1. Limited time only! Sign up for a free 30min personal tutor trial with Chegg Tutors
    Dismiss Notice
Dismiss Notice
Join Physics Forums Today!
The friendliest, high quality science and math community on the planet! Everyone who loves science is here!

Finding the coefficient of friction

  1. Mar 22, 2017 #1
    1. The problem statement, all variables and given/known data
    I am investigating a scenario where a pendulum with a bob attached is released from an angle and pushes a box to a certain distance. My goal is to find the coefficient of friction between the box and the surface it moved on.

    I have measurements for:
    - mass of the bob (125 grams)
    - mass of the box (267.3 grams)
    - length of the pendulum (27 cm)
    - distance the box moved from its initial position (3.2 cm)
    - angle of release (30 degrees above the x-axis)

    2. Relevant equations
    - 1/2(m)(v^2) = mgh
    - Vf = sqrt(2*g*h)
    - m1+v1=m2+v2
    - h = L(1-cos(theta))
    - Ff=μFN

    3. The attempt at a solution
    0.27(1-cos(30))= h = 3.62 cm
    sqrt (2*9.8*0.0362) = Vf = 0.84 m/s (speed before hitting the box)
    (0.125)(0.84)=(0.2673)(v2), v2 = 0.33 m/s (speed of the block)
    (9.8)(0.2673) = FN = 2.62 N

    I don't know where to go from here. If you could help me or point me in the right direction, I would greatly appreciate it, Thanks!
  2. jcsd
  3. Mar 22, 2017 #2
    Did you try solving through energy?
  4. Mar 22, 2017 #3
    I was only able to figure out the final velocity of the bob before it hit the box through energy.
    v = sqrt(2*g*h)
  5. Mar 22, 2017 #4
    Indeed. From there, you need to find the initial velocity of the box after the collision. Any ideas on how you might do it?
  6. Mar 22, 2017 #5
    Well, if I take conservation of momentum (m1+v1=m2+v2) into account then, (0.125)(0.84)=(0.2673)(v2). v2 being the velocity of the box after the collision, I calculated to be 0.33m/s.
  7. Mar 22, 2017 #6
    The equation for conservation of momentum is
    m1v1 + m2v2 = m1v1' + m2v2'
    If we assume the collision is elastic, then the second and third terms go away and you are left with
    m1v1 = m2v2'

    I didn't do the calculation, but will assume you got it correct. Now you have to take into account forces to find coefficient of friction. What will you do from here?
  8. Mar 22, 2017 #7
    I'm guessing by forces you mean forces such as driving force. So, W=FD. 1/2mv^2=W. (1/2mv^2) / d = F. (1/2(0.2673)(0.33^2)) / 0.032 = 0.455 N
    I'm not sure if this was a valid calculation, but if it is is the friction equal to 0.455N?
  9. Mar 22, 2017 #8
    By force, I mean sum up forces in both horizontal and vertical directions. Vertical is
    Fn - mg =0
    Fn = mg
    Horizontal is
    μFn = ma
    Agreed? If so, where could we go from here?
  10. Mar 22, 2017 #9

    Fn = mg, so the normal force equals (0.2673)(9.8) = Fn = 2.62 N
    μFn = ma, (Second Law) so (0.2673)(a)

    Then ma/mg = μ
    But how do I find the value of 'a'? The box moved 3.2 cm but way too quickly for me to record an accurate time.
  11. Mar 22, 2017 #10
    Ah, time is not even needed! The equation
    v2 = v02 + 2⋅a⋅Δx
    v... final velocity (0)
    v0... initial velocity (from your momentum calculation)
    a... acceleration
    Δx... change in distance
    will cover it. Just rearrange it to find a.
    As a side note, just make sure your distances are in meters (not cm) :smile:
  12. Mar 22, 2017 #11
    Ahh I see.
    So, the box stops at 0.032 meters so Vf = 0 but has an initial velocity of 0.33 m/s so Vi = 0.33 m/s
    Vf^2 = Vi^2 + 2ad
    0^2 = 0.33^2 + 2 (a) (0.032)
    -0.33^2 = 2 a (0.032)
    0.1089 = 2 a (0.032)
    ((0.1089 / 2) / 0.032) = a
    a = 1.70 m/s^2
    Did I do the calculation process correctly? If so, should I apply it to the 'a' in (μFn = ma) and then get a coefficient value?
  13. Mar 22, 2017 #12
    Looks correct to me.

    Also, I just realized that this question was placed in this general physics forum, while there is actually a separate forum for physics homework questions! I'm just letting you know for future reference and so you aren't confused if this thread get moved to another forum.
  14. Mar 22, 2017 #13
    Alright I'll be sure to post correctly next time! Thanks for your help, it helped a great load:smile:!
  15. Mar 22, 2017 #14
    No problem! Enjoy the forums...:biggrin:
  16. Mar 23, 2017 #15


    User Avatar
    Science Advisor
    Homework Helper
    Gold Member

    That is not correct. If the collision is perfectly elastic then v1-v2=v2'-v1'.
    If perfectly inelastic the masses coalesce, i.e. continue together:

    In practice, it will be partly elastic. Depending on the materials, you might be able to assume that the collision approximates one of the two extremes. What are the materials?
  17. Mar 23, 2017 #16
    Yes, that is also true. But please explain how the second and third terms do not cancel out.
    EDIT: I think I see what you're saying now. Second term cancels out for sure. The problem is, though, we don't even know what type of collision this is or what the materials are.
    Last edited: Mar 23, 2017
  18. Mar 23, 2017 #17


    User Avatar
    Staff Emeritus
    Science Advisor
    Homework Helper
    Gold Member

    Did you make these measurements, or are they given to you as an exercise.

    If you made the observations, what was the motion of the bob immediately after the collision?

    If they're given to you, was there any additional information regarding the nature of the collision or regarding the final state of the pendulum?
  19. Mar 23, 2017 #18


    User Avatar
    Science Advisor
    Homework Helper
    Gold Member

    The third term, m1v1' only goes to zero if the impinging object stops dead. That happens in perfectly elastic collisions if the two objects have the same mass. More generally, it can only happen if the impinging object is the lighter of the two, in exactly the right ratio to match the degree of inelasticity. Specifically, m1=m2R, where R is the coefficient of restitution.
  20. Mar 23, 2017 #19
    Me and my group made the measurements.
    The bob was released at an angle (30 degrees from rest / 60 degrees below x axis) and when released it hits the box, the box moves but the bob remains at the point where it was before moving it to an angle. So, like 90 degrees to the box after being hit.

    As for materials. The box was a tissue box. And the bob was, not sure how to describe this, filled with sand in a balloon. It looked like putty.
  21. Mar 23, 2017 #20


    User Avatar
    Science Advisor
    Homework Helper
    Gold Member

    Not exactly sure what you are saying there. Are you saying that the bob stopped dead at the point where the collision occurred? It did not continue to swing at all?
Know someone interested in this topic? Share this thread via Reddit, Google+, Twitter, or Facebook

Have something to add?
Draft saved Draft deleted

Similar Discussions: Finding the coefficient of friction