Angled forces and kinetic friction

[rasen58]

A 75 kg box slides down a 25.0° ramp with an acceleration of 3.60 m/s².

Find u_k between the box and ramp.

 

So I drew a free body diagram and angled it and then figured out almost all the values I need.

F_a is the force that the box already has as it's sliding down.

Fn is the normal force.

Fg is the force of gravity.

Fk is the force of kinetic friction.

 

The only one left is F_K but it seems like I'm doing it wrong. I thought it would be equal to F_gx + F_a but the answer says that I should be doing 311 - 270 and that is equal to F_K. Can anyone explain why that is? I thought it would just be the opposite of the other side of the x axis. Or did I draw my diagram wrong?

Edited July 21, 2014 by rasen58

[swansont]

F_a is the net force, right? What does it mean to be the net force?

[rasen58]

That's also what I was thinking, so I thought of doing F_a - F_gx but then I'd get a negative number, which I thought wouldn't make sense. But I just realized that it could actually make sense because it's just the opposite sign?

[studiot]

 

F_a is the force that the box already has as it's sliding down.

 

Where does this force come from.

That is : what provides it

or why does it exist?

 

And how does this fit in with Newton's Laws?

That is : which of Newton's laws says that a body 'has a force because it is moving'?

 

Swansont has already provided one strong hint.

 

IMHO it is more important that you properly understand this point than that you get the correct numerical answers to this problem (which you will do easily once you get the point).

 

Edited July 21, 2014 by studiot

[swansont]

That's also what I was thinking, so I thought of doing F_a - F_gx but then I'd get a negative number, which I thought wouldn't make sense. But I just realized that it could actually make sense because it's just the opposite sign?

 

Yes. These are vectors. You set the problem up with one direction being positive, the other being negative, so the sign gives you the direction.

 

How did you arrive at F_a - F_gx ?

[rasen58]

studiot: The force F_a comes from gravity pulling it down. And that relates to the second law which relates a force, mass, and acceleration.

 

swansont: Well I know that F=ma, and since we have a already (3.60 m/s²), I decided to calculate the force in the x direction only. So it would be F_x=ma_x.

And the force in the x direction is F_k or F_a+F_gx, so I used the second one. So that becomes (F_a+F_gx)=(m)(a) -> (F_a+311)=(75)(3.60) -> F_a+311=270 -> F_a=-41 N

[studiot]

 

studiot: The force F_a comes from gravity pulling it down. And that relates to the second law which relates a force, mass, and acceleration.

 

 

Yes and the body is accelerating

 

So it is not in equilibrium

 

So your free body diagram should not be an equilibrium diagram

 

This is what swansont (and your book) means by 'net force'

 

 

 

So can you write Newton's second law for the body in the direction of down the plane?

Remember that Newton's second law is a vector equation, so the force and the acceleration have direction.

Edited July 21, 2014 by studiot

[swansont]

studiot: The force F_a comes from gravity pulling it down. And that relates to the second law which relates a force, mass, and acceleration.

 

swansont: Well I know that F=ma, and since we have a already (3.60 m/s²), I decided to calculate the force in the x direction only. So it would be F_x=ma_x.

And the force in the x direction is F_k or F_a+F_gx, so I used the second one. So that becomes (F_a+F_gx)=(m)(a) -> (F_a+311)=(75)(3.60) -> F_a+311=270 -> F_a=-41 N

 

 

Yes, F=ma. That's the net force, which is the sum of all of the individual forces acting on the body. The net force, however, is not one of those individual forces.

 

You should never have F_net + F_other = … as a setup to a problem. It should always be of the form F_net = F₁ + F₂ + …

[studiot]

Perhaps the word resultant force instead of net force might make it more clear?

[rasen58]

studiot That makes a lot of sense now. So if it's not in equilibrium, then the left side of the x axis wouldn't equal the right side right? So I don't see how to make the equation if you can't determine the other side. The only thing I see is the F=ma and we have m of 75 and a of 3.6, so multiplying those would give you 270 N. But I think I already have that.

 

swansont Oh I think I see it now! Because it isn't in equilibrium like studiot said, the resultant force is equal to 270 N, which is the final force. So that means that because F_gx is going down at 311 N, but the F_k is pulling up at some number x, and the final resulting down force is 270, you can figure out x by doing F_a=F_gx+F_k, and that is 270 = 311 + F_k, so that means F_k is -41, but the negative makes sense because it's in the opposite direction, so it has a force of 41 N to the right.

