Clarification on Frictional Forces?

  • Thread starter Thread starter mneox
  • Start date Start date
  • Tags Tags
    Forces
AI Thread Summary
Frictional forces arise from the interactions between a body and the surface it rests on, with static friction opposing applied forces until a limiting value is reached. The static friction force can be calculated using the equation Ff = μsN, where μs is the coefficient of static friction and N is the normal force. Once the applied force exceeds this limiting value, the body begins to move, and the frictional force transitions to kinetic friction, represented by fk = μkN. In scenarios where the body is stationary, the applied force equals the static friction force, while in motion, the net force is determined by the difference between the applied force and the kinetic friction force. Understanding when to use static versus kinetic friction is crucial for accurately calculating frictional forces in different contexts.
mneox
Messages
35
Reaction score
0
Hi, I am requiring some assistance in getting some clarification on frictional forces.

I know that Ff = u(Fn).

How do you generally find the frictional force though? I noticed how sometimes Ff is the same as the applied force, and sometimes you find it by using Fnet = Fa - Ff.

Why is this? And how do I know when to use which? Thank you!
 
Physics news on Phys.org
Frictional force acts on a body due it's interactions with the surface on which it rests. So, when you apply a force on the body, there results an opposing force due to these interactions; the body being static all the while, the force is due to static friction. It is experimentally found that this force is proportional to the normal reaction acting on the body due to the the surface. Now when you go on increasing the applied force, then for a particular value of it, the frictional force reaches a limitting value,i.e, on applying further force the body will move. Thus,
fs=\musN
where fs is the force of static friction, N is the normal reaction, & \mus is the the coefficient of static friciton. \mus is defined only for the limitting value of the static frictional force, acting on the body at rest.
On applying a force greater than the limitting value of the frictional force, the body starts translating, and now the value of the frictional force reduces to smaller value. For this condition, \muk, the co-efficient of kinetic friction is defined, and the force is called the force of kinetic friction. We have,
fk=\mukN
So, when you're given the condition that the body is just about to move, force applied fa is,
fa=fs
But when the body is translating, the net force is given by,
Fnet=fa-fk
 
The rope is tied into the person (the load of 200 pounds) and the rope goes up from the person to a fixed pulley and back down to his hands. He hauls the rope to suspend himself in the air. What is the mechanical advantage of the system? The person will indeed only have to lift half of his body weight (roughly 100 pounds) because he now lessened the load by that same amount. This APPEARS to be a 2:1 because he can hold himself with half the force, but my question is: is that mechanical...
Hello everyone, Consider the problem in which a car is told to travel at 30 km/h for L kilometers and then at 60 km/h for another L kilometers. Next, you are asked to determine the average speed. My question is: although we know that the average speed in this case is the harmonic mean of the two speeds, is it also possible to state that the average speed over this 2L-kilometer stretch can be obtained as a weighted average of the two speeds? Best regards, DaTario
Some physics textbook writer told me that Newton's first law applies only on bodies that feel no interactions at all. He said that if a body is on rest or moves in constant velocity, there is no external force acting on it. But I have heard another form of the law that says the net force acting on a body must be zero. This means there is interactions involved after all. So which one is correct?
Back
Top