Finding the coefficient of friction

In summary, the bob pushes the box 3.2 cm with an initial velocity of 0.33 m/s. The coefficient of friction is 0.455 N.
  • #1
HarshK
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0

Homework Statement


I am investigating a scenario where a pendulum with a bob attached is released from an angle and pushes a box to a certain distance. My goal is to find the coefficient of friction between the box and the surface it moved on.

I have measurements for:
- mass of the bob (125 grams)
- mass of the box (267.3 grams)
- length of the pendulum (27 cm)
- distance the box moved from its initial position (3.2 cm)
- angle of release (30 degrees above the x-axis)

Homework Equations


- 1/2(m)(v^2) = mgh
- Vf = sqrt(2*g*h)
- m1+v1=m2+v2
- h = L(1-cos(theta))
- Ff=μFN

The Attempt at a Solution


0.27(1-cos(30))= h = 3.62 cm
sqrt (2*9.8*0.0362) = Vf = 0.84 m/s (speed before hitting the box)
(0.125)(0.84)=(0.2673)(v2), v2 = 0.33 m/s (speed of the block)
(9.8)(0.2673) = FN = 2.62 N

I don't know where to go from here. If you could help me or point me in the right direction, I would greatly appreciate it, Thanks!
 
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  • #2
:welcome:
Did you try solving through energy?
 
  • #3
Comeback City said:
:welcome:
Did you try solving through energy?
I was only able to figure out the final velocity of the bob before it hit the box through energy.
PE=KE
mgh=1/2mv^2
v = sqrt(2*g*h)
 
  • #4
HarshK said:
I was only able to figure out the final velocity of the bob before it hit the box through energy.
PE=KE
mgh=1/2mv^2
v = sqrt(2*g*h)
Indeed. From there, you need to find the initial velocity of the box after the collision. Any ideas on how you might do it?
 
  • #5
Comeback City said:
Indeed. From there, you need to find the initial velocity of the box after the collision. Any ideas on how you might do it?
Well, if I take conservation of momentum (m1+v1=m2+v2) into account then, (0.125)(0.84)=(0.2673)(v2). v2 being the velocity of the box after the collision, I calculated to be 0.33m/s.
 
  • #6
HarshK said:
(m1+v1=m2+v2)
The equation for conservation of momentum is
m1v1 + m2v2 = m1v1' + m2v2'
If we assume the collision is elastic, then the second and third terms go away and you are left with
m1v1 = m2v2'

I didn't do the calculation, but will assume you got it correct. Now you have to take into account forces to find coefficient of friction. What will you do from here?
 
  • #7
Comeback City said:
Now you have to take into account forces to find coefficient of friction. What will you do from here?

I'm guessing by forces you mean forces such as driving force. So, W=FD. 1/2mv^2=W. (1/2mv^2) / d = F. (1/2(0.2673)(0.33^2)) / 0.032 = 0.455 N
I'm not sure if this was a valid calculation, but if it is is the friction equal to 0.455N?
 
  • #8
HarshK said:
I'm guessing by forces you mean forces such as driving force. So, W=FD. 1/2mv^2=W. (1/2mv^2) / d = F. (1/2(0.2673)(0.33^2)) / 0.032 = 0.455 N
I'm not sure if this was a valid calculation, but if it is is the friction equal to 0.455N?
By force, I mean sum up forces in both horizontal and vertical directions. Vertical is
Fn - mg =0
Fn = mg
Horizontal is
μFn = ma
Agreed? If so, where could we go from here?
 
  • #9
Comeback City said:
By force, I mean sum up forces in both horizontal and vertical directions. Vertical is
Fn - mg =0
Fn = mg
Horizontal is
μFn = ma
Agreed? If so, where could we go from here?
Agreed!

Fn = mg, so the normal force equals (0.2673)(9.8) = Fn = 2.62 N
μFn = ma, (Second Law) so (0.2673)(a)

Then ma/mg = μ
But how do I find the value of 'a'? The box moved 3.2 cm but way too quickly for me to record an accurate time.
 
