Advanced Forces Question, Air Resistance and Friction.

1. Feb 21, 2012

Victorzaroni

1. The problem statement, all variables and given/known data

A car with a flat front is accelerating in the horizontal direction so that an 80 kg person on the front face of the car stays in equilibrium. The coefficient of friction between the person and the surface of the car is .8. The resistance force by the air on the person is Fair=320 N.

The minimum acceleration of the car should be:
(A) 25
(B) 16.5
(C) 12.5
(D) 8.5
(E) 4

2. Relevant equations

Ff=μFN

3. The attempt at a solution

I know a free body diagram would help for this one, but I'm confused as to the directions of the air and normal force.

Attached Files:

• photo-2.JPG
File size:
26.8 KB
Views:
114
Last edited: Feb 21, 2012
2. Feb 21, 2012

PhanthomJay

Picture yourself as that person on the car's face. If the car is moving from left to right, which way will the force of air (the 'wind' if you will) act on you?

Normal forces occur at contact points perpendicular to the surfaces. Does that help you determine it's direction? Also remember that acceleration is always in the direction of the net force acting on the object or person.

3. Feb 21, 2012

Victorzaroni

okay, so the air would push against the person, with an arrow towards the left, and the normal force would push in the other direction, towards the right. Is the force of the air equal to that normal force?

4. Feb 21, 2012

kushan

you might need to add a pseudo force as car is accelerating

5. Feb 21, 2012

PhanthomJay

If the air and normal forces were of the same magnitude, then the car and person would not be accelerating in the horizontal direction. There is no acceleration in the vertical direction, since the person does not slide down. You need to look in the y direction first to determine, using Newton 1, the friction force and thence the normal force required to prevent the person from falling down. What are the forces in the y direction? Then use Newton 2 in the x direction to find the min acceleration.

6. Feb 21, 2012

Victorzaroni

A pseudo force?

7. Feb 21, 2012

kushan

8. Feb 21, 2012

Victorzaroni

Okay so the frictional force required to keep him from falling off (y-direction) would be Ff=μmg, and FN-Fair=ma for the x-direction?

9. Feb 21, 2012

Victorzaroni

the frictional force? which works out to choice E, 4.0 m/s2

10. Feb 21, 2012

PhanthomJay

Ok let's start from scratch...........in the y direction, the frictional force must be equal to the person's weight, per Newton 1. And since Ff = uN, then N = ?? and a = ???.

(You noted that Ff = umg, but this is not correct, sorry, Ff = uN =mg).

11. Feb 21, 2012

Victorzaroni

oh now I see what your saying. So Ff=mg=(80)(9.8)=784, divide this by u to get Fn, so Fn=980. Fn-Fair=ma, plug in, a=8.25m/s^2, so choice B

12. Feb 22, 2012

PhanthomJay

The solution rounds off g to10 m/s^2, so a = 8.5m/s^2, which is choice D .

13. Feb 22, 2012

Victorzaroni

Thats what I meant lol.