How does static friction work?

Uh... shouldn't you be using kinetic friction if the object is already in motion? I thought static friction applied only to surface interfaces at which no relative motion occurs..

well thats what i thought as well but heres the question maybe i read it wrong

The floor of a railroad flatcar is loaded with loose crates having a coefficient of static friction of 0.20 with the floor. If the train is initially moving at a speed of 40 km/h, in how short a distance can the train be stopped at constant acceleration without causing the crates to slide over the floor?

Haven't done this stuff in a long time, but I think you need the mass of the crates as well. If you have that, here is the procedure I'd use.

1) Use the mass of the crates to find the force they exert normal to the floor.

2) Use the coefficient of friction to convert the normal force into the force required to move the crates.

3) Use Newton's law to convert calculated force from step 2 into deceleration value for the train.

4) Use acceleration and initial speed of train along the relationship between position/velocity/acceleration to find the shortest distance the train can stop.

You don't need the mass. If you draw a simple FBD, you see that the horizontal forces are μ_s*m*g in one direction and m*a in the other direction. So the mass drops out and you're left with a = μ_s*g. Once you have the acceleration, you can find the time required to stop and then the distance.

ok on the procedure, but m*a is NOT a force...

Dimensionally, it technically is...It's sometimes called the 'inertial force' (term coined by D'Alembert) even though it's not a real force.

dmb wrote:

ok on the procedure, but m*a is NOT a force...

why isn't m*a a force? weren't we all taught F = m*a? Second Law?

No, no. The crates on the flatcar have a static friction of 0.20. The flatcar is moving 40 km/h.

So, you are finding the distance the train can stop without the crates sliding. Since F=-(us)(N), N=normal force, this is what you do:

1. The force of friction cannot be less than the applied forward force, or the crates will move. So, find N to find the frictional force.

2. Set the applied force equal to the frictional force. So...

-(us)N = ma

N= mg, so...

-(us)mg = ma

The mass will cancel out, now solve for a:

-(0.20)(-9.81) = a

3. Now that you know the acceleration, you can find time because (change in velocity)/(change in time) = acceleration. So...

(40 km/h - 0 km/h)/(x-0) = a

*The train's initial velocity is 40 km/h but when it stops it becomes 0 km/h.

4. It asks for distance, which is (change in velocity)*(change in time).

And you get the answer. That's how it's done.

By the way, the answer should be 815.2 km. I think that's right.

i dont think so. i got about 940m and i'm not sure if that's right.

F(friction) = (mu)*F(normal)

F(deceleration) = m(crate)*a(deceleration)

the fastest the cart can decelerate is whatever the force of static friction will alow, so we can set these two equations together and solve for a(deceleration).

a(deceleration) = (mu)*g(acceleration of gravity)

and v/a = v/(v/t) = t this is the time it takes to decelerate to zero, or you can look at the equation

where the net velocity will be zero when the deceleration reaches the negative of the initial velocity.

v(initial)= a(deceleration)*t

t = v/a

and the distance for this travel is a(deceleration)*(t^2) = distance d.

d = a(dec.)*t*t = (mu)*g*t*t = mu*g*v*v/(a*a) = v*v/a

= (40*40)/(.2*9.8) ~ 1600/1.96 ~ 940m.

940m is half a mile and 40 km/hr is about 25 mph, so i think 940m is closer to the right answer than 815 km (which is like 744 miles - that's an extremely slow deceleration) but at the same time im not sure if my answer is correct. it makes sense in my head though.

its possible that

d = .5*a*t*t

so d might equal 940/2 = 470m which is a little more realistic i think.

Physics wrote:

well thats what i thought as well but heres the question maybe i read it wrong

The floor of a railroad flatcar is loaded with loose crates having a coefficient of static friction of 0.20 with the floor. If the train is initially moving at a speed of 40 km/h, in how short a distance can the train be stopped at constant acceleration without causing the crates to slide over the floor?

Your original question should have been, "How do wheels work?" When the wheels of a railroad car roll without slipping, there is no kinetic friction with the rail. The bottom of the wheel does not slide against the rail -- it goes down, and then comes up, like your foot hopefully does when you're running. If the wheels of a railroad car roll *while* slipping, for example if you lock the wheels because you're slamming on the brakes, then there is kinetic friction (a side comment: you know that the coefficient of kinetic friction is always lower than the coefficient of static friction, so braking so hard that you lock the wheels of your automobile or railroad flatcar causes you to break more slowly than you otherwise would -- similarly, you run faster if your feet aren't slipping). So you need to figure out how hard you can slam on the breaks without having the wheels start to slip (how much force you can apply), and from there you can figure out how quickly you can accelerate (or decelerate).

Units!

Lets convert the initial velocity of 40 km/h to 100/9 m/s (roughly 11.1 m/s). Thus it will be compatible with an acceleration in m/s^2.

Relevant here are the max static friction of 0.2 * mg in one direction and ma in the other direction (pseudo force from the train's deceleration).

We want to find where these two forces equal each other.

0.2 * mg = ma

m cancels out.

0.2 * g = a

(g is 9.8 m/s^2 here on Earth)

a = 1.96 m/s^2

Plug this into one of our kinematic equations:

v^2 = v_0^2 + 2a(x - x_0)

given:

v = 0 m/s

v_0 = 100/9 m/s

x_0 = 0 m

a = -1.96 m/s^2 (negative because we're now considering the acceleration of the train with respect to the ground, which is opposite of the inertial force on the boxes with respect to the train)

find x

0 = 10000/81 - 3.92x

x = 31.49 m

"what is a force" is somewhat of a deep question...

If you accept an axiomatic definition of force in mechanics, then Newton's Laws define the concept of force. If you don't then you're left with a very difficult problem.

Within the context of freshman physics, inertial terms are not forces because they do not arise as the result of an interaction between two (or more) bodies that obeys Newton's Laws -i.e. inerial terms are not part of an action reaction pair. So, you're not actually being thrown up against the wall in the gravitron - your body is trying to move in a straight line at a constant speed and it keeps colliding with the wall.

D'Alembert, Lagrange et al use "force" in a very loose way anyway. For example if your generalized coordinate is angular, then your generalized "force" is actually a torque...

Impossible for the plane to take off

I guess this is why we have a demand for engineers!

smell my watch wrote:

find x

0 = 10000/81 - 3.92x

x = 31.49 m

This guy got it right. I can't believe how retarded the rest of you are. Is that how you really learned physics? Did they just teach you to pick out your favorite equation and plug stuff into it to find the answer? The guy who said 800 km must have been joking because that's absolutely ridiculous.

Think about it logically. You have a maximum deceleration of of 1.96 m/s^2. That means the velocity function is v = initial velocity - a*t. The train will stop when v = 0, so from that you can find how long it will take the train to decelerate to zero velocity, t_stop. Then all you have to do is integrate the velocity function from 0 to t_stop with the initial condition that d=0 at t=0.

t_stop is somewhere around 5.6 seconds. With that information alone you should know that 940m is a tad too high.

Do you use Halliday and Resnick? Because I think I just did that problem in the physics section I tutor...

Wow, Halliday and Resnick is still the text of choice. I used that for freshman year of college in 1980.

This is precisely the reason why I went to law school.