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Friction Lab? Kinetic Friction?

  1. Nov 3, 2016 #1
    1. The problem statement, all variables and given/known data
    Hi, we had a Lab Activity on the Coefficient of Friction, but right now I am really confused and stuck. Since the teacher used an online resource, we are supposed to make two graphs based off the data and it's very unclear and unhelpful. :( I can't seem to find anything at all online that can solve this.

    We dragged a weight across two different surfaces using a force measure. We calculated the mass, the static force, the normal force, and the kinetic Friction. I'm having so much trouble right now with finding the Friction of the kinetic force! How am I supposed to get this? Should I just write down the kinetic force on the graph instead? I don't have the coefficient and I'm supposed to calculate it after....

    2. Relevant equations
    I know u=Ff/Fn, and Ff=u•Fn, but I don't have u??

    3. The attempt at a solution
    None yet. I'm very lost. :((
     
  2. jcsd
  3. Nov 3, 2016 #2
    Have you drawn a free body diagram?
     
  4. Nov 3, 2016 #3
    I did for the analysis questions....
     
  5. Nov 3, 2016 #4
    Let's see it.
     
  6. Nov 3, 2016 #5
    Ok. Actually, here is the whole thing if that clarifies things...
    untitled_by_shelilla-danavb6.jpg
    untitled_by_shelilla-danavak.jpg
     
  7. Nov 3, 2016 #6
    I think you'll find they key to this problem is to do with the applied force, just think about what the forces actually mean, and what they are doing. What does your friction equation you have written actually mean in terms of movement of the block?
     
  8. Nov 3, 2016 #7
    Well....that's the problem....frankly, I'm not sure what to do in order to get the Friction I guess... like, do I multiply the mass by the force of gravity and subtract it from the force applied? Is it the same as the static Friction? Should I use the coefficient from the static Friction to get the Kinetic? I tend to overthink things when I get stuck and it doesn't help at all that I can't find any similar references on how to solve it
     
  9. Nov 3, 2016 #8
    So, for static friction, Ff=uN, this the maximum amount of friction the block can exert on the floor (or vice versa), so this is the moment the block will begin to slide so force applied to make block move = maximum static friction.
    Now using similar logic, we now that coefficient of kinetic friction is (choose higher or lower) than coefficient of static friction, so this should mean to keep the block moving at roughly constant velocity the amount of applied force would need to be ______. And if it's not accelerating, all the forces must once again be balanced. Does this help you at all t start understanding the problem at hand?
     
  10. Nov 3, 2016 #9

    haruspex

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    I do not understand the numbers in your total mass column. In the first row, it is the same as the extra mass value. In the next two, it is 100g more than the extra mass, which seems right. In the fourth row, it is less than the extra mass(!)
     
  11. Nov 3, 2016 #10
    This is the free body diagram I had in mind.

    Friction.PNG

    W is the weight of the block and weights. N is the upward normal force exerted by the table on the bottom of the block. T is the tension in the wire applied by the spring scale. F is the friction force. This figure applies to all the cases, both static and kinetic, that you have been looking at. Since the block and weights are not accelerating in all cases, the force balances on the block (and attached weights) are:

    $$T=F\tag{horizontal}$$
    $$N=W\tag{vertical}$$

    So the spring is always measuring the friction force in all cases.

    Does this make any sense. Questions so far? (To be continued)
     
  12. Nov 3, 2016 #11
    Tbh I don't really know what I'm doing, sorry. I tried to just follow what my other group members wrote down (I was incredibly tired because I was fighting off a cold) and just left the columns that were already filled on the paper we were given the same. I think the teacher mentioned that we didn't have to do the extra mass, and one of the people my group said that since the Kinetic Friction required a formula, not to do it.
    I was really tired and out-of-it while we were doing this lab, so I don't remember everything clearly and didn't ask for everything to be clarified, so things are very confusing. Even more so because I had the day off school today but (I think) I have to hand it in tomorrow....sigh :(
     
  13. Nov 3, 2016 #12
    Thank you for showing me that diagram. It does, more or less, but um right now I'm finding it hard to calculate things/think right now without some sort of formula I can just paste on my work and use... is there anything you can think of that can just 'convert' it for me...?
     
  14. Nov 3, 2016 #13

    haruspex

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    Bizarre.
    The total mass is a key input to that formula. It's very easy to figure out. You are given the wooden block's mass as constant at 100g. If the extra mass on top is 200g, what is the total mass?
     
  15. Nov 3, 2016 #14
    I think so... struggling to think right now... does that mean then that the kinetic friction is just the negative kinetic force??
     
  16. Nov 3, 2016 #15
    Yes, well...we did it with metal weights on a slip of cardboard. I think the teacher just said to put down the mass of the weights....
    and yeah it was weird. I'm so pissed. The teacher just printed off some online lab and didn't even use the same materials.
     
  17. Nov 3, 2016 #16

    haruspex

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    Ah, that explains the 200g in the total mass column.
    But what about the 720g item? Is that because the extra masses you actually used were 200, 400, 600 and 720?
    Write the actual extra masses you used in both columns.
     
  18. Nov 3, 2016 #17
    Ugh, whatever. Thank you so much for all the replies and advice. I think I'm just going to ask my teacher the next time I see him.
     
  19. Nov 3, 2016 #18
    I'm not going to give you a formula. I can help you detive the formula, but you are going to have to work for it. Are you willing to try?
     
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