Coefficient of friction Problem

AI Thread Summary
The discussion centers on calculating the smallest coefficient of friction required to prevent slippage between a small box and a larger box when both are subjected to a horizontal force, resulting in an acceleration of 2.5 m/s². Participants clarify that the mass of the small box can be treated as a variable that cancels out in the calculations. Using Newton's second law, they derive the relationship between the forces acting on the boxes, leading to the conclusion that the coefficient of friction can be expressed as μ = a/g, where g is the acceleration due to gravity. Ultimately, the calculated coefficient of friction is 0.25, indicating the minimum friction needed to maintain the boxes' relative positions during acceleration. The conversation emphasizes the importance of understanding the forces involved without needing specific mass values.
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A small box resting on a larger box, which in turn sits on a table. When a horizontal force is applied to the larger box, both boxes accelerate together. The small box does not slip on the larger box.

If the acceleration of the pair of boxes has a magnitude of 2.5 m/s^2, determine the smallest coefficient of friction between the boxes that will prevent slippage.

I'm don't really know where to start on this one, it seems that the only useful information that I have (so far as I can see) is the accleration.
 
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Start with F=ma=uN. What is N? Hint -- the mass of the small box probably will cancel out...
 
looking frm the ground frame of refernce, the smaller block seems to be moving bcoz of friction, as there is no other force on it. ... (1)

k= coeff of friction,m=mass of smaller block, N=normal rxn on it,g=10 m/s2

balancing forces on small block in vertical direction,
mg=N

also, in horizontal direction,
ma=kN (due to (1))

where a = 2.5
therefore ma=kN
or ma=kmg
or a=kg
k= a/g = 0.25 .. answer
 
Hi there deepakalways, please refrain from showing full solutions. It is better to guide a poster through the process than simply providing answers.
 
highc said:
A small box resting on a larger box, which in turn sits on a table. When a horizontal force is applied to the larger box, both boxes accelerate together. The small box does not slip on the larger box.

If the acceleration of the pair of boxes has a magnitude of 2.5 m/s^2, determine the smallest coefficient of friction between the boxes that will prevent slippage.

I'm don't really know where to start on this one, it seems that the only useful information that I have (so far as I can see) is the accleration.
The coefficient of static friction gives the maximum force that static friction can impart. The actual force depends on the applied force. What is the force that the lower box applies to the upper?

AM
 
The force that the lower box applies to the other is the normal force equal and opposite to the force of gravity. Would that be 9.8 N*kg?
 
The lower box exerts two forces on the upper box: One is the normal force, which acts vertically upward and happens to equal the weight of the upper box (w = mg). The other force is the friction force, which acts horizontally. I believe that's the force that Andrew was asking you about. Hint: You know the horizontal acceleration of the upper box, so what must be the force on it? (Answer in symbols, not numbers.)
 
hi, new to the forums... am following along with you here. I also need help on this question.

Doc Al would the force be Kinetic Friction?
 
gonzalezphys said:
Doc Al would the force be Kinetic Friction?
Since the boxes do not slip, the friction will be static, not kinetic.
 
  • #10
highc said:
A small box resting on a larger box, which in turn sits on a table. When a horizontal force is applied to the larger box, both boxes accelerate together. The small box does not slip on the larger box.

If the acceleration of the pair of boxes has a magnitude of 2.5 m/s^2, determine the smallest coefficient of friction between the boxes that will prevent slippage.

I'm don't really know where to start on this one, it seems that the only useful information that I have (so far as I can see) is the accleration.

alright thanks! but I believe the original poster did not write the second part of the question.

c) If the acceleration of the pair of boxes has the magnitude of 2.5 m/s^2, determine the smallest coefficient of friction between the boxes that will prevent slippage.

how would i start this part?
 
  • #11
gonzalezphys said:
alright thanks! but I believe the original poster did not write the second part of the question.

c) If the acceleration of the pair of boxes has the magnitude of 2.5 m/s^2, determine the smallest coefficient of friction between the boxes that will prevent slippage.
:confused: Not only did the original poster write this part of the problem, it's word-for-word in the post that you quoted! In fact, that's the only part of the problem discussed in this thread. (Re-read the thread for tips.)
 
  • #12
LOL, I just realized, sorry I was talking on the phone while trying to post... :blushing:

i'm going to sit down now before I hurt myself
 
  • #13
Excuse me, but don't you need the mass to calculate the coefficient of friction, since the friction between two objects depends on the force they apply against each other, and to know the force you need the mass. Am I wrong on both points? But to find the coefficient of friction you need to know the normal force. Do you need the mass to calculate the coefficient of friction?:confused:
 
  • #14
To find the friction force, you'd need the mass. Luckily, this problem only asks for the coefficient of friction. It turns out that you don't need the mass to find that. (Try it and see.)
 
  • #15
I am sorry, but I still don't understand. The formula for the coefficient of static friction is us = FSmax/FN. FN is the normal force, which can only be calculated using the mass. Is this the way to conduct this problem solving?
 
  • #16
Hint: The friction force is the only horizontal force acting on the smaller box. Relate that force to the acceleration of the box. (Note that you only need the ratio of the two forces, not their absolute measures.)

In general: When you don't have a needed quantity, give it a symbol and carry on. Maybe you won't really need it. :wink:
 
  • #17
Am I correct in using the formula us = Fs / Fn?
 
