Finding Kinetic Friction on a Ramp: A Box Sliding Down at 1.20 m/s^2

[princesspriya]

Homework Statement

a.) a box slides down a 30 degree ramp with an acceleration of 1.20 m/s^2. Determine the kinetic friction between the box and the ramp.

Homework Equations

The Attempt at a Solution

the equation is uk=Fk/Fn but they did not provide me with atleast one force or one mass to figure out the Force normal or even force gravity so i am completely lost. please help.

 

[hotcommodity]

You won't need the mass to solve this problem. Let's just call the mass "m." You need one equation for the forces in the x-direction (let's assume the x-axis is parallel to the slope of the ramp), and one equation for the forces in the y-direction (with the y-axis perpendicular to the slope of the ramp). Can you show me how you would set up the force equations using Newton's Second Law?

 

[princesspriya]

well i know that Force downward-Force of fiction=mg; force downward is parallel to the ramp and so is force of fiction

 

[hotcommodity]

I'm not quite sure what you mean by "force downward," but I'll give you a more detailed hint. There are 3 forces acting on the box, the weight force W, the normal force N, and the frictional force f.

You will have two equations. One equation is the net force in the x-direction (along the slope of the ramp). The second equation will be for the forces acting in the y-direction, perpendicular to the ramp. Note that you will have to apply some trigonometry. Fill in the blanks:

\Sigma F_x = [ ] - [ ] = ma_x

\Sigma F_y = [ ] - [ ] = ma_y

When you fill in the blanks (the little boxes are blanks), note that you will have two weight terms (this is where the trig comes in). After you fill in these terms, think about the acceleration of the box in the x-direction, and the acceleration of the box in the y-direction (where is the acceleration zero?).

 

[princesspriya]

wouldn't there be four forces?
well for EFy=Fw-Fn=ma and EFx= Fd-Ff=ma

http://img516.imageshack.us/my.php?image=forcestp3.jpg

then Fn=mgcos30 and Fd=mgsin30 and Ff=uFn but i do not have the mass so that kind of where i am stuck

 

[hotcommodity]

wouldn't there be four forces?
well for EFy=Fw-Fn=ma and EFx= Fd-Ff=ma

http://img516.imageshack.us/my.php?image=forcestp3.jpg

Well if that is the diagram you're supposed to work off of, then yes, there would be four forces. I wasn't aware of the "downward force" force, was this a given force?

Assuming it is, this will change your force equations to:

\Sigma F_x = [ ] + [ ] - [ ] = ma_x

\Sigma F_y = [ ] - [ ] = ma_y

The force equations that you wrote are a start, and you're more than welcome to work off of those, but you're not taking any angles into account. It may help you to think about how you'd like to define your coordinate system.

 

[princesspriya]

well it slides down so it has to have the "downward force." i am sorry but its basically a self teaching class. but anyways i do not see why u added another blank for the EFx because would it just be Fd-Ff=ma

 

[hotcommodity]

Ok, I see what you mean by "downward force." You're saying that it is the x-component of the weight force. If I were you, I certainly wouldn't name components of forces, that will only make life harder than necessary :)

I'll show you how to set up the free body diagram and make life easy! Just give me a second.

 

[princesspriya]

okkies. lolz my teacher told me name it the downward force which i personally thought was stupid because its not quite going downwards but i guessed that's what everyone called it.

 

[hotcommodity]

I agree with you, haha. Ok, have a look see:

http://s62.photobucket.com/albums/h116/pepsi_in_a_can/?action=view&current=physhelp.jpg

Do you see how there are 3 forces acting. Also, so you see how the weight force has two components? You'll want to use that angle to figure out which weight component goes in the x-direction, and which one goes in the y-direction.

 

[princesspriya]

yea the one that goes in the x-direction is the so called "force downward" and the one that goes in the y direction is the "force normal"
but wouldn't i need mass in order to calculate the force of weight and then split it into its x and y components?

 

[hotcommodity]

Nope, the mass "m" will cancel out if we get our equations right. Now that you know which way each component of the weight force is pointing, see if you can fill in the blanks in the force equations below.

