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!

Mass is independent of tendency to break static friction?

  1. Jul 3, 2015 #1
    Hello,

    I'm trying to reconcile what seems to me to be a contradiction.

    Last week, I did a lab experiment where we stacked some metal bars in a tray and measured the force required to pull the tray into motion and break away from the force of static friction. As we predicted, the more mass we placed in the tray, the more force was required to move the tray. So clearly, to me, as you increase the mass, you increase the weight, thus you increase the normal force, and so force of static friction increases since Fn(mue) = Force of Friction, which means the force required to pull the tray to overcome Fn increases as mass increases. I hope this is correct (or else I'll have to redo my lab report).

    However, now that I'm learning about centripetal acceleration I'm going back to the linear static friction lesson from last week and wondering why the formula seems to lead to the masses cancelling out:

    [itex][F_{net} = ma] = [F_{friction} = ma] = [μ(F_{N}) = ma] = [μ(F_{G}) = ma] = [μ(mg) = ma]

    [μ(g) = a][/itex]

    In our lab, we used one surface type. So μ remained constant. We calculated the weight of the cart using the same constant for gravity, 9.8m/s^2.

    I definitely felt that has we added weight to the cart, I had to pull with more and more force to break the cart into motion. So why now that I'm learning about centripetal force am I learning that m actually cancels out of the equation?

    If mass definitely has an effect on the frictional force, then why does it cancel out of the equation and why is said not to be a factor?
     
    Last edited: Jul 3, 2015
  2. jcsd
  3. Jul 3, 2015 #2

    FactChecker

    User Avatar
    Science Advisor
    Gold Member

    It takes more force to overcome the static friction of a heavier mass. It also takes more force to cause the acceleration of a heavier mass. So static friction and acceleration stay in the same proportions regardless of mass.
     
  4. Jul 3, 2015 #3
    If friction is the only force acting on the object, it will generate a fixed acceleration of μg. However, to generate that acceleration, the heavier an object, the more frictional force would be required.

    Think of it like gravity. Gravitational force is more on heavier objects and is denoted by mg. However, the gravitational acceleration is constant regardless of object's mass. In the same way, the frictional force would be μmg, but the accleration is a constant μg.
     
  5. Jul 4, 2015 #4
    Okay so μg is constant regardless of the object's mass. Does that mean that when the object breaks free of static friction it will be accelerating at a rate of μg? That is, in order to slide, does the object need a net acceleration of μg?

    Or will the net acceleration on the object be a different rate than μg?

    For instance in an elevator at rest, gravity is pulling on you with an acceleration of 10 but you are not actually moving at 10. In fact you are not accelerating at all. If the Force of Gravity were a bit stronger than the opposing force holding the elevator stationary, then the elevator would accelerate downwards. For instance If the force downward were Fg= mg= 500kg(10) = 5000N. And the force upward was F= ma = 500(11) = 5500N. Then the net acceleration would be 11-10= 1m/s^2 upward. The "applied accelerations are 10m/s^2 and 11m/s^2, but you are actually travelling at an acceleration of 1m/s^2.

    So for friction, it seems you are saying that μg is like the acceleration downward even though the object is not actually accelerating at all. Then when there's a force applied that is greater than and in the opposite direction to the force of friction, then a net acceleration will occur.

    For instance, if μ =2, then the force of friction is mμg=m(a). So if there is an imbalance such that mμg<ma then Fnet= mμg -ma = 10kg(2)(10) - 10kg(21m/s^2) = 10kg(21-20)

    So the actual acceleration of the object is only 1m/s^2 even though forces with much larger accelerations, 20 and 21, are acting upon it.

    Then that means that in the equation μg=a, a does not represent the actual (net) acceleration of the body. But the acceleration applied in the direction opposite the frictional force. The net acceleration would be μg - a = a_net, and the net force on the object would be Fnet = m(μg-a). Yes?
     
  6. Jul 4, 2015 #5

    Nugatory

    User Avatar

    Staff: Mentor

    The F in F=ma is always the net force. Thus, you don't even start thinking terms of accelerations until after you've identified and added all the individual forces together to determine the net force.
     
