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Coefficient of Friction Questions I cant do

  1. Jan 17, 2008 #1
    Hi, this is my first post and i'm a bit rusty will mechanics so I need help with two questions i've come across

    Firstly on the question sheet I am given:
    a = F/m and w = mg and the static friction and kinetic friction equations

    also v[tex]^{2}[/tex] = v[tex]^{2}_{0}[/tex] + 2a[tex]\Delta[/tex]x

    How do I do these:

    1. The coefficient of friction between the bed of a truck and a box is 0.30.
    The truck is moving at 80km/h horizontally.
    What is the least distance in which the truck can stop if the box is not to slide?

    2. A car travelling at 30m/s on a horizontal road brakes suddenly to avoid hitting a pedestrian. What is the stopping distance if
    a) The car has ABS brakes, wheels do not slip ([tex]\mu[/tex]s=0.5)
    b) The car has no ABS, wheels lock ([tex]\mu[/tex]k=0.3)
     
    Last edited: Jan 17, 2008
  2. jcsd
  3. Jan 17, 2008 #2

    Tom Mattson

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    Staff Emeritus
    Science Advisor
    Gold Member

    Try to answer these questions, in order.

    1.) What is the maximum static frictional force that the truck bed can exert on the box before the box slides? Since you weren't given the mass of the box, you'll have to call it something (say m). As you will see, the value of m won't matter. m will cancel out eventually.
    2.) Using Newton's second law (a=F/m, which you cited) and your answer to 1.), what is the maximum acceleration that the box can withstand before sliding?
    3.) Given the numbers in the problem statement, and your answer to 2.), what is the minimum distance in which the truck can stop without the box sliding.

    We'll tackle the second question once you've got this one.
     
  4. Jan 18, 2008 #3
    Sorry I'm confused how you get to the point of cancelling m out

    I always get to this point:

    [tex]f_{s,max} = \mu_{s}F_{n}[/tex]

    [tex]f_{s,max} = \mu_{s} w[/tex]

    [tex]f_{s,max} = \mu_{s} mg[/tex]

    [tex]f_{s,max} = \mu_{s} \frac{F}{a}g[/tex]

    [tex]f_{s,max} = 0.3 \times 9.8\frac{F}{a}[/tex]

    Since I don't have F or a; I get stuck. What am I doing wrong?
     
  5. Jan 18, 2008 #4
    You are looking for [itex]a[/itex], so not having it is OK.

    You do have F! You have it twice.

    Up to [itex]f_{s,max} = \mu_{s} m g[/itex] you are calculating a theoretical maximum force, [itex]f_{s,max}[/itex]. When you move to the next equation, making the substitution [itex]m = \frac{F}{a}[/itex] you are still dealing with this theoretical maximum force but now you are also dealing with the deceleration, [itex]a[/itex], that this theoretical maximum force can produce.

    The substitution assumes the acceleration [itex]a[/itex] and calculates the force that must be producing it. [itex]F[/itex] and [itex]f_{s,max}[/itex] are the same force. Substituting [itex]F = f_{s,max}[/itex] will lead to the solution.

    With this insight, can you do it more elegantly?
     
  6. Jan 19, 2008 #5
    Oh ok :D, I've managed to now get this:

    [tex]ma = \mu_{s}mg[/tex]
    [tex]a = \mu_{s}g[/tex]
    [tex]a = 0.3 \times 9.8[/tex]
    [tex]a = 2.94[/tex]

    therefore:

    [tex]\Delta x = \frac{v^{2} - v^{2}_{0}}{2a}[/tex]

    [tex]\Delta x = \frac{0^{2} - 80^{2}}{2 \times 2.94}[/tex]

    [tex]\Delta x = 1088.44[/tex]

    however the answer I am given is 83.7m, what have I doing wrong this time?
     
  7. Jan 19, 2008 #6
    You have mixed your units, using both hours and seconds for time.
     
  8. Jan 19, 2008 #7
    And meters and kilometers for distance.
     
  9. Jan 21, 2008 #8
    Silly me, I overlooked that, thank you.

    I think I've now done question 2 correctly now.

    2a)

    [tex]ma = \mu_{s}g[/tex]
    [tex]a = 0.5 \times 9.8[/tex]
    [tex]a = 4.9[/tex]

    so

    [tex]\Delta x = 91.84[/tex]

    2b)

    [tex]ma = \mu_{k}g[/tex]
    [tex]a = 0.3 \times 9.8g[/tex]
    [tex]a = 2.94[/tex]

    so

    [tex]\Delta x = 153.06[/tex]

    however the answer given for 2b is slightly different (152.9). Have I missed something small here?
     
  10. Jan 21, 2008 #9
    The answer is 152.9 if g is 9.81.

    Some would say the answer should be 200 m because the problem data is given to one signifcant figure and the answer should not exhibit greater accuracy. Makes sense; spurious accuracy is, er, spurious.

    If you want to give answers correct to 5 significant figures (as you did) it's safest to state that you are assuming the problem data is accurate enough to justify it: "assuming the coefficient of friction give as 0.3 is actually 0.30000 ..."
     
  11. Jan 21, 2008 #10
    Thats wonderful, I've finally done it, Thank You all very much.
     
  12. Jan 22, 2008 #11
    Ok I've now come across a new question where my answers are different to the answers given:

    11. Two blocks attached by a string slide down a 20o slope. The lower
    block has mass m1=0.25kg and coefficient of kinetic friction 0.2.
    The upper block has mass m2=0.8kg and coefficient of kinetic friction 0.3.
    Find
    a) the acceleration of the blocks and the
    b) the tension T in the string.

    I manage to get this:

    [tex]a = \mu_{k}g cos\theta[/tex]

    lowerblock:
    [tex]a = 0.2 \times 9.81 \times cos20[/tex]
    [tex]a = 1.8437[/tex]

    upperblock:
    [tex]a = 0.3 \times 9.81 \times cos20[/tex]
    [tex]a = 2.7655[/tex]

    however the answer provided is a = -0.809 m/s2, how do you go about that.

    Also what ways do I use to find out the Tension in the string?
     
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