What is the Frictional Force on a Car Moving Down a 14% Incline?

[biotech4me]

Homework Statement

A 1,970-kg car is moving down a road with a slope (grade) of 14% while slowing down at a rate of 3 m/s^2. What is the direction and magnitude of the frictional force? (define positive in the forward direction, i.e., down the slope)?

Homework Equations

∫=μ*m*g*cosθ
Fn = m*g*cosθ
F= m*a

The Attempt at a Solution

i know m = 1970kg
a = -3m/s^2
g=-9.8m/s^2
So i converted the grade using arctan(14/100) = ~8°
then did tan(8°) = .14 = μ

Fn = (1970)(9.8)cos(8) = 19118N
19118*.14 = 2676.5N = ∫

thats as far as i could get but i know i missed a step somewhere. I did a similar problem and got it right but that was with a constant velocity. not sure where to throw that in here. would m*a be the normal force in this case?thanks in advance I've been reading these threads for a while and the community seems very helpful

 

[bossman27]

1.

The Attempt at a Solution

i know m = 1970kg
a = -3m/s^2
g=-9.8m/s^2
So i converted the grade using arctan(14/100) = ~8°
then did tan(8°) = .14 = μ

Fn = (1970)(9.8)cos(8) = 19118N
19118*.14 = 2676.5N = ∫

thats as far as i could get but i know i missed a step somewhere. I did a similar problem and got it right but that was with a constant velocity. not sure where to throw that in here. would m*a be the normal force in this case?

Why did you assume $tan(\theta)$ is the friction coefficient? That would mean that the friction coefficient is always just the grade, independent of what causes the friction. Obviously, that can't be the case.

To solve this problem, we don't even need to work with the normal force or friction coefficient, since they are both contained within $F_{friction}$ and we know all of the other necessary variables.

Here's how I worked it, where $F_{s}$ is the force "down the slope," or parallel to the slope surface:

$F_{s} = mgsin(\theta) = (1970)(9.8)sin(8) = 2686.88 N$

It is only accelerating parallel to the plane's surface, so:

$F_{s} - F_{f} = F_{net} = ma$

$2686.88 - F_{f} = 1970(-3) = -5910$

$F_{f} = 8596.88 N$

Then if you did want the friction coefficient, you would just divide that answer by the normal force. You would get:
$\mu \approx .45$

Edit: technically, I should have written $F_{s} + F_{f} = F_{net} = ma$, so that $F_{f}$ would have a negative sign since I defined "up the slope" as the negative direction, but you get the idea.

 

[AJ Bentley]

Hint:
Force = mass * acceleration
(Be careful what value you take for the acceleration. Remember acceleration is a vector)

 

[PhanthomJay]

Welcome to PF!

Homework Statement

A 1,970-kg car is moving down a road with a slope (grade) of 14% while slowing down at a rate of 3 m/s^2. What is the direction and magnitude of the frictional force? (define positive in the forward direction, i.e., down the slope)?

Homework Equations

∫=μ*m*g*cosθ
Fn = m*g*cosθ
F_net= m*a

The Attempt at a Solution

i know m = 1970kg
a = -3m/s^2
g=-9.8m/s^2
So i converted the grade using arctan(14/100) = ~8°
then did tan(8°) = .14 = μ

No, this method of calculating μ only applies when the object is moving at constant velocity.

Fn = (1970)(9.8)cos(8) = 19118N
19118*.14 = 2676.5N = ∫

thats as far as i could get but i know i missed a step somewhere. I did a similar problem and got it right but that was with a constant velocity. not sure where to throw that in here. would m*a be the normal force in this case?


thanks in advance I've been reading these threads for a while and the community seems very helpful

You have the correct equation for the normal force, but you don't need it in this problem. Apply Newton's 2nd law parallel to the incline after first identifying all the forces acting on the car. The friction force is one of the forces acting up the incline, what is the other force acting on the car down the incline? To solve for the friction force, you don't need to calculate the friction coefficient first...you can solve for it later if the problem asks for it. Watch signage.

 

[biotech4me]

sorry i have difficulty explaining thought processes.

i did (m*g*sin)/(m*g*cosθ) after eliminating the m and the g from both sides I am left with sinθ/cosθ. which is the same as tanθ


Fs+Ff=Fnet=ma

where Fs is the force of x so Fs = (1970)(-9.8)sinθ =-2686.88 (got something different here)
then Fnet = 1970(-3) = -5910N

5910- (-2686.88) so i got 8596.88? i don't even know anymore. going on 6 hours trying to wrap my head around this. starting to get delirious i will be back with fresh eyes in the early morning

 

[bossman27]

sorry i have difficulty explaining thought processes.

i did (m*g*sin)/(m*g*cosθ) after eliminating the m and the g from both sides I am left with sinθ/cosθ. which is the same as tanθ

(m*g*sinθ)/(m*g*cosθ) is simply the ratio of force parallel to the slope to force perpendicular to it. It makes sense that this would be equal to the grade(i.e. also the ratio of the height/length of the slope) but it doesn't really do anything for you in this problem since you already know the angle of the slope.

where Fs is the force of x so Fs = (1970)(-9.8)sinθ =-2686.88 (got something different here)
then Fnet = 1970(-3) = -5910N

5910- (-2686.88) so i got 8596.88? i don't even know anymore. going on 6 hours trying to wrap my head around this. starting to get delirious i will be back with fresh eyes in the early morning

You are actually correct. That was my mistake, I boneheadedly used the grade instead of the angle. I guess you aren't the only one who needs some sleep here.

If you check back, my previous reply will be corrected.