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!

The wording "just begin to move" for the force of static friction?

  1. Aug 9, 2014 #1
    I am confused with the wording for a static friction problem; this particular wording also lead me to a conceptual question on the force of static friction. It all started with a question in my text book that I figured out how to solve for the right answer. The question states, a box of m is sitting on an incline of 45 degrees and it requires an applied force F up the incline to get the box to begin to move. What is the maximum coefficient of static friction? The answer is (F-mgsin45)/mgcos45 by setting the equation as Applied force = mgsinpheta + Force of static fraction. However, the reasoning for the answer states that,"when the object "just begins to move" upwards the forces in both direction are equal and the force of static friction is at its maximum."

    My questions:

    1.)What does it mean when it says just begins to move or get the box to begin to move, does it mean just before it moves? I think the wording just begins to move means that it has moved(better word is budged) and it got out of the frictional groove that kept it stuck.

    2a.)For example, if the applied force was greater than friction static force, the object would have budged out of that groove with the surface, for that moment, with an acceleration. If the force was equal to the friction static force, it would have gotten out of that groove with zero acceleration, but this is where I am confused: I keep thinking there would be a constant velocity that is not zero that budged the object, at the moment, out of the groove of the surface.

    2b.)I know that the object should not be moving because an object at rest will have 0 velocity, but what about that initial budge?

    Please, I need help understanding this concept?

    Thanks Physicsforums!

    Sincerely,
    Jonathan
     
  2. jcsd
  3. Aug 9, 2014 #2

    Simon Bridge

    User Avatar
    Science Advisor
    Homework Helper
    Gold Member
    2016 Award

    Welcome to PF;
    1. you are correct - they mean in the instant just before it moves or exactly at the point where movement begins. Once movement is underway you no longer have a static friction problem.

    For an object mass m at rest on a flat surface, and we apply horizontal applied force ##\small{\vec F = F\hat\imath}##, the velocity ##\small{\vec v(t) = v(t)\hat i}## is calculated as follows: $$ v(t)=\begin{cases}
    0 & : F\leq \mu_s mg\\
    \frac{F-\mu_k mg}{m}t & : F > \mu_s mg
    \end{cases}$$

    2a. If the object is initially stationary, and there is zero acceleration, it's velocity does, indeed, remain constant... at zero.

    2b. what "initial budge"?
    In friction experiments, you sometimes start something sliding by giving it a slight tap.
    The closer the object is to overcoming the static friction, the smaller the tap that is needed.

    Note:
    - the applied force is always equal to the static friction force ... the static friction force varies to a maximum level whichis where the friction no longer holds.
    - you appear to have a common misconception about how friction works that may help you understand the model
    Friction is not caused by surface roughness.
    - IRL you will never have the situation where the applied force is exactly equal to the maximum possible static friction and the friction is not so well behaved at the critical point anyway. What you are learning is a calculation method.
     
  4. Aug 9, 2014 #3
    At the moment the force overcomes static friction, does the velocity of the object gradually begins to increase or it "jumps" to instantaneously begin its movement with some velocity already greater than zero?

    I of course expect the answer will be gradual velocity increase, but how can we tell, what would be the way to prove it or explain it?
     
  5. Aug 9, 2014 #4

    AlephZero

    User Avatar
    Science Advisor
    Homework Helper

    Newton's 2nd law: force = mass x acceleration.

    If the extermal applied force increases a tiny amount to start the object moving, the resultant force on the object changes from 0 to Fstatic - Fdynamic. So there is an instantaneous change in the acceleration, but the velocity increases gradually from 0.

    This is exactly the same as an object with no friction, where you "suddenly" apply a force F and it starts to move.
     
  6. Aug 9, 2014 #5

    Simon Bridge

    User Avatar
    Science Advisor
    Homework Helper
    Gold Member
    2016 Award

    The equation in post #2 assumes that the applied force overcomes static friction at t=0.
    Notice that friction is overcome when the applied force is greater than the max static friction - not when it is equal to.

    If I put the applied force
    $$F(t)=\begin{cases}\frac{\mu_smg}{t_0}t &:& t\leq t_0\\
    \mu_smg &:& t>t_0 \end{cases}$$ ... and the object is initially stationary, then F exactly matches the maximum static friction at t=t0. If it kept increasing, then the object would start accelerating from zero velocity. However, the way I wrote that out, technically the object can stay like that forever. (IRL other things intervene.) The slightest knock will set the object moving.

    This slight knock has to take place over a very small time δt and results in a small change in velocity δv.

    The resulting equations are:

    $$a=\begin{cases}0 &:& t\leq t_0\\
    \frac{\delta v}{\delta t} &:& t_0<t\leq t_0+\delta t\\
    (\mu_k-\mu_s)g &:& t>t_0+\delta t\end{cases} \\
    \implies v=\begin{cases}0 &:& t\leq t_0\\
    \frac{\delta v}{\delta t}(t-t_0) &:& t_0 < t \leq t_0+\delta t\\
    (\mu_k-\mu_s )(t-t_0+\delta t)g + \delta v&:& t>t_0+\delta t\end{cases}
    $$ ... this assumes the bump is produced by a force that is approximately constant - which is not normally the case (the F-t graph of a small bump usually looks like an inverted parabola). But it is a fair for our purposes because I want you to consider the situation that δt becomes very small ...

    Then consider that in real life there are always some small vibrations in the equipment constantly giving very tiny bumps too small to see.
     
