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Help with a problem (Friction)

  1. Aug 1, 2006 #1
    I'm currently taking grade 11 physics, and I can't seem to figure this question out for the life of me. I know this is only my first post, but I've tried to look every, asked a bunch of people and can't figure it out. I don't have a teacher since I'm doing it by Learn-At-Home. I'm hoping someone here can help me.

    a) Cathy takes the bus home from work. In her hand she holds a 2.0kg cake box, tied together with a string. As she ascends the bus, the box accelerates upward at a rate of 2.5m/s^2. What is the force being exerted on the string?

    b) Cathy sets the box on the seat beside her. The bus accelerates from rest to 60.0km/hr(16.7m/s) in 4.0s, and the box begins to slide. What is the coefficient of static friction(µ) between the box and the seat of the bus?

    c) A taxi suddenly cuts in front of the bus, causing the bus driver to slam on the brakes. The bus driver reduces speed from 60.0km/hr(16.7m/s) to 20.0km/hr(5.56m/s) in 1.5s. Does Cathy's cake slide forward?
     
  2. jcsd
  3. Aug 1, 2006 #2

    berkeman

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    -a- What is the fundamental equation that relates the Force F, the mass m, and the acceleration a of an object?

    -b- Using some of the info from -a-, what is the maximum horizontal force that friction can exert on an object? If you know the horizontal acceleration of the object just before it breaks loose, what does that tell you about the coefficient of static friction? What can you say about the coefficients of static and dynamic friction usually?

    -c- Once you get -a- and -b-, you should understand enough to get -c-, I believe.

    Show us some of your work, and maybe we can figure out what you are missing.
     
  4. Aug 1, 2006 #3
    a. F=ma

    b. I'm not sure about the mazimum force, but I'd assume it can't be any higher than the accelleration, or the friction itself would cause it to move. The acceleration is.. 4.2m/s^2 just before it breaks loose. What it tells me, I'm not sure. It's less than 1? The coefficient of static friction is less than the coefficient of dynamic friction.

    c. Yes, I'm pretty sure I can get this one, but the 2nd one is really stumping me!
     
  5. Aug 1, 2006 #4

    berkeman

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    -b- Are you familiar with the equation [tex]F = \mu N[/tex] ? What does that equation indicate?

    And BTW, the coefficient of static friction is usually higher than the dynamic one. Think about how something starts slipping -- it sticks up until the point where the static friction is overcome by too big of a sliding force, and once sliding, the thing keeps sliding. That's because the max coefficient is usually the static one, and once that's overcome, the dynamic friction is not enough to re-stop the object until the sliding force is reduced some. (Quiz question -- reduced by what ratio?)
     
  6. Aug 1, 2006 #5
    Yes, I know that that is the equation that I'm supposed to use. Sorry, I meant to put that static is higher.

    Anyways, my textbook gives me that the formula should be [tex]F_{f} = \mu F_{N}[/tex]. I know that the normal force is equal, but negative to the force of gravity. (2.0)(-9.8)<-[Because it's going down] = 19.6N. So, the Normal force is 19.6N. However, for the equation [tex]F_{f} = \mu F_{N}[/tex] I only have the Normal Force. How do I solve for the force of friction?

    (QUIZ: I have no idea what the ratio is :|. Taking phycics in Learn-At-Home was a bad idea. :uhh:)
     
  7. Aug 2, 2006 #6

    berkeman

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    That force is the maximum horizontal force that can be accommodated before the object slips. It is a scalar equation, not a vector equation, so the maximum frictional force is not in the same direction as the normal force due to gravity. Does that help?
     
  8. Aug 2, 2006 #7
    Not really. I know that the Normal force is in a different direction, but I'm not really sure how I'm supposed to solve for it, unless I use a different equation. I only know what I was given in the book, but I'm still missing too many variables. I'm beginning to think I'm just missing something extremely simple that would explain this entire problem to me.
     
  9. Aug 2, 2006 #8

    Doc Al

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    Hints for b:

    Assume that at the given acceleration, the box barely begins to slide.

    What is the acceleration of the box?
    How great a force is required to provide that acceleration?
    What force is providing that acceleration?
    What must the coefficient of static friction be?
     
  10. Aug 2, 2006 #9
    :| So i just use F=ma to find out the force acting on the box during said acceleration, and that force must be pretty much the same as the force that causes the box to move since it's at that point that it moves?!
     
