A 2.5 kg block and a 3.1 kg block slide down a 30 degree incline.

  • #1
Homework Statement

A 2.5 kg block and a 3.1 kg block slide down a 30 degree incline. The coefficient of kinetic friction between the 2.5 kg block and the slope is 0.23; between the 3.1 kg block and the slope is 0.51. Determine the (a) acceleration of the pair and (b) force the lighter block exerts on the heavier one.

The attempt at a solution

I need help with this problem. I started to do it and I figured that since they are together they would be accelerating together therefore I decided to treat it as one system to try to calculate acceleration. When I did this I came up with some outrageous answer. Could someone help me set this up?

Can you add the to coefficients of kinetic friction together and then use that number?
 

Answers and Replies

  • #2
dynamicsolo
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Since the 3.1 kg. block has the greater coefficient of friction with the surface of the incline, the only way this problem makes sense is to have the blocks in contact with the 3.1 kg. block downhill from the 2.5 kg. block.

You will need to make a free-body diagram of each block, which are each now subject to four forces: weight, normal force, kinetic friction, and the contact force. Three of these forces are arranged the same for both blocks. What does the contact force on each block look like? How do the magnitudes compare?

In answer to your last question, the coefficients of friction do not add together...

This raises the interesting question, "How do you find the contact force?" The answer is: you don't! If you work out the forces along the inclines on each block, calling the contact force, say, F_c, you'll end up with two equations in two unknowns, F_c and a. Solve for F_c in each equation and set the two results equal; we don't actually care what F_c is, much as we usually don't care what the tension T is in an Atwood's-machine problem, or the like...

The result should give a reasonable acceleration, which in fact should have the same value for both blocks.
 
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  • #3
The contact forces would be against each other correct? So those would be in opposite directions in my diagram. How do i figure out what that force is?
 
  • #4
dynamicsolo
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All you need to know is that the contact force points in opposite directions on each block. You do not need to evaluate the contact force. It is parallel to the incline on each block, since it acts perpendicularly to their contact surfaces, which are in turn perpendicular to the inclne.

For each block, write the equation for the force components acting parallel to the incline, and just call the contact force something (F_c is good). You will find that you have two similar looking equations, each of which can be solved for F_c. Set them equal to each other and you will be able to solve for a.
 
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  • #5
I'm really having trouble trying to set up an equation for this. Can you give me a better idea of how to set it up?
 
  • #6
dynamicsolo
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How much work have you done so far with inclined plane problems? You need to know how to break force vectors into components perpendicular and parallel to the incline in order to work out the force equations clearly. Have you done this before?
 
  • #7
Look at the attachment it has the two diagrams I drew. I can't seem to put together an equation for it. I haven't done a whole lot of these problems before.
 

Attachments

  • #8
dynamicsolo
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The attachment probably won't be cleared for hours, if history is any guide.

We'll look at each force in turn and how it acts on each block.

1) The weight forces, Mg and mg, are straight down, so what angle do they make to the incline? What is the component of weight that is perpendicular to the incline? parallel to the incline?

2) The normal force for each block points perpendicular to the incline, so it has no direct influence on the sliding of the blocks. But it is equal in magnitude to the perpendicular component of the weight, and we'll need that in order to find

3) the kinetic friction forces, which are parallel to the incline (since they oppose the motion of the blocks, which way do they point?). What is the relation of kinetic friction, f_k , to the normal force?

4) The contact force between the blocks is equal and opposite on each block (by Newton's Third Law). As we've said, the contact force on each block is parallel to the incline.

So the forces on each block parallel to the incline are:

1) the parallel-to-incline component of the weight of the block;

2) the kinetic friction ; and

3) the contact force.

The sum of these on each block gives the net force parallel to the incline, which will tell us something about the acceleration of each block (which is the same for both).

That's the set-up for the two equations.
 
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  • #9
1) The weight forces, Mg and mg, are straight down, so what angle do they make to the incline? What is the component of weight that is perpendicular to the incline? parallel to the incline?

The angle would be 30 degrees? Perpendicular would be sine? Parallel would be cosine?
 
  • #10
dynamicsolo
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The angle would be 30 degrees? Perpendicular would be sine? Parallel would be cosine?
The weight force (gravity) points straight downward. The incline is tilted 30 degrees to the horizontal. So the angle between the weight forces and the incline surface is...?

As for which trig function goes with which component, keep in mind that sine goes with the component "opposite" that angle, and cosine goes with the "adjacent" component.
 
  • #11
I also have a question regarding this problem.

It seems as though the people who have replied are addressing about how you would approach this question as a whole.

I found the acceleration of this system but I want to know how to do part b of this question.

I labeled the 2.5 kg block to be m1 and 3.1 to be m2 and the values of kinetic friction respectively mu1 and mu2

To find the acceleration, I set up the equation m1gsin(theta) + m2gsin(theta) - mu1*m1gsin(theta) - mu2*m2gsin(theta) = (m1 + m2)a

I found the acceleration.

How do I answer part b with this answer?
 
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