# Homework Help: HELP .hardest question ever

1. Nov 2, 2005

### physicsfreak

HELP.....hardest question ever

a car on a horizontal road makes an emergency stop such that all four wheels lock and slide.
the coefficient of friction is 0.40. The whells are 4.2m apart
the center of mass is 1.8m behind and 0.75m above the front axle. The car weighs 1.1x10^4 N.
Calculate acceleration
Fn and Ff on each wheel

Can someone pleaseeeeee help me solve this question

2. Nov 2, 2005

### ranger

You didnt even try. Show what you have done. Thats the rules which you agreed to.

Last edited: Nov 2, 2005
3. Nov 2, 2005

### seeker03

"a car on a horizontal road makes an emergency stop such that all four wheels lock and slide." the coefficient of friction is 0.40. The car weighs 1.1x10^4 N.
<-------------------|------------>
-X +X
this means that the car is going to slid along the road, with a force four tenth's of its weight opposing its velocity forward. in other words, the only acceleration is going in the negative X direction and is equal to the force normal times the coefficient. This is eqaul to (11,000)*(4/10)N. This is true no matter how the weight is distributed, because
1) Ffriction = (Mu Friction (static or kinetic))(Force Normal)
2) Force Normal = W

The whells are 4.2m apart
the center of mass is 1.8m behind and 0.75m above the front axle.

This just means that you draw a triangle with 2 points being the front axle and the back axle, with length 4.2m, along with a point 1.8m back from the right side point and .75m up. This gives you a triangle, admitedly not a 90 X Y triangle, that you can solve the angles for using distance formula (to find side length) and using the Law of Cosines. Distribute the weight, and divide by two to find weight on each individual wheel within each of the two sets. Then, multiply the Fn by the Friction coefficient to find Ff.

Calculate acceleration
Fn and Ff on each wheel

Hope that helps a bit, although i dont know how to calculate acceleration of the car without any more information

4. Nov 10, 2005

### physicsfreak

i'm not sure but i think i could use torque to find the normal force on each wheel
F1 = front wheel F2 = back wheel
using the front wheel as the pivot point
sum of torque = 0 = F1(0) + (-1.1x10^4)(1.8) + F2 (4.2)
F2 = 4714.3 N F1 = 6258.7 N
Is that right?......does the 0.75m have to do with anything?
For force of friction:
Friction force of front wheel would be 0.4 x 6285.7 = 2514.28 N
friction force of back wheel would be 0.4 x 4714.3 = 1885.7 N

could you tell if i am doing it right and how to find acceleration.. maybe tell me some concept of finding acceleration in this situation.
Thank you
YOU are awesome!!

5. Nov 10, 2005

### physicsfreak

Thank you so much Seeker......from your explanation i understand the question a whole lot more

6. Nov 10, 2005

### lightgrav

you'd be fine, if the car was not accelerating.

But it is, isn't it?

Because this is NOT a static scenario, you REALLY WANT
to take torques around the c.o.m. and set them to zero.
(it is no longer your choice of axis ... without HUGE extra effort)

the 0.75 IS important ...
you know that a really tall thing is more likely to fall over.

7. Nov 11, 2005

### BerryBoy

I agree with lightgrav, because the force of friction is applied at the bottom of the wheel (which is below the centre of mass), there will be a torque of the system that will cause the NORMAL FORCE at the front to increase.

Torque is the way forward here.

Acceleration - Assuming C.O.M doesn't move up/down during this skid, then acceleration is due to the force of friction due to weight alone. Basically, you have to assume the car's suspension is broke!

At the end of the day, if there was a NORMAL FORCE greater than the WEIGHT, the car would start to fly! So acceleration, is the simplest to work out in this scenario.

a = (co-efficient of kinetic friction)(Weight of car)/Mass of car

Regards,
Sam

8. Nov 16, 2005

### physicsfreak

so.... i found out the acceleration to be
F=ma
-4400=1122a
a=-3.92m/s

and Force of friction to be
Ff= mu Fn
Ff= (0.4)(1.1x10^4)
Ff=4400N

now i am having problems with Fn now that the centre of mass needs to be incorporated. I know that I need to use torque to figure it out, but I just don't understand how to put C.O.M. into torque.
Sorry, I just begin taking physics and I don't understand a lot of concepts. So, if anyone could help me, it would be greatly appreciated. Thank You All