  • #10
HarshK said:
But how do I find the value of 'a'? The box moved 3.2 cm but way too quickly for me to record an accurate time.
Ah, time is not even needed! The equation
v2 = v02 + 2⋅a⋅Δx
where
v... final velocity (0)
v0... initial velocity (from your momentum calculation)
a... acceleration
Δx... change in distance
will cover it. Just rearrange it to find a.
As a side note, just make sure your distances are in meters (not cm) :smile:
 
  • #11
Comeback City said:
Ah, time is not even needed! The equation
v2 = v02 + 2⋅a⋅Δx
where
v... final velocity (0)
v0... initial velocity (from your momentum calculation)
a... acceleration
Δx... change in distance
will cover it. Just rearrange it to find a.
As a side note, just make sure your distances are in meters (not cm) :smile:
Ahh I see.
So, the box stops at 0.032 meters so Vf = 0 but has an initial velocity of 0.33 m/s so Vi = 0.33 m/s
Vf^2 = Vi^2 + 2ad
0^2 = 0.33^2 + 2 (a) (0.032)
-0.33^2 = 2 a (0.032)
0.1089 = 2 a (0.032)
((0.1089 / 2) / 0.032) = a
a = 1.70 m/s^2
Did I do the calculation process correctly? If so, should I apply it to the 'a' in (μFn = ma) and then get a coefficient value?
 
  • #12
HarshK said:
Did I do the calculation process correctly?
Looks correct to me.
HarshK said:
If so, should I apply it to the 'a' in (μFn = ma) and then get a coefficient value?
Exactly!

Also, I just realized that this question was placed in this general physics forum, while there is actually a separate forum for physics homework questions! I'm just letting you know for future reference and so you aren't confused if this thread get moved to another forum.
 
  • #13
Comeback City said:
Looks correct to me.

Exactly!

Also, I just realized that this question was placed in this general physics forum, while there is actually a separate forum for physics homework questions! I'm just letting you know for future reference and so you aren't confused if this thread get moved to another forum.

Alright I'll be sure to post correctly next time! Thanks for your help, it helped a great load:smile:!
 
  • #14
HarshK said:
Alright I'll be sure to post correctly next time! Thanks for your help, it helped a great load:smile:!
No problem! Enjoy the forums...:biggrin:
 
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  • #15
Comeback City said:
If we assume the collision is elastic, then the second and third terms go away and you are left with
m1v1 = m2v2'
That is not correct. If the collision is perfectly elastic then v1-v2=v2'-v1'.
If perfectly inelastic the masses coalesce, i.e. continue together:
m1v1+m2v2=(m1+m2)v1'.

In practice, it will be partly elastic. Depending on the materials, you might be able to assume that the collision approximates one of the two extremes. What are the materials?
 
  • #16
haruspex said:
That is not correct. If the collision is perfectly elastic then v1-v2=v2'-v1'.
If perfectly inelastic the masses coalesce, i.e. continue together:
m1v1+m2v2=(m1+m2)v1'.

In practice, it will be partly elastic. Depending on the materials, you might be able to assume that the collision approximates one of the two extremes. What are the materials?
Yes, that is also true. But please explain how the second and third terms do not cancel out.
EDIT: I think I see what you're saying now. Second term cancels out for sure. The problem is, though, we don't even know what type of collision this is or what the materials are.
 
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  • #17
HarshK said:

Homework Statement


I am investigating a scenario where a pendulum with a bob attached is released from an angle and pushes a box to a certain distance. My goal is to find the coefficient of friction between the box and the surface it moved on.

I have measurements for:
- mass of the bob (125 grams)
- mass of the box (267.3 grams)
- length of the pendulum (27 cm)
- distance the box moved from its initial position (3.2 cm)
- angle of release (30 degrees above the x-axis)

Homework Equations


- 1/2(m)(v^2) = mgh
- Vf = sqrt(2*g*h)
- m1+v1=m2+v2
- h = L(1-cos(theta))
- Ff=μFN

The Attempt at a Solution


0.27(1-cos(30))= h = 3.62 cm
sqrt (2*9.8*0.0362) = Vf = 0.84 m/s (speed before hitting the box)
(0.125)(0.84)=(0.2673)(v2), v2 = 0.33 m/s (speed of the block)
(9.8)(0.2673) = FN = 2.62 N

I don't know where to go from here. If you could help me or point me in the right direction, I would greatly appreciate it, Thanks!
Did you make these measurements, or are they given to you as an exercise.

If you made the observations, what was the motion of the bob immediately after the collision?

If they're given to you, was there any additional information regarding the nature of the collision or regarding the final state of the pendulum?
 
  • #18
Comeback City said:
Yes, that is also true. But please explain how the second and third terms do not cancel out.
EDIT: I think I see what you're saying now. Second term cancels out for sure. The problem is, though, we don't even know what type of collision this is or what the materials are.
The third term, m1v1' only goes to zero if the impinging object stops dead. That happens in perfectly elastic collisions if the two objects have the same mass. More generally, it can only happen if the impinging object is the lighter of the two, in exactly the right ratio to match the degree of inelasticity. Specifically, m1=m2R, where R is the coefficient of restitution.
 