    Last edited: Aug 9, 2014
  7. Aug 9, 2014 #6
    For most of these static problems I would just assume the velocity jumps from 0 to a certain value, and that should be OK. For other problems, an assumed constant acceleration of the COM can be used to determine the rise in velocity of the object.

    For even more advanced situations, the release of stress in the material and how that affects the movement of the object is needed. An object held in place by the static friction will be under shear stress by a body force such as gravity on an incline, or even by a point force, or other type of force, acting somewhere upon object.

    If perchance the static friction is overcome, or diminished, the strain at the bottom of the object has to decrease before any movement does occur. Initially the bottom of the object will move, and as the strain diminishes upward, the rest of the object will begin to move. The acceleration of the COM of the object may not actually jump from zero to a certain value but have a more complicated curve.
     
  8. Aug 9, 2014 #7
    Sorry Simon and Alpha, I just have two more questions: 8@...

    1.Does just begin to move mean that the applied force is at the point of overcoming the force of static friction?

    2.And lastly, if the applied force does overcome the force of static friction, does the object have an instantaneous velocity and acceleration at the point before we involve the force of kinetic friction?

    Thanks Simon Bridge and Alpha Zero! It is the best feeling to finally understand a concept after wrestling with it for a long time. I am starting to slowly understand. I especially loved the friction misconception website. You guys are the best!
     
  9. Aug 10, 2014 #8

    haruspex

    User Avatar
    Science Advisor
    Homework Helper
    Gold Member
    2016 Award

    Why? Suppose the force is gradually increased until it overcomes static friction, i.e. it is infinitesimally greater than maximum static friction. The applied force F then is effectively Mg sin θ + μs Mg cos θ. The net force up the plane is now F-μk Mg cos θ = (μsk) Mg cos θ. The acceleration is (μsk) g cos θ. No sudden jump in speed.
    If, as other posts have suggested, the motion is initiated by an additional tap then F at that time is some indeterminate value between Mg sin θ + μk Mg cos θ and Mg sin θ + μs Mg cos θ. There will in this case be a sudden change in speed, depending on the magnitude of the tap, and the acceleration anything from 0 to (μsk) g cos θ.
     
  10. Aug 10, 2014 #9
    Yes, although looks to me its subject to Zeno's paradox and the problem of infinite divisibility. Also, instantaneous change in acceleration sounds unnatural to me in the same way instantaneous change in velocity is, only not as much.
     
  11. Aug 10, 2014 #10
    Just begin to move is precisely defined by the moment when object's velocity becomes greater than zero, but it's still vague. It's hard to talk about that moment as it can not be pointed at specifically because of the problem of infinite divisibility - you can always say: "it was not then, it was a little bit sooner". To make it easier to think about it one should look at time intervals instead of time instants.


    Electric and magnetic, and gravity fields, which are responsible for friction, as well as other types of "collisions", have their gradients, so every interaction is always kind of like between rubber balls, it's spring-like. Therefore I'd say there can not exist such thing as instantaneous velocity or instantaneous acceleration.
     
    Last edited: Aug 10, 2014
  12. Aug 10, 2014 #11

    Simon Bridge

    User Avatar
    Science Advisor
    Homework Helper
    Gold Member
    2016 Award

    Bottom line: there is no hard-and-fast definition of what the statement means. You have to interpret it by using the context.
    In the context of a secondary school or freshman college assignment or exam question, they are usually referring to the instant before motion has started - as per the examples I gave you. However, you will have to use your judgement on this.

    That depends on the accuracy you need.
    I have actually already covered these questions in earlier posts.
    The bottom line is that the physical models you are being taught are very simple - so you can expect that they may not be a good match for reality in the fine details. In the model you have already, the velocity starts at zero and increases smoothly. You add in extra maths for each additional complication you want to account for.

    etc. In other words - the treatment depends on the situation.

    i.e. if you gradually tilt a surface until the object starts sliding?
    Are we using a rigid-body approximation or do we model all objects as a kind of stiff jelly?

    I suspect the context of post #1 is a secondary school, Newtonian physics, level.
    There are other models.
     
  13. Aug 10, 2014 #12
    As i said in the rest of my post, the problem will dictate whether the acceleration is important to determine, depending upon the complexity of solution chosen. or needed.

    In fact, one can come up with situations where the response of the object does not follow normal simple solutions and the dynamics from the instant that static friction is overcome warrant a consideration. As an example, for an object being held on an incline, the point of application of the force can set up a moment to attempt a rotation of the object, and the bottom can move either up or downwards from release of static friction, from the unstable situation where static frcition is just being balanced.

    As you have pointed out, a deeper analysis gives a deeper solution to what is happening.
     
    Last edited: Aug 10, 2014
  14. Aug 10, 2014 #13
    Thank you everyone! I figured it out! All of you have helped me to go deeper into this concept. Wow, sooo deep!
     
  15. Aug 10, 2014 #14
    I now have doubts about what I said about instant velocity and spring-like interaction.

    Let two neutrons move with half the speed of light on their course for head-on collision. At the very instant their velocity vector reverses back to the direction they came from (bounce off), is their velocity zero or already half the light speed in reverse?
     
Know someone interested in this topic? Share this thread via Reddit, Google+, Twitter, or Facebook




Similar Discussions: The wording "just begin to move" for the force of static friction?
  1. Static friction and work (Replies: 39)

Loading...