational force). But you're missing the torque caused by the wall. You must bring that one in too. After a bit of substitution, you should be able to use that to find the frictional force.

    Newton's second law states that the sum of all forces equal mass times acceleration.
    m \vec a = \sum_i \vec F_i
    Similarly, the same applies to all all torques and angular accelerations:
    I \vec \alpha = \sum_i \vec \tau_i
    And since in this problem, the system is in static equilibrium (nothing is rotating around and accelerating angularly, i.e. [itex] \vec \alpha = 0 [/itex]),
    0 = \sum_i \vec \tau_i
    So if you pick a point, say the point where the ladder meets the floor, the moment (aka. torque) produced by gravity and the moment produced by the wall must sum together to zero (assuming the moments are taken about the same point).
    Last edited: Oct 31, 2011
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