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Friction with two bodies

  • #1
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Homework Statement


A block of mass m = 1.85 kg rests on the left edge of a block of larger mass M = 6.65 kg. The coefficient of kinetic friction between the two blocks is 0.310, and the surface on which the 6.65kg block rests is frictionless. Aconstant horizontal force of magnitude F = 20.10 N is applied to the 1.85 kg block, setting it in motion as shown in the figure, part (a). If the distance L that the leading edge of the smaller block travels on the larger block is 2.80m, how long will it take before this block reaches the right side of the 6.65 kg block, as shown in the figure, part (b)? Use g = 9.81m/s2

Then, answer:

How far does the 6.65kg block move in the process?

Homework Equations


μkN=F


The Attempt at a Solution


I've drawn the FBDs for both masses.

In the first case, we have that for the smaller mass, there is friction acting in the opposite direction to motion and the force applied, weight and the normal force. For the other case, the larger mass, we have the frictional force (opposite to the one that the smaller mass experiences), the normal force and the weight.

I've thought that the accelerations MUST be different as they aren't moving with the same acceleration (they aren't connected). I keep getting that the net force on the smaller mass is:

ma = F - μkN

And then, the value of the acceleration would be 7.77m/s2. Then i solved for time and i got 0.849s but apparently it's wrong.

In the other case, i calculated the acceleration of the other block which is just the frictional force. The acceleration turns out to be 0.846m/s2 and the distance should be 0.305m

I keep getting wrong answers for these exercises. What am i doing wrong? Thank you very much.
 

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Answers and Replies

  • #2
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The acceleration of the top block relative to the ground relative to the ground is 7.77 m/s^2, but this is not the acceleration of the top block relative to the bottom block, since the bottom block is also accelerating relative to the ground. What is the acceleration of the top block relative to the bottom block?
 
  • #3
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It should be the differences of the two accelerations, thus making it 6.924m/s2. Then the new time would be 0.899s.

And what would be the answer in the other case? I'm guessing there's no relative motion, so we should have that the distance is 0.342m (with the new time obtained). Or does it undergo relative motion as well?
 
  • #4
collinsmark
Homework Helper
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Homework Statement


A block of mass m = 1.85 kg rests on the left edge of a block of larger mass M = 6.65 kg. The coefficient of kinetic friction between the two blocks is 0.310, and the surface on which the 6.65kg block rests is frictionless. Aconstant horizontal force of magnitude F = 20.10 N is applied to the 1.85 kg block, setting it in motion as shown in the figure, part (a). If the distance L that the leading edge of the smaller block travels on the larger block is 2.80m, how long will it take before this block reaches the right side of the 6.65 kg block, as shown in the figure, part (b)? Use g = 9.81m/s2


Homework Equations


μkN=F


The Attempt at a Solution


I've drawn the FBDs for both masses.

In the first case, we have that for the smaller mass, there is friction acting in the opposite direction to motion and the force applied, weight and the normal force. For the other case, the larger mass, we have the frictional force (opposite to the one that the smaller mass experiences), the normal force and the weight.

I've thought that the accelerations MUST be different as they aren't moving with the same acceleration (they aren't connected). I keep getting that the net force on the smaller mass is:

ma = F - μkN
That's the right equation for the small block. :approve:

And then, the value of the acceleration would be 7.77m/s2.
You made a rounding error. It's not a huge error but it is significant. You might want to redo that calculation.

Then i solved for time and i got 0.849s but apparently it's wrong.
The approach you are using is as if the larger block remained motionless. But it's not motionless; it's moving too! :smile: The total distance that the smaller block moves is greater than 2.80 m.

In the other case, i calculated the acceleration of the other block which is just the frictional force. The acceleration turns out to be 0.846m/s2
That is the correct acceleration for the larger block. :approve:

and the distance should be 0.305m
I'm not sure where that came from. :uhh:

I keep getting wrong answers for these exercises. What am i doing wrong? Thank you very much.
Form an equation for the position of the small block, and form a different equation for the position of the large block, both as a function of time, t. Don't forget to include L in there somewhere. That will take the form of the initial position of one of the two blocks. (The two blocks don't have the same initial position, but they do have the same final position.)

Then set the two equations equal to each other (since they have the same final position) and solve for t.
 
  • #5
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4,208
It should be the differences of the two accelerations, thus making it 6.924m/s2. Then the new time would be 0.899s.

And what would be the answer in the other case? I'm guessing there's no relative motion, so we should have that the distance is 0.342m (with the new time obtained). Or does it undergo relative motion as well?
There aren't two separate problems here. There is only one. Figs. a and b are Before and After. Fig. a is the picture at time t = 0. Fig. b is the picture at 0.899 seconds.
 
  • #6
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There aren't two separate problems here. There is only one. Figs. a and b are Before and After. Fig. a is the picture at time t = 0. Fig. b is the picture at 0.899 seconds.
I'm so sorry! I forgot to include the other part of the problem:

How far does the 6.65kg block move in the process?
 
  • #7
collinsmark
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It should be the differences of the two accelerations, thus making it 6.924m/s2. Then the new time would be 0.899s.
There aren't two separate problems here. There is only one. Figs. a and b are Before and After. Fig. a is the picture at time t = 0. Fig. b is the picture at 0.899 seconds.
Well, it's not exactly 0.899 seconds. There is a rounding/precision error there too.
 
Last edited:
  • #8
collinsmark
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I'm so sorry! I forgot to include the other part of the problem:

How far does the 6.65kg block move in the process?
you can determine that later if you wish. :smile: But I wouldn't worry about it now. Form two equations for that distance, one for the small block, and another for the large block. Then set the equations equal to each other and solve for the time t. Of course then you have t and the problem is solved. But if you wish, you can go back and plug your newly found t into either one of those equations to find the distance the block moved. :smile:
 
  • #9
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you can determine that later if you wish. :smile: But I wouldn't worry about it now. Form two equations for that distance, one for the small block, and another for the large block. Then set the equations equal to each other and solve for the time t. Of course then you have t and the problem is solved. But if you wish, you can go back and plug your newly found t into either one of those equations to find the distance the block moved. :smile:
Alright! I'll fix up my rounding error in the acceleration, but is the problem done right now?

Edit: When two objects are accelerating, in this case, we always have to consider the acceleration of an object relative to the other one right? (sort of a "relative motion" problem?)
 
  • #10
collinsmark
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Edit: When two objects are accelerating, in this case, we always have to consider the acceleration of an object relative to the other one right? (sort of a "relative motion" problem?)
You always have to consider velocities relative to each other. But I advise being careful with relative accelerations.

I concede it works well in this particular problem. Finding the difference in acceleration between the small block and the large block, and then using that to find the time over relative distance L will lead you to the correct answer, for this particular problem.

But you can get the same answer using my previous advice of equating two different formulas, one for the large block and one for the smaller block, all relative to the ground. (As it turns out, when doing the algebra, there is a [a1 - a2] factor that comes about.) But with my advice, you don't need to consider the relative accelerations from the beginning, when you are setting up your equations.

I caution against solving problems by jumping straight into relative accelerations because later on, when you are dealing with Einstein's special relativity, that approach might get you in trouble (by that I mean wrong answers). Velocities are relative, yes. But avoid jumping into relative accelerations if you can.
 
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