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Homework Help: Can anyone check my working on the friction upcoming test question

  1. Jan 17, 2014 #1
    1. The problem statement, all variables and given/known data
    Question 1

    Which statement is true of the coefficient of static friction?

    It is the force resisting motion.

    It is proportional to the static friction force.

    It is a dimensionless quantity expressing the ratio of limiting friction to normal reaction.

    It is the ratio of friction to normal reaction.

    Question 2

    The ratio of limiting friction to the normal reaction is known as

    angle of friction.

    angle of repose.

    coefficient of static friction.

    coefficient of kinetic friction.

    Question 3

    Which of the following statements is NOT true about friction?

    Kinetic friction is larger than static friction.

    Limiting static friction is dependent on normal reaction acting on the body.

    Coefficient of friction denotes surface roughness of the bodies in contact.

    Frictional force acts tangential to the surfaces in contact.

    Problem sum


    2. Relevant equations

    3. The attempt at a solution

    1. 3rd option
    2. 3rd option
    3. 1st option.

    Problem sums:

    Where W is the skier's weight:
    Force pressing skier agaist slope = W×Cosθ
    Friction force in motion = 0.07×W×Cosθ ...... (1)
    Skier's weight component acting down the slope = W.Sinθ ...... (2)
    (1) and (2) are equal giving:
    0.07×W×Cosθ = W.Sinθ from which:
    Tanθ = 0.07 so that θ = arc-tan(0.07) = 4.0˚

    Weight of ladder = 18×9.8 = 176.4 N
    Weight of man = 90×9.8 = 882.0 N
    Put the man at the top of the ladder
    Vertical component of force between ladder and ground (normal force) =Fn= 174.6 + 882 = 1058.4 N
    To find reaction of wall against the ladder take moments of all applied forces about bottom end of ladder:
    (176.4×2 + 882×4).sin(30˚) = Rw×4.Cos(30˚) (Rw = reaction of wall)
    Rw = (176.4×2 + 882×4).sin(30˚) / [4×Cos(30˚)]
    (176.4×2 + 882×4).½ /(2.√3) = 560.2 N
    The horizontal component of the reaction force between ladder and ground = Fh = Rw since all horizontal forces on the ladder must balance and there are only these two!
    Necessary friction coefficient = Fh/Fn = 560.2/1058.4 = 0.529
  2. jcsd
  3. Jan 17, 2014 #2
    It all looks right to me.

    You should note that in the second problem, you can more easily (in my opinion) retrieve the coefficient of static friction by taking moments about the top of the ladder instead of the bottom.

    [itex] \sum M_{t} = 0 = W_{L}*(L/2)*sin(30) - F_{y}*L*sin(30) + F_{x}*L*cos(30) [/itex]

    [itex]\sum F_{y} = 0 = F_{y} - W_{L} - W_{M} [/itex]

    [itex] F_{x} = μF_{y}[/itex]

    where [itex] F_{y} [/itex] and [itex] F_{x} [/itex] act on the bottom of the ladder

    Solving for [itex] F_{y} [/itex] and [itex] F_{x} [/itex] in the force and friction equations, them plugging into the moment equation, you can solve for [itex] μ [/itex] more easily
  4. Jan 18, 2014 #3
    It seem that the ski question answer may be wrong according to the result i get, any clue why?
  5. Jan 18, 2014 #4


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    Science Advisor
    Homework Helper
    Gold Member

    Looks ok to me.......
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