 

I get it now! Thanks! F_a as the result makes sense. Thanks a lot.

[studiot]

 

 

 

 

 

 

 

studiot That makes a lot of sense now. So if it's not in equilibrium, then the left side of the x axis wouldn't equal the right side right? So I don't see how to make the equation if you can't determine the other side. The only thing I see is the F=ma and we have m of 75 and a of 3.6, so multiplying those would give you 270 N. But I think I already have that.

swansont Oh I think I see it now! Because it isn't in equilibrium like studiot said, the resultant force is equal to 270 N, which is the final force. So that means that because F_gx is going down at 311 N, but the F_k is pulling up at some number x, and the final resulting down force is 270, you can figure out x by doing F_a=F_gx+F_k, and that is 270 = 311 + F_k, so that means F_k is -41, but the negative makes sense because it's in the opposite direction, so it has a force of 41 N to the right.

I get it now! Thanks! F_a as the result makes sense. Thanks a lot.

 

 

 

I am glad you are beginning to see the light, but I sorry to tell you that my comments only addressed the beginning of the problem and you have quite a way to go to fully sort it.

 

It would make your life a lot easier for future problems if you tried out the questions I was suggesting, in the order I suggested them.

 

There are three and only three forces acting on the body.

All three of these may be replaced by a single resultant.

 

Two of these are a right angles.

Your question asks for the coefficient of friction, which is the ratio of two of them.

 

It is, however very, very important to realise that you either have the resultant acting or the system of three forces, but not both.

It is common to show the resultant dashed or in another colour or distinguished in some other way because of this.

 

The resultant is the sum of all the forces acting (in this case 3) it replaces them as an alternative.

 

This leads to the theorem that if the resultant is zero then the forces are in equilibrium, or that the acceleration of the body is zero, by Newton’s second Law.

You need this fact to complete the problem.

 

Turning to your use of x and y directions.

 

It is conventional to use x for the horizontal direction and y for the vertical.

 

So if an examiner sees them, that is what he will think.

 

The acceleration given is parallel to the plane, not vertical ie not along the x or y axis.

 

Can you tell me what is the acceleration perpendicular (normal) to the plane?

 

You then have two accelerations, one parallel to the plane and one perpendicular to the plane and you can write two Newton’s second law equations to obtain the two forces you require to find the coefficient of friction.

 

How are we doing ?

 

 

[rasen58]

There is no acceleration perpendicular to the plane right?

[studiot]

 

There is no acceleration perpendicular to the plane right?

 

 

Yes that is correct.

 

This leads to a very simple Newtons second law equation, that enables you to calculate one of the two forces involved in the coefficient of friction.

 

If the person who plonked -1 on my post#11 would like to state their difficulty, I will try to explain.

Edited July 28, 2014 by studiot

[rasen58]

Yes, and that is what I already did. Thanks.

[studiot]

Acceleration is a vector quantity.

 

What I was trying to do was to get you to understand what you were doing so that you could present it in an appropriate way.

 

The final arithmetic is trivial.

 

So if N is the Normal reaction and F the frictional force

 

The mass of 75kg is accelerating down a 25 degree slope.

 

Resolving acceleration perpendicular to the slope

 

Mass x acceleration = 75 x 0 = 75gcos25 - N by Newton’s second Law (Often called N2)

 

Resolving parallel to the slope Mass x acceleration = 75 x 3.6 = 75gsin25 - F by N2

 

This allows you to properly calculate [math]{\mu _k}[/math]

 

and has the advantage that if you make a mistake you have shown the examiner you know what you are doing so will gain all the available method marks

 

If you work this correctly you will note something interesting about the mass, which means that you do not need to calculate the actual values of the forces.

 

You should get an equation

[math]{\rm{acceleration}}[/math] = [math]g\sin \theta [/math] - [math]{\mu _k}g\cos \theta [/math]

ie independent of the mass.

 

What value did you make the coefficient of friction?

 

Interestingly this equation also tells us that for the body to slide down the slope

 

[math]\sin \theta [/math] > [math]{\mu _k}\cos \theta [/math]

 

If on the other hand

 

[math]\sin \theta [/math] < [math]{\mu _k}\cos \theta [/math]

 

Then the body will not move.

Further the frictional force will be

[math]mg\sin \theta [/math]

 

Which is less than the frictional force when moving.

That is the full limiting friction is not developed.

Edited July 28, 2014 by studiot

[rasen58]

Oh wow, thanks, That was detailed.