I asked a question on frictional force an NO ONE could answer... Why?

Question

The problem was simple, I was only looking for the frictional force.

I gave two equations where the answer could be derived.

This is the problem:
A 1500-kg car is stationary on a ramp that makes an angle of 35 degrees with the horizontal. What is the magnitude of the force parallel to the ramp that keeps the car stationary?

I can't solve it. Please help.

So the answer is 8431.57 N? Is the formulae µ = Ff/Fn, µ = 4/3 correct? [If it is the answer would be 1.333]. Are they related?

Sorry, mr. bryan g, well I asked because our physics prof gave it to us and said we MUST get the correct answer because he's allowing us to consult as many well-educated physics people as we want. And no one's like that around here.

But mr. VB guy, the problem IS simple, I just could not get the frictional force because I thought the coefficient of friction is needed.

Answer 1

If the ramp is at angle 35 degree,

The normal reaction is mg cos 35. Fn = mg cos 35 = 12041.5 N

The frctinal force is now µ mg cos35. Ff = µ mg cos35 = µ*12041.5

The force pulling it down the ramp is mg sin 35.

The force that is keeping the car in equilibrium must be equal but opposite of mg sin 35.

mg sin 35 = 1500 x 9.8 x sin35 = 8431.6 N.

This 8431.6 N is inclusive of the frictional force and the force applied on the car to keep it equilibrium.

If P is the applied force then (P +µ*12041.5) = 8431.6 N.

From the above equation µ can be found only if P is given.

If we assume that P = 0.

then the car is under equilibrium only by the action of frictional force then

µ can be found by the simple formula µ = tan 35 = 0.7.


The above formula is derived as follows.

Since applied force P = 0, we have

Frictional force is µ mg cos35 and this equal to mg sin 35.

Therefore µ = tan 35.

The frictional force = 0.7 x 1500 x 9.8 x cos 35 = 8431.6

Answer 2

Find the portion of the force pushing the car down the ramp by multiplying 9.8 times 1500, and then multiplying this times sin(35). Since the car is stationary and the only force acting to "push it up" the slope is friction, this will be the answer.

Answer 3

It is equal to the partial force from gravity in the opposite direction.

1500-kg * g * sin(35)

Where g is the acceleration due to gravity. For something to be stationary, the forces must cancel out. if the car is restricted to the ramp, (which is to be assumed, because it is stationary) then the force down the ramp (gravity) must be equal to the force up the ramp (friction)

I can't believe you all are having such trouble with this!

Answer 4

Your question would not encourage self belief that the guy framing it really is conscious what they are talking about. "the motorcar stopped after a distance of 40 4.29 meters" no you'll degree a breaking distance to 4 decimal places, the position would 0 be? at the same time as the motive force sees the pedestrian?- How did you recognize? at the same time as he starts to brake? yet even the gap travelled in the course of the time taken to bypass a foot onto the brake will be over 0.01 meters. It basically tells us that it really is a poorly conceived question. "The Vertical portion of the rigidity is 88,2 N" What rigidity? No rigidity has been aspect out previously so this lower back sounds like a made up question the position the asker would not comprehend their own question. If it ability the downward rigidity of the motorcar through its mass, why not basically supply the autos weight? also 88,2 N even if it really is meant to be 88.2 N is loopy mild for a vehicle. even if that's the some stupid French way of writing 88200N it isn't a regarded format and is loopy too heavy. "at the same time as the horizontal portion of the rigidity is sixty 3,5 N" What rigidity? at the same time as? the web ahead rigidity appearing on the motorcar previously it starts breaking? The rigidity with which it hits the pedestrian? And lower back, the cost is basically stupid for something to do with a vehicle. the full question reads like it became made up through someone who's basically assigning numbers to issues with out any authentic theory of what they propose, what information is needed or a thanks to exhibit themselves obviously. it truly is why you received't get any 'exciting' solutions. all and sundry who's conscious what they are talking about received't supply this question a 2d look.

Answer 5

geez.
It's mg*sin(35).
Technically the static friction force must be this value
or larger.
You can see this from the force diagram. Remember, everything
must equal or else it will move. Since parallel to the ramp, the downward force is mg*sin(35), there must be an equal force up the ramp.
Now, since Ffric = mu*FN
mu = mgsin / mgcos = sin35/cos35 ~ 0.7
The coefficient of static friction is about 0.7

Does this help?

Answer 6

very easy, Use F=uR, F=minimum acting force to move car u=coefficient of friction R= normal rxn, R=mg cos 35, then substitute n get ans.
Hope u enjoy study physics.

Answer 7

If this is simple, why can't you answer it?

These people have got a point. If this is part of your homework, you should be doing it yourself, not just putting it on Y! Answers. Besides, you said it was simple.

Answer 8

m=mass
g=9.8m/s^2
theta=angle of incline=35deg
fs=static frictional force
* means multiply
downhill force=m*g*sin*theta
thus, for the car to be stationary, fs=m*g*sin*theta
fs=1500*9.8*sin35
fs=8431.57N

I checked my answer with a friend it was confirmed to be correct, good luck.

Answer 9

You can't solve it because there isn't enough information in the problem statement. You need to know the coefficient of friction.

-Guru

Answer 10

at 35 deg. the car is not "stationary" it starts rolling, unless breaks applied, or air in the tire released. or other tricky tricks at work, like ....35 deg to the side not forth/back...etc...