  • #19
Me and my group made the measurements.
The bob was released at an angle (30 degrees from rest / 60 degrees below x axis) and when released it hits the box, the box moves but the bob remains at the point where it was before moving it to an angle. So, like 90 degrees to the box after being hit.

As for materials. The box was a tissue box. And the bob was, not sure how to describe this, filled with sand in a balloon. It looked like putty.
 
  • #20
HarshK said:
the bob remains at the point where it was before moving it to an angle. So, like 90 degrees to the box after being hit.
Not exactly sure what you are saying there. Are you saying that the bob stopped dead at the point where the collision occurred? It did not continue to swing at all?
 
  • #21
haruspex said:
Not exactly sure what you are saying there. Are you saying that the bob stopped dead at the point where the collision occurred? It did not continue to swing at all?

Wait I have a revision, the collision was inelastic. Asked a friend to confirm. So I believe haruspex may be right when he said m1v1+m2v2=(m1+m2)v1'
 
  • #22
So with this in mind, I'm guessing I will have to re-calculate the velocity of the box, the acceleration and coefficient value.

m1v1+m2v2=(m1+m2)v1'
(0.125)(0.71)+(0.2673)(0) = (0.125+0.2673)(v1'), v1' = 0.23m/s

(vf^2-vi^2)/2d = a
(0^2 - 0.23^2) / 2(0.032) = a, a =-0.83 m/s

-0.83/9.8 = μ
μ = 0.085

Not sure I did this right. But that value does not seem correct to me. We also did an experimental procedure, we used a spring scale to pull the box on the surface which showed a reading of about 0.71N. (0.71 / 0.2673 x 9.8) = μ = 0.271

I'm confused :(
 
  • #23
haruspex said:
If perfectly inelastic the masses coalesce, i.e. continue together:
m1v1+m2v2=(m1+m2)v1'.
So with this in mind, I'm guessing I will have to re-calculate the velocity of the box, the acceleration and coefficient value.

m1v1+m2v2=(m1+m2)v1'
(0.125)(0.71)+(0.2673)(0) = (0.125+0.2673)(v1'), v1' = 0.23m/s

(vf^2-vi^2)/2d = a
(0^2 - 0.23^2) / 2(0.032) = a, a =-0.83 m/s

-0.83/9.8 = μ
μ = 0.085

Not sure I did this right. But that value does not seem correct to me. We also did an experimental procedure, we used a spring scale to pull the box on the surface which showed a reading of about 0.71N. (0.71 / 0.2673 x 9.8) = μ = 0.271

I'm confused :(
 
  • #24
I think most of the collision was elastic but initially was inelastic for a very small distance.
 
  • #25
HarshK said:
Wait I have a revision, the collision was inelastic. Asked a friend to confirm. So I believe haruspex may be right when he said m1v1+m2v2=(m1+m2)v1'
That equation is only valid if the collision is perfectly inelastic.
HarshK said:
I think most of the collision was elastic but initially was inelastic for a very small distance.
I don't think collisions work that way. It can only be one type of collision throughout.
 
  • #26
Comeback City said:
That equation is only valid if the collision is perfectly inelastic.

I don't think collisions work that way. It can only be one type of collision throughout.
Yes and no. If the impinging mass is the smaller, which it is here, it may rebound somewhat. Since it is a pendulum it might then swing forwards again and strike the box a second time, and maybe more, especially since the box is being slowed by friction.
If this happens, momentum is not conserved. Each back and forth swing of tne pendulum converts negative momentum to positive.
With hindsight, the experimental set-up ought to include a bar just above the bob at the point where the bob hits the box. The string hits the bar at about the same instant that the bob hits the box, so the bob is prevented from a second impact.

(And it is quite common for impacts to be biphasic. If a stone hits a metal plate at low speed it will bounce off, no permanent impression on the plate. At higher speeds, the initial phase of the impact may still be elastic but as it proceeds the elastic limit of stress will be exceeded.)
 
  • #27
HarshK said:
Not sure I did this right.
If it was completely inelastic then the two bodies move forward together. Since friction is slowing the box, the bob should stay with it for a while. Thus the friction is limited by the weight of the box, but it has to decelerate both masses. In short, you cannot write -a=μg.
 
  • #28
haruspex said:
Yes and no. If the impinging mass is the smaller, which it is here, it may rebound somewhat. Since it is a pendulum it might then swing forwards again and strike the box a second time, and maybe more, especially since the box is being slowed by friction.
If this happens, momentum is not conserved. Each back and forth swing of tne pendulum converts negative momentum to positive.
With hindsight, the experimental set-up ought to include a bar just above the bob at the point where the bob hits the box. The string hits the bar at about the same instant that the bob hits the box, so the bob is prevented from a second impact.
I agree completely with this possibility, but I don't think the author of this problem intended for anything that complicated. I think its just asking about the one collision (be it elastic or inelastic).
 
