Μk sliding friction of a block down an Inclined Plane

[yawtsok]

Homework Statement

The data in my lab suggests that the mass of the block, M, does not have an effect on the inclination of the plane needed to have the block slide down with uniform velcoity.

But its asking why this is so? And I do not know why! Someone please help me ASAP.

Also, it wants we to derive the equation μk = tan θ.

There is a right triangle pictured with the right angle on the left side and the hypotenuse towards the right. Taking h to be the opposite side and L as the hypotenuse, I divided h by l to get the tangent. Than sin^-1 of that value to get the angle. Then tan θ to get
μk. But that's not exactly "deriving" it, right? How should I derive it then? Someone please help quickly ! :(

Homework Equations

μk = tan θ

The Attempt at a Solution

 

[cepheid]

Welcome to PF!

The relation you are trying to derive is one that holds under specific circumstances, namely that the friction force is just enough to prevent the block from accelerating down the plane. Draw a free body diagram of a mass, including all of the forces it would experience while on an inclined plane. Using Newton's 2nd law, you can figure out how the forces must be related to each other in order for this condition of static equilibrium to hold. Solve for mu.

As for the independence of the result from mass -- can you think of a reason why this might be? What does the frictional force that acts on the block depend upon (other than mu?)

EDIT: because of the circumstances I talked about at the beginning of this post, namely that tan(theta) represents the maximum angle before which a block will begin sliding, it makes more sense to use the coefficient of static friction here. Are you sure it was not supposed to say mu_s = tan(theta)?

 

[yawtsok]

Welcome to PF!

The relation you are trying to derive is one that holds under specific circumstances, namely that the friction force is just enough to prevent the block from accelerating down the plane. Draw a free body diagram of a mass, including all of the forces it would experience while on an inclined plane. Using Newton's 2nd law, you can figure out how the forces must be related to each other in order for this condition of static equilibrium to hold. Solve for mu.

As for the independence of the result from mass -- can you think of a reason why this might be? What does the frictional force that acts on the block depend upon (other than mu?)

EDIT: because of the circumstances I talked about at the beginning of this post, namely that tan(theta) represents the maximum angle before which a block will begin sliding, it makes more sense to use the coefficient of static friction here. Are you sure it was not supposed to say mu_s = tan(theta)?

hey, thanks so much for the quick reply.

Yes, it actually says muse of k and not s. Answering your previous question, I guess it depends on the material? Am i right or wrong. lolz

How should i derive the equation muse of K = tan (theta) ? :((

 

[cepheid]

Yes, it actually says muse of k and not s.

Okay. In that case, what tan(theta) represents is the maximum angle for which the block will slide at a constant speed if it is already sliding, and will stay still if it is already still. EDIT: to clarify, for angles higher than this, if it is already moving, then it will accelerate. But it will not necessarily begin to move if it is still, since static friction generally requires a larger force to overcome than kinetic friction.

Answering your previous question, I guess it depends on the material? Am i right or wrong. lolz

No. μ_k is what is determined by the two materials that are in contact. I asked you what ELSE the friction force depends upon OTHER THAN μ_k.

How should i derive the equation muse of K = tan (theta) ? :((

By following the steps I suggested in the first paragraph of my previous post! Remember, this condition only holds true when all of the forces are balanced (equilibrium). And by the way, the name of the Greek letter μ is 'mu', which is pronounced 'mew.' I don't know where you're getting 'muse' from.

 

[yawtsok]

Okay. In that case, what tan(theta) represents is the maximum angle for which the block will slide at a constant speed if it is already sliding, and will stay still if it is already still.



No. μ_k is what is determined by the two materials that are in contact. I asked you what ELSE the friction force depends upon OTHER THAN μ_k.