\Sigma F_x = [ ] - [ ] = ma_x

\Sigma F_y = [ ] - [ ] = ma_y

 

[princesspriya]

wouldn't it still be "Fd"-"Ff"=MAx and for the net force in Y i m kinda lost.

 

[hotcommodity]

wouldn't it still be "Fd"-"Ff"=MAx and for the net force in Y i m kinda lost.

Yes that would be the correct equation for the net force in the x-direction, but you'll do yourself a favor if you deal in terms of the weight W, instead of Fd. You understand the frictional force acting in the x-direction just fine, now let's look at the weight component in the x-direction (and forget about the "downward force," I know your teacher defined it, and shame on him/her, because it doesn't support any physics concept whatsoever).

So now you've figured out the net force in the x-direction:

\Sigma F_x = Wsin30 - Ff= ma_x

What other component of the weight should act in the y-direction, and what other force acts in the upward y-direction (you've already told me). Fill in the blanks for the forces in the y-direction.

 

[princesspriya]

ooo kk it would be W-Wcos30=MAy

 

[hotcommodity]

Very very close! Note that you decided to make the friction act in the negative x-direction, which means that the weight component Wcos30 will be positive in our equation. In other words, you flipped the coordinate system that I defined in my picture, which means that the positive y-axis will point into the ramp.

Also, you said that another "W" is acting in the y-direction. Take one more look at that diagram and tell me if this is true. If not, tell me which other force is acting in the y-direction (perpendicular to the slope of the ramp).

 

[princesspriya]

well the force that is opposite of Fn is acting in the y direction but they would cancel out each other, which would make sense since the forces in y direction would have to cancel in order to keep the box in place so that it would not go through the ramp into the Earth or up in the air. right?

 

[hotcommodity]

Keep in mind that we're adding up all of the forces in the y-direction. This is one of the ideas behind the force equations. So now you've noted that we have Wcos30 acting in the positive y-direction, and the normal force Fn acting in the negative y-direction. So we have:

\Sigma F_y = Wcos30 - Fn = ma_y

You've also stated that these forces "cancel out," which is another way of saying that the block does not accelerate in the y-direction, right? So if the acceleration in the y-direction is zero, a_y = 0, we can plug this into the above equation to simplify. Now we have:

\Sigma F_y = Wcos30 - Fn = 0

Can you solve for the normal force? How does the normal force fit into our equation for the forces in the x-direction?

 

[princesspriya]

well Fn=Wcos30 but that would have nothing to do with the x direction. and for the Net force in x wouldn't it be WsinX-Ff=1.20m. since 1.20 is the acceleration.

 

[hotcommodity]

well Fn=Wcos30 but that would have nothing to do with the x direction. and for the Net force in x wouldn't it be WsinX-Ff=1.20m. since 1.20 is the acceleration.

Yes, excellent observation. So now we have Fn = Wcos30 and our force equation for the x direction:

Wsin30 - Ff = (m)(1.20)

What is the weight force W equal to in terms of mass and gravitational acceleration (g)? Also, what is the frictional force Ff equal to in terms of the normal force and the given coefficient of kinetic friction?

And after knowing the above, what substitutions can you make into the force equation above?

 

[princesspriya]

well it would be 9.81msin30-u/fn=1.20m and that changes into 9.81msin30-u/(9.81mcos30)=1.20m. but we still cannot solve for m or u because there are two unknowns.

 

[hotcommodity]

well it would be 9.81msin30-u/fn=1.20m and that changes into 9.81msin30-u/(9.81mcos30)=1.20m. but we still cannot solve for m or u because there are two unknowns.

Oh you're right, this whole time I thought we were solving for \mu_k. The thing is, we can solve for \mu_k, because the mass will cancel out in the force equation for the x-direction. But as you pointed out, we don't know the mass "m," and so we can't determine the weight W, we don't know the normal force N, and we don't know the frictional force f_k. We're left with two equations and three unknowns, and we need at least as many equations as we have unknowns. Double check the problem to see if the mass, weight, normal force, or \mu_k is a given quantity, if not, you cannot solve for the frictional force.

 

[princesspriya]

hehe i understood it. i realized that you can cancel out the mass :) thanks a lot for the helP!

 

[hotcommodity]

You're most certainly welcome :)