# Friction braking of a bar.

1. Dec 17, 2008

### capterdi

Hi,

Suppose a round steel bar of 25 mm diameter and 72 m lengh, which is traveling at 3.7 m/sec. I need to know which is the distance needed for the bar to brake by frictioning against a flat steel surface. Coefficient of friction is 0.3 and density of steel 7.85 ton/m3.

Thank you.

2. Dec 17, 2008

### Topher925

Do you need help finding an answer or do you just want someone to do this for you?

I would start by determining the mass of the rod and then finding the Lagrangian of the system.

3. Dec 17, 2008

### capterdi

I gladly would like someone to solve this problem for me, so I can see which are the formulas involved, and also the solution.

Thanks

4. Dec 17, 2008

### minger

It seems to me that you can just find the weight of the bar. Assuming the bar is horizontal, the weight will also be the normal force. Multiply said normal force by the coefficient of friction to obtain the friction force. Divide by the mass to get your acceleration. Divide the velocity by the acceleration to get the time needed.

5. Dec 17, 2008

### capterdi

Minger,

Ok...let me try and see what I get...

Thanks

6. Dec 18, 2008

### uart

The stopping time is,

$$t = \frac{v}{\mu g}$$

And the stopping distance is,

$$d = v t - 0.5 \mu g t^2$$

If you dont need to know the time then you can calculate the stopping distance directly from the equation,

$$d = \frac{v^2}{2 \mu g}$$

Where $\mu$ is the coefficient of friction, v is the initial velocity and g is the gravitational aceleration aceleration (approx 9.8 m/s^2).

Last edited: Dec 18, 2008
7. Dec 18, 2008

### capterdi

uart,

Ok...I can understand those equations, and have all needed data to solve them. Thanks a lot.

8. Dec 18, 2008

### capterdi

Sorry...didn't notice before: The mass isn´'t involved in the equations? I suppose it's not the same distance and time for a mass of 1 kg than for 1,000kg.

9. Dec 18, 2008

### minger

You should have derived these using my approach. You would have arrived at the same point, but with the knowledge of where they come from. The friction force again is the coefficient times the normal force:
$$F_f = \mu m g$$
This will be the only force acting on the bar, so we can relate it with the famous equation:
$$F_f = \mu m g = m a$$
Of course the masses will cancel out and you get the relationship between acceleration and coefficient of friction. From here, substitute the definition for acceleration:
$$\mu g = \frac{v}{t}$$
and of course simply solve for time. So, theoretically in an ideal world, yes, both a 1kg and a 1000kg bar moving at the same initial velocity will stop at the same time.

10. Dec 18, 2008

### uart

Yes it actually is the same, the mass cancels out in the calculations. Think of it like this, the 1000kg object will experience 1000 times the frictional force compared with the 1kg object, but the deceleration (a=F/m) will be the same.

Last edited: Dec 18, 2008
11. Dec 18, 2008

### capterdi

Right.

Thanks minger & uart for your help.