By following the steps I suggested in the first paragraph of my previous post! And by the way, the name of the Greek letter μ is 'mu', which is pronounced 'mew.' I don't know where you're getting 'muse' from.

the friction force depends on the normal force as well. right? so how is the normal force affecting it? about the muse my professor says it that way haha so i thought that was how it's spelt. :( i don't know how to draw a free-body diagram. but i am guessing there is a normal force from the plane acting on the block as well as the force of friction acting opposite from the motion and the force of gravity acting down on the block. :(

 

[yawtsok]

does the normal force balance out the weight of the block? :(

 

[cepheid]

the friction force depends on the normal force as well. right? so how is the normal force affecting it?

Yes! The normal force is what I was looking for there. So, why is the maximum inclination angle independent of mass? Intuitively you would think that since more massive objects have more weight, they would require a shallower angle in order to remain still (all other things, like the material, being kept the same). However, since the friction force depends upon the normal force (which also depends on mass), we end up with the result that although the more massive objects have a larger gravitational force pulling them down the plane, they also have a larger friction force available to prevent them from sliding down it. And it is larger by exactly the same factor. In other words, these two effects cancel each other out, and the max angle needed to stay still is the same regardless of mass. When you do your calculations, you'll notice that the mass cancels from both sides of your equation -- the theory will be in agreement with your experimental data.

about the muse my professor says it that way haha so i thought that was how it's spelt. :(

That's kind of unusual. Ok.

i don't know how to draw a free-body diagram. but i am guessing there is a normal force from the plane acting on the block as well as the force of friction acting opposite from the motion and the force of gravity acting down on the block. :(

Drawing a free body diagram for an object just means drawing that object by itself, removing any of the other elements in the system (hence "free"), and then drawing all of the forces that are acting ON that object. The reason for isolating the body is to make sure that the only forces that are included are the forces that are exerted ON that object. It removes the clutter and helps one avoid getting confused by other forces that might be exerted, for example BY the object on its surroundings. However, since this is a fairly simple system, such confusion is unlikely, and if you want to include the inclined plane in the diagram as well, that's fine. In fact, in this case it will probably help. You are correct that in this situation, your inventory of all of the forces that are acting on the mass will include 1. the friction force 2. the normal force 3. the component of the gravitational force (weight) that acts along the plane 4. the component of the weight that acts perpendicular to the plane.

Since the body is in static equilibrium, the forces have to balance in both the "along the plane" and "perpendicular to the plane" directions. These requirements will give you equations representing relations amongst 1, 2, 3, and 4. The result you are looking for will follow from these relations.

 

[yawtsok]

thank you so much but the block is actually moving with constant velocity and not in static equilibrium!

 

[cepheid]

That's still called static equilibrium. Remember Newton's second law. If the block's velocity is constant, then by definition it is NOT accelerating, right? If it is NOT accelerating, then by Newton's second law, the NET force on it is ZERO, which means that the forces on it must all be balancing out in the way I described.

C'mon man, try a little harder! THINK. Draw the picture. Set up the force balance equations.

 

[yawtsok]

That's still called static equilibrium. Remember Newton's second law. If the block's velocity is constant, then by definition it is NOT accelerating, right? If it is NOT accelerating, then by Newton's second law, the NET force on it is ZERO, which means that the forces on it must all be balancing out in the way I described.

C'mon man, try a little harder! THINK. Draw the picture. Set up the force balance equations.

lol, physics does not come so naturally to me. what are force balance equations? p.s. I'm not a man.

 

[cepheid]

lol, physics does not come so naturally to me. what are force balance equations? p.s. I'm not a man.

Right, sorry, I was just being colloquial. I'm talking about the equations you get when you consider the fact that the forces have to be in balance. The equations referred to by the boldface word 'equations' in my quote below.

Since the body is in static equilibrium, the forces have to balance in both the "along the plane" and "perpendicular to the plane" directions. These requirements will give you equations representing relations amongst 1, 2, 3, and 4. The result you are looking for will follow from these relations.

