How can Superman minimize the braking distance?

[TheChemist_]

Homework Statement

So here is the Problem: A big truck (m=50t) is going at 100km/h. Now the truck can't brake anymore and Superman comes to aid. He (has a mass of m=100kg) tries to stop the truck while he is wearing shoes with a rubber-texture and a friction coefficient of µ = 0.8. Assume that F(brake) (the deceleration applied by superman) of superman is anti-parallel to the movement of the truck.
a) How long is the braking distance?
b) How can Superman minimize the braking distance? Reminder: The road should still be intact and remember that Superman is also super-strong. A simple sentence is enough

Homework Equations

The Attempt at a Solution

I have thought about this problem but I am still very insecure if my attempt can be correct. I assumed that I can use the formula for s (distance) from the "Velocity-indipendent-friciton" we learned which would be:

s(T) = 1/2*(v₀^2/(g*µ)) T is the moment when the truck stops

v₀ = 27.77m/s (100km/h)
g = 9.81 m/s
µ = 0.8

I used the formula s(t) = v₀*t - 1/2(µ*g*t^2) and inserted T = v₀ / (µ*g)

I really need advice on this one and I don't know if my thought process is correct.
Thx

 

[BvU]

Hello Chemist,

I suspect the ink of your symbol for the friction factor has bleached considerably in this post . Let's use $\ f \$, defined by the maximum friction force $F_{\rm friction, \; max}$ (or just shorthand $F\$) and the normal force $\ N\$ through $F_{\rm friction, \; max} = f N\$.

You correctly use the SUVAT equations for uniformly accelerated motion$$
s(t) = {1\over 2} * a t^2 $$ and $v = v_0 - a t \Rightarrow T = (v_0 - 0)/a$ where $F_{\rm f} = f mg = ma \Rightarrow a = fg \$.

However, what's superman doing in this story ? does he stand in front of the truck and skids over the road ? Then $F_{\rm f} = f * 100 g$ and that has to brake the whole 50.1 ton (meaning your formula doesn't fly -- two masses involved instead of one) ?

 

[haruspex]

I assumed that I can use the formula for s (distance)

In case it is not clear from BvU's post, your assumption is wrong. That formula only applies in certain circumstances. Better to go back to first principles and consider forces.
The question is not well phrased. As b) implies, there is something Superman can do to minimise the distance. What that is should become evident when you consider the forces. But it is not clear whether a) is to be answered with or without assuming Superman does that. My guess is your are supposed to assume he does not.

 

[TheChemist_]

OK, first of all let me thank you for your fast reply, and secondly: Yeah superman stands in front of the truck and is skids over the road. Is my formula still correct or how do I approach it correctly if not?
(Probably not as I assumed because there are two masses...
but could one say, that Superman and the truck can be considered just ONE mass from the moment of braking onwards?)

 

[TheChemist_]

Hello Chemist,

I suspect the ink of your symbol for the friction factor has bleached considerably in this post . Let's use $\ f \$, defined by the maximum friction force $F_{\rm friction, \; max}$ (or just shorthand $F\$) and the normal force $\ N\$ through $F_{\rm friction, \; max} = f N\$.

You correctly use the SUVAT equations for uniformly accelerated motion$$
s(t) = {1\over 2} * a t^2 $$ and $v = v_0 - a t \Rightarrow T = (v_0 - 0)/a$ where $F_{\rm f} = f mg = ma \Rightarrow a = fg \$.

However, what's superman doing in this story ? does he stand in front of the truck and skids over the road ? Then $F_{\rm f} = f * 100 g$ and that has to brake the whole 50.1 ton (meaning your formula doesn't fly -- two masses involved instead of one) ?

I am sorry that the symbol for the friction factor has been bleached away...
Ok so you are saying that I could firstly find out what the acceleration a (or in this case the deceleration) is and then insert that into the SUVAT equation to find s?

 

[haruspex]

could one say, that Superman and the truck can be considered just ONE mass from the moment of braking onwards?)

for some purposes but not for others. Please treat this as a multiple body problem: draw a free body diagram for each and consider the forces on each.

 

[TheChemist_]

I have to say that Superman is just standing there and not applying any extra force.
Ok, and what if I consider an approach over the energies of the two bodies, like this:
E(truck) = 1/2 * m(truck)* v^2(truck)
and
E(superman) = F(braking)* s where F(braking) = m*g*µ and s = braking distance (yeah I know F*s is normally Work but let's just call it E(superman))

then I have: E(truck) = E(superman)

1/2* m(truck)* v^2(truck) = m(sup)* g* µ* s

so:

s = m(truck)* v^2(truck) / (m(sup)*g*µ*2)

I would then get s = 868.06m

What about this?

 

[TomHart]

s = m(truck)* v^2(truck) / (m(sup)*g*µ*2)

I would then get s = 868.06m

Are you sure you did that calculation right?

