Using your common sense to solve a physics problem.

As any jerk scientist would tell you, do the math and you'll find the answer to this question.edited to add: took me 5 minutes, but then I'm only a jerk engineer.This message has been edited by DrJones*, 09-23-2004 05:05 PM

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*not an actual doctor

There isn't anything wrong with hands on experience. But you should understand that your first sentence works both ways. For all of your 24 years of hands on experience and common sense you might still miss something that I would see through my knowledge, honed by 4 well 7 but I meandered into history and anthropology as well years in a university. I pity the person that doesn't respect education. Sure I've met jerk engineers, but they're vastly outnumbered by jerk tradespeople/regular citizens who think the "college boy" doesn't understand whats going on. Common sense is great, but unfortunatley isnt all that common. Well then you're getting substandard engineers if they're working for that much. I don't know any engineer who started at less than $45K a year and thats in Canadian dollars.Have you figured out why the mass of the car wasn't given yet?This message has been edited by DrJones*, 09-23-2004 05:46 PMThis message has been edited by DrJones*, 09-23-2004 06:33 PM

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*not an actual doctor

Well its closer to $35K in exchange. I shouldn't have added the "Canadian Dollars" line as it is irrelevant. The canadian dollar has as much buying power in Canada as the US dollar does in the states. So an engineer making C$30K here is just as crazy as an engineer making $30K there.This message has been edited by DrJones*, 09-23-2004 07:01 PM

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*not an actual doctor

Well there are multiple solutions to this problem depending on the type of Aluminium and what I-beam design you use. The first combinination that worked for me (and I'll admit that I might be off as I'm a mechanical engineer not a civil and I haven't done this type of problem since my second year, which was long long ago):Aluminium 1100-H14
I Beam W310x143This might not be the best or most economical combo, it was just the first one that worked for me. And once again I add the disclaimer that I am not a civil engineer so I fully accept that I might have screwed up somewhere.But I dont think the purpose of this thread is to be a dick-measuring contest (or the appropriate female situtation).And once again you do not require the mass of the car to solve the initial problem. What Lam supplied in steps 1, 2 and 3 is more than enough. Start to set up the problem and you should see why the mass of the car is irrelevant.This message has been edited by DrJones*, 09-24-2004 02:39 AM

Nope. Acceleration due to gravity is a constant 9.81 m/s² here on earth no matter what angle the object is at. Check the dimensions (ie units of measurement) used in your formula:U_k is dimensionless
G your acceleration due to gravity is measured in m/s²
D is distance measure in m
v is velocity measured in m/sso inputing these units into your formula you getv=(2U_k*m/s²*m)²doing the math you get v=(2U_k m²/s²)²which works out to v=4U_k²(m⁴/s⁴)so your final velocity would have units of m⁴/s⁴ which just aint right.edited multiple times for spelling and making sure all the HTML tags are working.This message has been edited by DrJones*, 09-24-2004 02:45 AM

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*not an actual doctor

rotational energy has for some reason slipped out of my brain but rotational momentum (about a point O) is:H_O= r(cross)mv
r= position vector from point O to the object
m= mass of object
v= velocity vector of the object.
(cross) is of course the vector cross product operation.

That's they way I did it. 33 mph, he was speeding before he hit the brakes.

See you're mixing things up here. Gravity will always apply a force on an object that is directly straight down. The object will apply forces to the surface that its on. The direction and magnitude of these forces will depend on the angle of the surface. It's vector math. These resultant forces will control your acceleration down the hill. No its not. The units do not match up, it cant be correct. Its also not the formula you originally gave.g=m/s²
d=m
so 2U_k*g*d=(unitless)(m/s²)(m)=(unitless)(m²/s²)m²/s² is not the unit that we measure velocity with. Your formula is absolutely not right.This message has been edited by DrJones*, 09-24-2004 02:52 PM

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*not an actual doctor

The length of the skid is given as 30m. The U_k is a constant 0.45. You can't change the rate of gravity. Gravity is a constant 9.81 m/s² here on earth.edited to add: I'll type out a complete step by step solution tonight after I see Shaun of the Dead.This message has been edited by DrJones*, 09-24-2004 04:07 PM

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*not an actual doctor

Ok, now I see that you're on the right track you're just wording it in a confusing manner. I'll put up my step-by-step solution in about 3 hours unless theres any objections?

Starting from Part 2.U_k=0.45 Coefficient of kinetic friction.tan(d) = a/b in this case a= 8 and b= 100
so tan^-1 (8/100)= d
d=4.57^o(1) 2*a*(x₂ - x₁) = v₂² - v₁²the acceleration of the car a=is currently unknown
initial position x₁=0
end position x₂=30m
initial velocity v₁= currently unknown, also the final answer we are looking for
final velocity v₂=0So looking at equation (1) we see that we need to find the acceleration of the car during its braking before we can find the initial velocity.We now need a formula that incorporates acceleration.(2) F_T=ma
F_T= force exerted on the car during its braking. This force will be the sum of the frictional force and the component of the gravitational force that is acting along the surface.
m= mass of the car
a= the acceleration of the car during the brakingWe now need to figure out the forces on the car. Which are:The force of gravity F_G=m*g acting directly downwards
This force can be broken down into a component acting along the surface and a component acting against the surface.Force acting along the surface (3) F_P= m*g*sin(d)
Force acting against (90^o to the surface this is also known as the Normal Force) the surface F_N=m*g*cos(d)The frictional force will act along the surface opposite to the direction of motion. When determining the frictional force the force acting against the surface is multiplied by the coefficient of friction (either static or kinetic), so:(4) F_F=U_k*F_N=U_k*m*g*cos(d)So now we have the forces we need to figure out F_T.
If we set the direction that car is travelling as positive we find that:F_T=F_P-F_FAnd here begins the counterintuitive stuff. We don't need to find the value of F_T. We can sub in equation (2). So we getm*a=F_P-F_F
We then sub in (3) and (4) for F_P and F_F and we get:m*a=m*g*sin(d)-U_k*m*g*cos(d)The mass of the car just drops out (say A-2=5 and 10*A-10*2=10*5 once you do the math you'll find that A=7 in both equations), this is why you didn't need to know the mass of the car.a=g*sin(d)-U_k*g*cos(d)we now can find the acceleration of the car during braking. We throw in the appropriate values and get:
a=-3.62 m/s²We now can go all the way back to equation (1) and find v₁.(1) 2*(-3.62)*(30-0) = 0 - v₁²and doing the math we getv₁=14.73 m/s=32.95 mph.The guy was speeding when he hit the brakes.This message has been edited by DrJones*, 09-25-2004 01:19 AMThis message has been edited by DrJones*, 09-25-2004 01:23 AM

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*not an actual doctor