  • #18
To find the coefficient of friction using that formula, the mass is necessary. Since another variable's value is needed, either the normal force or the mass must be known. Incidentally, mass also needs to be known to calculate the normal force. I found that I could not answer this question using this formula, but it is the only formula the book gives, so I must be doing something wrong. I just can't see a solution to this problem.
 
  • #19
Okay. So I use the formula and plug in the acceleration value making this:

us = Fs / (2.5 m/s2) (mass)

In accord to my last post, no solution can be made without additional information.
 
  • #20
randomwinner said:
Am I correct in using the formula us = Fs / Fn?
You'll need that formula.

randomwinner said:
To find the coefficient of friction using that formula, the mass is necessary. Since another variable's value is needed, either the normal force or the mass must be known. Incidentally, mass also needs to be known to calculate the normal force. I found that I could not answer this question using this formula, but it is the only formula the book gives, so I must be doing something wrong. I just can't see a solution to this problem.
What's Fs? Hint: Use Newton's 2nd law.
What's Fn?

You do not need to know the mass. Just call it "m". Try it!
 
  • #21
randomwinner said:
Okay. So I use the formula and plug in the acceleration value making this:

us = Fs / (2.5 m/s2) (mass)
This substitution is incorrect.
 
  • #22
What? F = ma
 
  • #23
randomwinner said:
What? F = ma
Nothing wrong with that formula, but how did you use it to substitute for Fn?
 
  • #24
I see now where I went wrong
 
  • #25
In this case though, the normal force would be the downward force, equal ma, in which a = 9.8 m/s2. Still need to know the mass though.
 
  • #26
randomwinner said:
In this case though, the normal force would be the downward force, equal ma, in which a = 9.8 m/s2. Still need to know the mass though.
Call the mass "m" and continue. Answer my questions from post #20.
 
  • #27
Fn = m (9.8 m/s2)
so i get:

Fs = us (m) (9.8m/s2)

I take when you say "use Newton's second law" you are referring to the formula's and not the fact that an unbalanced force will cause an object to accelerate.

Are you saying that I should be able to figure out the force applied by knowing the acceleration?
 
  • #28
randomwinner said:
Fn = m (9.8 m/s2)
so i get:

Fs = us (m) (9.8m/s2)
So far, so good.

I take when you say "use Newton's second law" you are referring to the formula's and not the fact that an unbalanced force will cause an object to accelerate.
I don't understand the distinction.

Are you saying that I should be able to figure out the force applied by knowing the acceleration?
Absolutely. At least in terms of the unknown mass.
 
  • #29
Doc Al said:
Absolutely. At least in terms of the unknown mass.

Now that is impossible. It all comes back to mass. A certain force value will not make all things accelerate at the same rate. It depends on the mass.
 
  • #30
randomwinner said:
Now that is impossible.
Not only isn't it impossible, it's easy. Reread exactly what I've been saying.
It all comes back to mass.
Sure does.
A certain force value will not make all things accelerate at the same rate. It depends on the mass.
That's true, that's why the force will be in terms of the mass. Luckily, we don't need to know the actual mass to solve this problem.

Can you answer these?
Your mass is "m", what's your weight?
Your mass is "m", your acceleration "a", what's the net force on you?

If you can answer those questions, use the same idea to answer the ones I've been asking:
What's Fs? Hint: Use Newton's 2nd law.
What's Fn?
Call the unknown mass "m".
 
  • #31
randomwinner said:
Now that is impossible. It all comes back to mass. A certain force value will not make all things accelerate at the same rate. It depends on the mass.
Doc Al is trying to get you to see that while mass is certainly a part of the analysis, you don't need to know what it actually is. To see this, you have to do the analysis.

You know what the horizontal accelerating force must be applied to the top block by the bottom one if it accelerates at 2.5 m/sec^2. So write it out.

You know what determines the maximum force that the lower block can apply to the top block. Write that out.

You are asked to assume that this maximum force is applied to the top block when the acceleration is 2.5 m/sec^2. So what kind of relationship exists between the forces? Put that relationship between the expressions for the forces and you will solve the problem.

AM
 
  • #32
So are you saying that it wouldn't make a difference if the box was 20 or 50 kg, as long as it was accelerating at 2.5 m/s squared? That the certain value for the coefficient of friction would be enough to hold both boxes of different weight to the larger box if they were accelerating at 2.5 m/s squared?
 
Last edited:
  • #33
Doc Al said:
Luckily, we don't need to know the actual mass to solve this problem.

Wouldn't you need a higher coefficient of friction to hold a smaller box in place while accelerating?
 
  • #34
randomwinner said:
So are you saying that it wouldn't make a difference if the box was 20 or 50 kg, as long as it was accelerating at 2.5 m/s squared? That the certain value for the coefficient of friction would be enough to hold both boxes of different weight to the larger box if they were accelerating at 2.5 m/s squared?

randomwinner said:
Wouldn't you need a higher coefficient of friction to hold a smaller box in place while accelerating?
Why don't you do the analysis and find out?
 
  • #35
Well, this is what it looks like.

The applied force on the box would have to equal the force of static friction.

So FA = 2.5 m/s2 x mass
And FS = us x 9.8 m/s 2 x mass

Therefore I can cancel out the mass which leaves me with:

2.5 m/s2 = us x 9.8 m/s2

Then I would divide and then have my answer. Correct?
 
  • #36
Exactly correct.

Fs = ma = μmg
Thus, μ = a/g.
 
  • #37
All right. Thanks for the help!
 

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