  7. Jul 4, 2015 #6
    I think you're confusing static and kinetic friction. Initially if the object is at rest on a flat surface it is in equilibrium with no net force. As begin to apply a force on the object static friction opposes your applied force until you reach the maximum static friction possible. The maximum static friction possible is μmg. Then when you apply a force greater than that maximum the object is set in motion with an acceleration of (Fapplied -μmg)/ m. As soon as I let go of the object, the only net force acting on it will be kinetic friction. It is this force that produces the acceleration μg which is the same regardless of mass. That is the rate at which the object will slow down.


    This is analogous to the example of tossing a ball up in the air. Once tossed, the only net force acting on the ball is gravity and it produces an acceleration, g, which is the same regardless of mass.
     
  8. Jul 4, 2015 #7

    I have 2 situations.

    Situation 1) I'm given a known mass and I measure the force required to set the object in motion. And then I calculate μ.

    For example, I start with a 1kg mass. Max static friction is measured with a force meter to be, say, 5N. Normal force is 1kg(10m/s^2) = 10N. fs = μ(FN), so μ = 5N/10N = .5.

    If I wanted to know the acceleration, I would have to realize that any motion will depend on the coefficient of kinetic friction , as umath1 advised. So if the coefficient of kinetic friction was μ=.2 , then the net force on the object would be 5N - μ(FN) = 5N - .2(10N) = 5N - 2N = 3N.

    So acceleration would be f/m = 3N/10m/s^2 = .3m/s^2

    If the coefficient if kinetic friction happened to be μ= 0, then
    acceleration would be (5N - μ(FN))/m = (5N - 0N)/1kg = 5m/s^2

    Situation 2). I am given a mass and coefficient of friction ( I believe it must be static friction).

    Lets say the mass is 1000kg and μ = .5. If I want to find the maximum acceleration without skidding, I can look at the situation in two ways.

    A). I can look at the mass as a car on wheels where the vehicle will travel at the indicated acceleration since the tires are not experiencing friction even though the car is in motion.

    Or

    B) I can look at the mass as a block, in which any movement will be a sliding motion that has overcome the static friction force.


    In perspective A, the equation μmg=a indicates that a is the actual rate of acceleration the vehicle is traveling, even though the vehicles tires have not yet overcome the static friction.

    In perspective B, the block is not analogous to the vehicle, but rather seem more analogous to the tires.

    So, [μmg = ma] = [.5(1000kg)(10m/s) = 1000kg(a)] = a = 2m/s^2

    In perspective A, the car is already in motion and travelling at a rate that is approaching 2m/s^2 before skidding.

    However, in perspective B, the block is not in motion at anytime until after 2m/s^2.

    But what does that mean?

    With the car you can actually measure the vehicles acceleration. With the block, there is no apparent acceleration occurring from 0m/s^2 to 2m/s^2, yet 'a' is the variable that determines whether the block will be set into motion.

    So it seems like although it takes exactly the same acceleration to overcome the force of static friction in each case, the acceleration is visible in the scenario with the car and hidden (as it is hidden in the vehicle's tires) in the scenario with the block.

    So the 'a' does not actually translate to motion in any object until after the object exceeds the acceleration μg.

    I think I was confused by the vehicles motion vs the tire's motion.

    Supposing μg = 1.999.../s^2 so the object begins sliding at, say, 2m/s^2. This does not mean the object actually accelerates at this rate because kinetic friction is already acting on the object from the moment it begins to move.

    So my question is A) is Everthing I said so far correct.

    B) Is the acceleration of an object at the moment it breaks free from static friction μg and then does it slow down very rapidly to a net acceleration of ((μs)(mg)- (μk)(mg))/m = (fs-fk)/m = F/m = a_net?

    Or is the acceleration always a_net from the instant it is in motion?
     
Know someone interested in this topic? Share this thread via Reddit, Google+, Twitter, or Facebook




Similar Discussions: Mass is independent of tendency to break static friction?
  1. Static friction (Replies: 6)

  2. Static Friction (Replies: 15)

  3. Static friction (Replies: 3)

  4. Static friction ? (Replies: 4)

Loading...