 

[TheChemist_]

Are you sure you did that calculation right?

Holy moly, I think I messed something up big time
ok it should be s = 24099m

But is the assumption correct?

 

[TomHart]

But is the assumption correct?

Well, that's a good question. Where was superman before he tried stopping the truck? Did he fly and land in front of the truck at the same speed, or was he stopped in the road in front of the truck with his hands held out and the truck hit him pushing him backward? (I think I would opt that he was moving the same speed as the truck when he positioned himself in front of it.) But once he gets positioned in front of the truck trying to stop it with his feet skidding on the ground, doesn't his own mass have to be included as part of the overall kinetic energy?

 

[TheChemist_]

Well, that's a good question. Where was superman before he tried stopping the truck? Did he fly and land in front of the truck at the same speed, or was he stopped in the road in front of the truck with his hands held out and the truck hit him pushing him backward? (I think I would opt that he was moving the same speed as the truck when he positioned himself in front of it.) But once he gets positioned in front of the truck trying to stop it with his feet skidding on the ground, doesn't his own mass have to be included as part of the overall kinetic energy?

No, it says that Superman is standing there and "waiting" for the truck and then stopping it. (although that would't work because the truck would be totally smashed, but it's physics for not physics students :P)

 

[TomHart]

although that would't work because the truck would be totally smashed

A big truck hitting a 100 kg man at 100 km/hr would certainly dent up the front end, but it wouldn't demolish the truck. So if it really is a collision between the truck and superman (and an inelastic collision since they stuck together), that collision should also be taken into account before your work/energy calculation. Any ideas how to manage that?

 

[TheChemist_]

A big truck hitting a 100 kg man at 100 km/hr would certainly dent up the front end, but it wouldn't demolish the truck. So if it really is a collision between the truck and superman (and an inelastic collision since they stuck together), that collision should also be taken into account before your work/energy calculation. Any ideas how to manage that?

Well in the text it says that we should not consider any inelastic collision energy...it sounds strange I know, but it's simplified

 

[TomHart]

Well in the text it says that we should not consider any inelastic collision energy...it sounds strange I know, but it's simplified

Okay, that settles that question.
But back to the work/energy calculation: What if you had a man on a bicycle going 100 km/h who lost his brakes and a big truck positioned itself in front of him and locked up the truck brakes to help stop the man on the bicycle. Would the friction force from the truck's tires only be used to stop the bicycle, or would it also be needed to help stop the truck also?

 

[haruspex]

1/2* m(truck)* v^2(truck) = m(sup)* g* µ* s

Yes. Do you see how this differs from your original equation, which did not involve masses? This will be important in answering part b.

ok it should be s = 24099m

Looks about right.

 

[BvU]

Are we ready now to address part (b) ? Because skidding for 24 km is bound to end up prematurely and sup needs to think of a more effective way to avoid disaster !

 

[TomHart]

Are we ready now to address part (b) ?

I know this isn't my problem, but I still have not gotten past part (a).

then I have: E(truck) = E(superman)
1/2* m(truck)* v^2(truck) = m(sup)* g* µ* s

I think the equation should be:
(0.5)(m_truck + m_superman)v_truck² = m_supermangμs
Can someone please confirm or deny?

 

[BvU]

Posts #11 and #13 dealt with that small detail: we'll ignore it. A reasonable thing to do if e.g. a friction coefficient is known to 1 decimal only.

But confirm your expression I can: provided supie lands gently at the front bumper with 100 km/h Post #11 says not so...

My ${1\over 2}Mv^2 = \mu\, mg\, s$ gives 24.58 km for 50 ton, 0.2% more for 50.1 . How did Chemi find 24.1 km ? (**)

(I see the friction coefficient $\mu$ is back in post #1 -- maybe the bleach was only in my computer)(**) Ah, I see, 27.77 m/s instead of 100/3.6 and 10 m/s² instead of 9.81 .

 

[TomHart]

provided supie lands gently

I didn't know you and he were that close that you can address him on a nickname basis. For me it's Mr. Superman.

How did Chemi find 24.1 km ?

Hmmm, I don't know; I guess I never checked his math. I just now tried using g = 10 ms^-2 and it comes out to 24.1 km.

Thank you for your response.

 

[TomHart]

maybe the bleach was only in my computer

And just so you know, yes, I think it was only on your computer because I saw your original "bleach" comment and verified that I could see the μ.

 

[TheChemist_]

Sry, since I solved the problem over energy I didn't really bother coming back. For those who wondered: Yeah I used g = 10 ms^-2

This thread can now be closed! (can I close it)

Thank you everybody for helping me out!

 

[haruspex]

Sry, since I solved the problem over energy I didn't really bother coming back. For those who wondered: Yeah I used g = 10 ms^-2

This thread can now be closed! (can I close it)

Thank you everybody for helping me out!

OK, but what did you answer for part b?