  11. Aug 2, 2006 #10

    berkeman

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    Sounds like you're starting to get it!
     
  12. Aug 3, 2006 #11
    I can't thank you guys enough. I have never been so frustrated in my life as I have over this question!
     
  13. Sep 14, 2006 #12
    oddly enough, im taking the same course and having trouble with the same question, (dunno if i got it right).

    for b) i got myou = Ff/fN
    = -22.52N/-11.26N
    = 2

    does this sound accurate or am i missing something? (note: i have like a half page of calculations on this if more info is needed).

    ~Amy
     
  14. Sep 14, 2006 #13

    Doc Al

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    I suspect you are missing something. :wink:

    Show your work and we can take a look. (Don't just give equations, give your reasoning also.)
     
  15. Sep 14, 2006 #14
    the only thing i can think to change is make the gravity acceleration a negative.

    but here are the equations:

    acceleration = (16.667m/s/4s)
    = 4.17 m/s ^2

    Fnet = ma
    = (2 kg)(4.17 m/s^2)
    = 8.34N

    Fn = Fnet - Fg ((2kg)(9.80) = 19.6N = Fg)
    = 8.34N - 19.6
    = -11.26N

    Ff = mFn
    = (2kg)(-11.26N)
    = -22.51

    myou = Ff/fN
    = -22.52N/-11.26N
    = 2

    :smile:

    ~Amy
     
  16. Sep 14, 2006 #15

    Doc Al

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    OK.

    OK.

    Here's where you fell off the boat. (Note: Fnet acts horizontally; Fn & Fg act vertically.)

    Hint: There is only one (horizontal) force acting on the box. What kind of force is it?

    A general tip: Usually it's best to solve things algebraically (using symbols) until that last step when you plug in numbers. Sometimes things cancel out or simplify conveniently.
     
  17. Sep 15, 2006 #16
    the kinetic force.

    hows this:

    myou = a/g
    = 4.17 m/s^2/ 9.8 N/kg
    = 0.425

    :biggrin:

    ~Amy
     
  18. Sep 15, 2006 #17

    Doc Al

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    What is "kinetic force"?

    That's it.
     
  19. Sep 18, 2006 #18
    thanks :biggrin:

    for c) i got -7.41 for acceleration. (but didnt use the - for the equation since a negative myou probably isnt possible). or should i make it a negative anyways?

    so 7.51/9.8 = 0.756. i assume this is right. so the cake does slide. should i say that it slides in the opposite direction as it slid in -b-?

    and there's a lab assignment involving tissue boxes, a ruler, and an elastic band.. anyhoo, it says to determine how weight, speed and surface area affect friction. so im suppose to alter the surface area of the tissue box for the surface area part.

    it mentions nothing about different surfaces like (like carpet vs hard flooring). but one of the questions asks "calculate the average kinetic coeffient of friction for the two surfaces you used". when they say 'surfaces' do they mean different surface areas or different surfaces?

    any insight will be appreciated.

    also, does anyone know what approx a normal sized tissue box weighs? (dont have a small accurate scale here).

    and so if the elastic band stretchs twice it's normal lenth, the force = 2? :zzz:

    ~Amy
     
  20. Sep 18, 2006 #19

    Doc Al

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    Yes, that's the acceleration of the car. The minus sign just means it's slowing down instead of speeding up. (Not sure what you meant with your comment about a negative [itex]\mu[/itex].)

    Yes, the box slides. But what is your reasoning? You did some kind of calculation, but I don't know what that was supposed to be.

    The two surfaces are the surfaces that are rubbing against each other, producing the kinetic friction that you are studying. For example: tissue box & table surface.

    Why would you think that?
     
  21. Sep 18, 2006 #20
    for the calculation for the myou = a/g = -7.41/9.8 = -0.756

    it slides because 0.756 is bigger than 0.425 (as in -b-).

    :blushing:

    k.. was just reading my book and it says something to the tune of "take the mass of the object multiplied by accelaration of gravity (9.8 m/s^2) and divide the amount of stretching of the elastic band (using gravity i guess) = spring constant". and then i take spring constant times the amount of stretching in my experiment = the force in newtons. :smile:

    if i knew the weight of the tissue box, i'd be good to go.

    ~Amy
     
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