I’m doing some impromptu physics tutoring and ran into a question I don’t recall the answer to.
Suppose I have a box, and I’m pulling on it in the positive x-direction. I know that gravity acts on the box downward, and the normal force (in this case equal to gravity) acts upwards, and then there’s a force due to friction acting opposite my force. I know that friction force is equal to the coefficient of friction times the normal force. If my applied force exceeds that friction force, then I move the object. (Right?)
The question I ran into is - what if the friction force moving opposite the applied force exceeds my applied force? Based on my force diagram, then the object should be moving opposite my applied force, which is obviously incorrect, so there must be another force acting in the same direction as my applied force to precisely “counteract” the frictional force (in other words, this force must be equal to the frictional force minus my applied force).
Can someone give me a quick elaboration? I know there’s static and kinetic forces, and I suspect the answer is somewhere in there, but an internet search didn’t help me and I don’t have a physics textbook to consult (and, since it’s the summer, the student I’m working with doesn’t either.)
If your applied force exceeds that friction force you accelerate the object. If it’s equal you are moving at a constant speed (ignoring other forces like air resistance), and if it’s lower you are decelerating.
That’s all moving friction* mind you. Stationary friction could be zero, or anything between zero and the force needed to overcome the stationary friction.
*the simplified, actually tractable, physics class version.
The friction force is equal to or less than the coefficient of friction times the normal force. It has only the magnitude and direction require to prevent the box from moving, so it will never be larger than the force you’re applying to the box.
An easy way to think about this is: what would the frictional force be, and in which direction is it, if you’re not pulling the box at all? Obviously the frictional force is zero, because it has nothing to resist.
And also because if the computed final force ends up negative, the resulting acceleration is also negative… so the mass winds up being moving backwards relative to the normal force input. Which would look funnier than hell.
Right - that’s why I knew there was something else at work, since the box isn’t accelerating away from me if I try to pull it. I missed that the frictional force is less than or equal to, and thought it was just equal to. I guess it’s like the normal force in that regard, right - the normal force is “just enough” to balance out all the other vertical forces (assuming the box isn’t moving up and down).
Glad it was simpler than I thought - I was worried there was some other force acting on the box that combined with the frictional force to balance it all out. Thanks!
This is important, though it might not be for your tutoring situation. IIRC static friction is higher than sliding friction so the force need to move the object from stationary is less than that to keep it moving at a constant velocity. But that class was 45 years ago so I’m a little rusty.
There are two kinds of dry friction*, static and kinetic. Static friction applies when two surfaces are at rest with respect to each other, and kinetic friction applies when two surfaces are already sliding against each other. Kinetic friction is the easiest: It’s always given by f_k = \mu_kN, and always opposes the direction of the relative motion. It’s quite possible for kinetic friction to be larger than the force pushing the box along the floor, and for the acceleration to hence be negative: What this means is just that the box is slowing down (remember, kinetic friction only applies when the box is moving, so it’s always possible for the box to slow down).
Static friction is more complicated, but that complication is hidden by the fact that the formula looks almost exactly the same: f_s \le \mu_sN. So long as the forces involved are small enough, static friction will always be just big enough to counter whatever other forces you have that need countering, to keep the surfaces at rest relative to each other. But if the forces it’s trying to counter get too big, static friction won’t be able to keep up any more, and so the object will start sliding (and now you need to use kinetic friction).
This shouldn’t be too unfamiliar, since normal force and tension forces also work this way: Sit a brick on a table, and the normal force by the table on the brick is just large enough to counteract the brick’s weight… unless the brick is too heavy, in which case the table breaks. Hang a bucket from a rope, and the tension in the rope is just large enough to counteract the bucket’s weight… unless the bucket is too heavy, and the rope snaps. It’s just that we do more experiments in physics labs involving overcoming friction than we do involving breaking tables and snapping ropes.
And in fact, if we’re dealing with dry friction at all, \mu_s can’t be less than \mu_k. If it were, then if we put the right amount of force on an object, it wouldn’t be possible for the object to be either moving or stationary. Though it is possible for the two coefficients of friction to be the same, or very, very close to the same.
*(and a bunch of other kinds of non-dry friction, but we don’t usually talk about those, and dry friction is an approximation anyway)