Draw the picture. Note that the forces have to balance in the direction along the plane, because the block is not accelerating along the plane. This gives you an equation (because it means that two of the four forces have to have magnitudes that are equal to each other). More specifically, the friction force acts "along" the plane, because it always opposes the motion. A component of the weight mg also acts in the direction along the plane. You can figure out what this component is in terms of mg and the inclination angle using trigonometry. You should now be able to write down the equation representing the force balance in this direction (namely that these two forces add up to zero i.e. they are equal in magnitude).

Now consider the condition that the forces acting in the direction perpendicular to the plane must be balanced (since the block is not accelerating in that direction either). The normal force acts perpendicular to the plane by definition. There is also a component of the weight that acts perpendicular to the plane. Once again you need to use trigonometry to figure out this component. Equate these two forces.

Finally, you need to consider the relationship between the friction force and the normal force. Short of doing your homework for you, which we don't do here, I cannot be more explicit than this.

 

[yawtsok]

Right, sorry, I was just being colloquial. I'm talking about the equations you get when you consider the fact that the forces have to be in balance. The equations referred to by the boldface word 'equations' in my quote below.



Draw the picture. Note that the forces have to balance in the direction along the plane, because the block is not accelerating along the plane. This gives you an equation (because it means that two of the four forces have to have magnitudes that are equal to each other). More specifically, the friction force acts "along" the plane, because it always opposes the motion. A component of the weight mg also acts in the direction along the plane. You can figure out what this component is in terms of mg and the inclination angle using trigonometry. You should now be able to write down the equation representing the force balance in this direction (namely that these two forces add up to zero i.e. they are equal in magnitude).

Now consider the condition that the forces acting in the direction perpendicular to the plane must be balanced (since the block is not accelerating in that direction either). The normal force acts perpendicular to the plane by definition. There is also a component of the weight that acts perpendicular to the plane. Once again you need to use trigonometry to figure out this component. Equate these two forces.

Finally, you need to consider the relationship between the friction force and the normal force. Short of doing your homework for you, which we don't do here, I cannot be more explicit than this.

Lol, I don't think he wants me to dig too deep into it. Perhaps saying that the Normal force balances the mass of the block out so mass is not a factor in determining Mu of K is enough.

 

[yawtsok]

is that enough cepheid ?

 

[cepheid]

Lol, I don't think he wants me to dig too deep into it. Perhaps saying that the Normal force balances the mass of the block out so mass is not a factor in determining Mu of K is enough.

I'm sorry, but what you've said here is not correct, and doesn't make much sense. You should re-read my explanation of why mass doesn't matter in post #7. More massive objects, in addition to having more weight, also experience more friction. The latter is due to the higher normal force between them and the surface they are sitting on.

is that enough cepheid ?

No, it really isn't. To "derive" a result means to mathematically show that that result is true. In other words, you should be able to arrive at that result algebraically, as a natural consequence of what you already know (the established principles that you start out with). In this case that major starting point is Newton's second law.

If you set up the equations in the way I suggested, then after a very small number of algebraic manipulations of those equations, the result that μ_k = tanθ WILL come out.

EDIT: Here is an idea of what I'm talking about:

Condition 1 - sum of forces acting parallel to the plane = 0

[1] (component of weight pulling down the plane) + (friction) = 0

Of course, one of these forces will be negative, since there are two directions "along the plane". For simplicity, you could say that the "down the plane" direction is positive, and the "up the plane" direction (in which friction points) is negative.

Condition 2 - sum of forces acting perpendicular to plane = 0

[2] (component of weight pushing into the plane) + (normal force) = 0

Again, these two forces have opposite sign.

[1] & [2] are the two equations that you get by considering the force balancing. I've expressed them in words, rather than symbolically. Now, you need to figure out these two components of the weight -- the parallel and perpendicular components. The full weight is mg, and the components will be related to this by trigonometric ratios involving the inclination angle. If you draw a picture (a right triangle) of the weight vector resolved into parallel and perpendicular components, it should be obvious what the components are.

That's it. Now I've really held your hand and guided you through it.