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Homework Help: Is a coefficient of friction of 7.7 on mousetrap car reasonable?

  1. Oct 19, 2014 #1
    1. The problem statement, all variables and given/known data
    We're told to find the coefficient of friction for a mousetrap car that we've made, using values that we attained ourselves. We released the arm of the mousetrap onto a scale to find Fa → 1.0*10^2 Newtons.
    This is how we found Fa:

    m = 1.020kg (←mousetrap held back and hammer released onto scale, value recorded from scale)
    a = 9.8m/s/s (←factoring in gravity, which affects mousetrap's "release weight" on scale)
    Fa = ?

    Fa = ma
    = 1.020kg * 9.8m/s/s (Fg)
    = 9.996 Newtons
    = 10. Newtons (← rounded to correct amount of sig digs.)

    Using that, we solved for acceleration :

    Vi = 0.0m/s
    t = 1.56s (time to complete 3m)
    d = 3.0m
    a = ?

    d = Vi(t)+0.5a*(t^2)
    a = (3m)/(0.5)(1.56^2)
    a = 2.46 m/s/s
    = 2.5m/s/s (← rounded to correct amount of sig digs.)

    With that Ff and Fnet was solved for :

    Fa = 10. Newtons
    m = 0.128kg
    a = 2.5m/s/s
    Ff = ?

    Ff was needed later so rather than doing m*a to find Fnet, we found Ff first to find Fnet:

    Fnet = ma
    Fa + Ff = ma
    Ff = 0.128kg(2.5m/s/s)- 10. Newtons
    Ff = -9.68 Newtons
    = -9.7 Newtons (← rounded to correct amount of sig digs.)

    And now Fnet:

    Fnet = Fa-Ff
    = 10. Newtons - 9.7 N
    = 0.3 Newtons

    At this point, everything seemed reasonable, but then we needed to solve for the coefficient of friction.

    2. Relevant equations
    μk = Fk/Fn

    3. The attempt at a solution
    μk = Fk/Fn
    = -9.7 Newtons / (0.128)*(9.8)
    = 7.73
    = 7.7 (← rounded to correct number if sig digs.)

    The question is, does 7.7 make sense for a coefficient of friction, and does everything else look sound?. The mousetrap car we made seemed to be of low friction , except between the ground and the wheel, where we wanted to maximize friction (balloons were used).

    Last edited: Oct 19, 2014
  2. jcsd
  3. Oct 19, 2014 #2


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    Staff: Mentor

    Check the calculation of Fa. 0.1*10 (approximate values) gives something close to 1, not 10.

    Why does your mass value change between the calculations?

    What are Ff and Fnet?
    It is not meaningful to add or subtract horizontal and vertical forces like you do there.

    7.7 is not your coefficient of friction, that is an order of magnitude too large.
  4. Oct 19, 2014 #3
    The mass value changes because we were told to measure the Fa of the mousetrap using a scale (release mousetrap onto scale record mass value). So there are two masses

    The 0.128kg mass value is the weight of the car, the mass from the release of the mousetrap should actually be 1.020kg. It's been fixed.

    Ff is -9.7 Newtons
    Fnet is 0.3 Newtons.

    What do you mean by that?
    Last edited: Oct 19, 2014
  5. Oct 19, 2014 #4


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    Yes I can read that, but what are those quantities? Where do you define them?

    What does "the mass from the release of the mousetrap" mean?

    If you introduce some quantity, you should define and explain it. Otherwise the calculations are just a collection of numbers and it is really hard to understand what you are doing (even more so if there is an error - because 7.7 is not correct).
  6. Oct 19, 2014 #5
    Oh, okay.

    What I'm assuming Ff is is that it's the energy that's lost when converting the mousetrap's energy to turning power at the wheels. Fnet is the net force that the car actually moves with.

    We held the mousetrap upside down, put the scale underneath and let go of the trap, so that the hammer on the mousetrap would hit the scale. The value that we saw on the scale (1020g) was recorded as a mass. Using that mass with the acceleration of gravity (which we assume affected the reading), we found the force applied.

    Thanks for your help so far.
    Last edited: Oct 19, 2014
  7. Oct 19, 2014 #6


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    How can Ff be an energy, if it is a force?

    Well this is the force of the spring mechanism itself. It is not the force used to accelerate the trap - otherwise your acceleration would be much higher (actually, the wheels would just slip or the car would tip over).
  8. Oct 19, 2014 #7
    I just meant that Ff is whatever force is lost to get the car moving.

    I'm not sure I understand what you mean by that, are you saying the Fa is incorrect? If so, how would I find the correct Fa so that I can use that to find the correct Ff, Fnet and coefficient of friction?
  9. Oct 19, 2014 #8


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    Based on the mass of the car and its acceleration, you can calculate the horizontal force between car and ground - the force necessary to accelerate it.
    Based on the mass of the car and gravity, you can find the vertical force.

    If all wheels are slipping, the ratio gives you the coefficient of dynamic friction. If they are not, the ratio is a lower limit on the coefficient of static friction - it could be better but the trap is not strong enough to reach that limit.

    There is no need to know details of the acceleration mechanism in the trap.
  10. Oct 19, 2014 #9
    If you meant Fnet when you said "the force necessary to accelerate it" then Fnet calculated using m*a returns 0.32 Newtons, which is (approximately) what I got from Fnet calculated using Fa and Ff. Unless you mean that m*a will return Fa?

    Will that return the applied force?

    So then:
    9.8*.128 + Ff = Fnet
    1.25 + Ff = 0.32 Newtons
    Ff = 0.32 - 1.25

    Ff = 0.93??

    uk = Ff / Fn
    uk = 0.93/(.128*9.8)

    = 0.74??

    We were told that the applied force is the reading on the scale from the mousetrap.
  11. Oct 19, 2014 #10
    If not the applied force from the mousetrap, how would I find the force applied?
  12. Oct 19, 2014 #11


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    I mean "force that accelerates the car forwards". No need to introduce so many different notations and different forces.

    It will return the vertical force between car and ground.
  13. Oct 19, 2014 #12
    I understand that 7.7 is the incorrect coefficient of friction. But I still don't know what I did wrong. The vertical force between the car and the ground is the normal force. No? I think I used that correctly when I incorrectly calculated the coefficient of friction.

    "I mean 'force that accelerates the car forwards'. No need to introduce so many different notations and different forces."
    So how would I apply that value to an equation? Would that not be Fnet? If that is Fnet, how do I find the other values? I need to find Fnet, Fa, and Ff.

    Thanks again for your help. :)
  14. Oct 20, 2014 #13


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    As far as I understand your definition of Fnet, yes. It also agrees with Fk that you used (but did not define).
    There is no need for other values.
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