Finding the coefficient of friction

[HarshK]

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!

 

[Comeback City]

Did you try solving through energy?

 

[HarshK]

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)

 

[Comeback City]

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?

 

[HarshK]

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.

 

[Comeback City]

(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?

 

[HarshK]

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?

 

[Comeback City]

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
F_n - mg =0
F_n = mg
Horizontal is
μF_n = ma
Agreed? If so, where could we go from here?

 

[HarshK]

By force, I mean sum up forces in both horizontal and vertical directions. Vertical is
F_n - mg =0
F_n = mg
Horizontal is
μF_n = 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.

 

[Comeback City]

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)

 

[HarshK]

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)

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?

 

[Comeback City]

Did I do the calculation process correctly?

Looks correct to me.

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.

 

[HarshK]

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!

 

[Comeback City]

Alright I'll be sure to post correctly next time! Thanks for your help, it helped a great load!

No problem! Enjoy the forums...

 

[haruspex]

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?

 

[Comeback City]

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.

 

[SammyS]

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?

 

[haruspex]

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.

 

[HarshK]

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.

 

[haruspex]

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?

 

[HarshK]

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'

 

[HarshK]

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 :(

 

[HarshK]

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 :(

 

[HarshK]

I think most of the collision was elastic but initially was inelastic for a very small distance.

 

[Comeback City]

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.

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.

 

[haruspex]

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.)

 

[haruspex]

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.

 

[Comeback City]

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).

 

[MagnificentLiver]

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 mv_i² – FΔx = ½ mv_f²

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

N=W=mg F_f=μmg F_fΔx = μmgΔx

½ mv_i² – μmgΔx = ½ mv_f² (remember V_f=0)

½ mv_i² – μmgΔx = 0

½ mv_i² = μmgΔx (we only have y components) (masses cancel)

½ v_i² = μgΔx

μ= ½ v_i² / gΔx (check unit analysis ( our units should cancel))

μ= ((.5)(.33)²)/((9.8)(.032)) = .174

 

[Comeback City]

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.

 

[MagnificentLiver]

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

 

[Comeback City]

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.

 

[haruspex]

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.

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?

 

[MagnificentLiver]

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.

 

[Comeback City]

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.