While I am asking more than one question in this thread they are all small concept tests that you can answer with a yes or no, and is all related to me understanding kinetic and static friction.

What I know:

Force of friction = Mu*N.

Static friction = The friction that keeps the object from moving.

Kinetic friction = The friction that is pushing against the motion of the object.

What I don't think I understand/or what I think is the problem:

I don't think I know what Mu(s), and Mu(k) really are, and I might be mixing them (or know if they are even one).

What I don't understand:

According to a physics teacher the force of static friction /= Mu(s)*N except sometimes.

How is that?

Qustion number 2 in the picture. http://gyazo.com/ad3f358af9f36c1ddcb2e8b4d93188c2

If an object is standing still it's static so how can the answer to the question be false?

The same box as in question 1. Question 2. http://gyazo.com/b5869084b618e96e6a1fe491274295f4 The answer is yes.

I think this one is yes because for the block to move the tension in the string needs to be bigger than the maximum static friction.

The same box as in question 1. Question 3. http://gyazo.com/52af207d0c0b55453cc200b875d3e179

Question 4. http://gyazo.com/52af207d0c0b55453cc200b875d3e179 It must be moving right? But not accelerate.

  • $\begingroup$ why didnt you use the trusted image service imgur provides that is inlined, I dont know what gyazo is. Re "How is that": the thing is, its an heuristic that works most of the time, there is no fundamental reason why it should be strictly proportional, but it makes sense that it is linear at least for small driving forces and experimentally it matches $\endgroup$ – Bort Jul 26 '15 at 19:58
  • $\begingroup$ You can search Gyazo up. It is completely safe and one of the most popular screen-shot capturing programs out there. $\endgroup$ – David Lund Jul 26 '15 at 20:03

The coefficient of friction is the ratio of the frictional resistance force to the normal force. The coefficient of static friction is the ratio of the maximum amount of friction that must be overcome to start an object moving, to the normal force. The coefficient of kinetic friction is the ratio of the amount of friction that must be overcome to keep an object moving at constant velocity, to the normal force.

The product of the coefficient of static friction and the normal force is the maximum friction force that must be overcome by an applied force in order to start the block moving. If the block is stationary, the applied force is less than the maximum required, but could be any amount between zero and the maximum.

If tension on the string is greater than the minimum kinetic frictional force, then the object is moving. As the minimum kinetic frictional force is only what's necessary to maintain constant velocity, there will be acceleration up to a point.

Here's an explanation of static and kinetic friction: http://hyperphysics.phy-astr.gsu.edu/hbase/frict2.html. The graphs plotting frictional resistance against applied force should make this more clear.

  • $\begingroup$ What is the the force between 0N until the object moves is called then if the static force of friction is = mu(s)*N only when the object is starting (or RIGHT before it is starting) to move. $\endgroup$ – David Lund Jul 26 '15 at 20:15
  • $\begingroup$ Thank you. That text cleared up a lot I think. I think I thought Mu(s) was all the force between 0N and until something moved, and Mu(k) was all the force after it was moved. $\endgroup$ – David Lund Jul 26 '15 at 20:31
  • $\begingroup$ @David Lund: There is no specific name, to my knowledge, for the various amounts of force which can be applied between zero frictional resistance and the threshold of motion. After the threshold has been crossed, even if applied force increases, kinetic friction will stay the same through a range of low speeds. Of course, the speed of the object will increase, but kinetic frictional force increases only at higher speeds. $\endgroup$ – Ernie Jul 27 '15 at 0:42
  • $\begingroup$ I don't understand why they put this stuff on hold as too broad. Reading the few (and my comments about the) questions and you'll clearly understand what I don't understand. This guy clearly understood what I needed help with, and answered everything within a few paragraphs. The questions are all connected. It's like asking, what is 2+2,4+4, and 5+5. You can answer all these questions by answering the general idea of adding. $\endgroup$ – David Lund Jul 30 '15 at 13:01

Not the answer you're looking for? Browse other questions tagged or ask your own question.