  • #29
If we assume the bob came to rest as it hit the box then we know the initial velocity of the box as 0.33 m/2 and the final velocity as 0 m/s (rest).
From this can we not take Kinetic(initial) + Work = Kinetic(final) T1 + U = T2. U=FΔx
½ mvi2 – FΔx = ½ mvf2

FΔx => FfΔx Ff is going in the -x direction so in our equation it will be (-), remember Ff = μN

N=W=mg Ff=μmg FfΔx = μmgΔx

½ mvi2 – μmgΔx = ½ mvf2 (remember Vf=0)

½ mvi2 – μmgΔx = 0

½ mvi2 = μmgΔx (we only have y components) (masses cancel)

½ vi2 = μgΔx

μ= ½ vi2 / gΔx (check unit analysis ( our units should cancel))

μ= ((.5)(.33)2)/((9.8)(.032)) = .174
 
Last edited:
  • #30
MagnificentLiver said:
FΔx => FfΔx Ff is going in the -x direction so in our equation it will be (-), remember Ff = μN

N=W=mg Ff=μmg FfΔx = μmgΔx
Is "N" representing normal force? If so, you cannot say it is equal to "W" (which I believe is Work(?)). Simple dimensional analysis doesn't allow this.
 
  • #31
Comeback City said:
Is "N" representing normal force? If so, you cannot say it is equal to "W" (which I believe is Work(?)). Simple dimensional analysis doesn't allow this.
N=Normal
W= Weight
U= Work
We have the weight of the box (W) pushing down on the surface and the normal reaction for pushing up (N) . Will assume up is positive.
ΣFy=0
N-W=0
N=W
W=mg
 
  • #32
MagnificentLiver said:
N=Normal
W= Weight
U= Work
We have the weight of the box (W) pushing down on the surface and the normal reaction for pushing up (N) . Will assume up is positive.
ΣFy=0
N-W=0
N=W
W=mg
My mistake. I thought W was representing Work in your equation.
 
Last edited:
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  • #33
Comeback City said:
don't think the author of this problem
I believe this is a lab, not an abstract problem. Complications need to be considered, especially since the experimenters may have taken insufficient note of exactly what happened.
MagnificentLiver said:
If we assume the bob came to rest as it hit the box
That's a major assumption.
How about considering the various extremes and getting the range of possible coefficients?
 
  • #34
haruspex said:
I believe this is a lab, not an abstract problem. Complications need to be considered, especially since the experimenters may have taken insufficient note of exactly what happened.

That's a major assumption.
How about considering the various extremes and getting the range of possible coefficients?
I agree it is a major assumption, but without actually seeing the lab we are unsure of the complete reaction between the two objects. I was just trying to use a broad approach to show how μ would be found if the pendulum reacted in that situation.
 
  • #35
haruspex said:
I believe this is a lab, not an abstract problem. Complications need to be considered, especially since the experimenters may have taken insufficient note of exactly what happened.
Good point. Your idea of a contraption preventing further collisions was a good one in that case.
 
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Related to Finding the coefficient of friction

What is the coefficient of friction?

The coefficient of friction is a measure of the amount of resistance between two surfaces in contact with each other. It is a dimensionless number that describes the ratio of the force required to move one surface over the other to the force pressing the two surfaces together.

Why is it important to find the coefficient of friction?

The coefficient of friction is important because it helps us understand the interactions between different surfaces and how much force is needed to overcome the frictional resistance. This information is crucial in designing and engineering various systems and structures, as well as predicting and preventing potential accidents.

How is the coefficient of friction measured?

The coefficient of friction can be measured using a variety of methods, including the inclined plane method, the spring scale method, and the drag method. These methods involve measuring the force required to move an object over a surface and using this information to calculate the coefficient of friction.

What factors affect the coefficient of friction?

The coefficient of friction can be affected by several factors, including the nature of the surfaces in contact, the roughness of the surfaces, the amount of force pressing the surfaces together, and the presence of any lubricants or contaminants. Temperature and humidity can also have an impact on the coefficient of friction.

How can the coefficient of friction be reduced?

The coefficient of friction can be reduced by using lubricants, such as oils or greases, between the two surfaces. Additionally, using smoother or more slippery materials for the surfaces in contact can also help reduce the coefficient of friction. In some cases, changing the design or surface texture of the objects can also help decrease the coefficient of friction.

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