Average Force of two people pushing away on carts

[Tasdel]

Homework Statement

What is the Average Force between m1 and m2. Show all work.
Note that the masses are both at rest, and begin to push off of each other in opposite directions.
Their velocity is measured when they reach 20cms distance. (I calculated the velocity when they reached this point as well. Please let me know if I did that correctly as well using the kinematics formulas d=1/2(0)at^2 and vf=vi+at. Thanks.

m1=148.4 kg, velocity in negative direction = 1.47 m/s, acceleration = 5.47 m/s^2
m2=91.4kg, velocity in positive direction = 2.35 m/s, acceleration = 2.35 m/s^2
Distance both carts traveled 20cm (0.2m) velocity of each cart recorded once the cart passed this point.
Time of contact as m1 and m2 push away from each other 0.78s
2. The attempt at a solution

FΔt=Δ(mv)
F=Δ(mv)/Δt
F=(m1+m2)(vtotal)/0.78s
F= ((148.4+91.4)*(2.35-1.47))/0.78
F=(239.8)*(0.88)/0.78
F=270.54N

Is this correct to find the average force? or should I find the forces out individually on each m1 and m2.
Thanks for your time!

 

[kuruman]

Their velocity is measured when they reach 20cms distance. (I calculated the velocity when they reached this point as well.

That's what's important. What are these numbers?

You are assuming that when the two accelerate, they do so at constant acceleration. That is far from being the case. Remember Newton's 3rd Law. The force on each person is equal and opposite to the other person. Yet, you claim that the 148.4 kg person has a higher acceleration than the 91.4 kg person! How can that be? Forget the kinematic equations and think impulse.

 

[Tasdel]

That's what's important. What are these numbers?

You are assuming that when the two accelerate, they do so at constant acceleration. That is far from being the case. Remember Newton's 3rd Law. The force on each person is equal and opposite to the other person. Yet, you claim that the 148.4 kg person has a higher acceleration than the 91.4 kg person! How can that be? Forget the kinematic equations and think impulse.

Hi there. I accidentally put the wrong acceleration of cart 2 acceleration is 13.8m/s^2 so sorry.

I was given that in a second there are 240 frames. at 65 frames, the 148kg cart reaches the 0.2m. so i gathered:
65/240 = 0.27s so I plugged in.. d=vi*t+(1/2)at^2
2(0.2m=(1/2)a(0.27)^2)
0.4m=a(0.27)^2
0.4m=0.0729a
a=5.47 m/s^2

 

[Tasdel]

So basically I do have distance.. so maybe do FΔT=(m1+m2)*(d/t)
(d/t) because v=d/t?
F*(0.78)=(239.8)*(0.2/0.27)
F=((239.8)*(0.74))/0.78
F=227.73N

Does this work?? Thanks again.

 

[kuruman]

I suggest that you forget about the details of the acceleration because you don't know what they are and you cannot assume that the acceleration is constant. The kinematic equations do not apply here. All you can calculate is the average force that each person exerts on the other. You know the impulse equation F_avg.Δt = m Δv.
For each cart, you need to:
(a) Find the time Δt that they are in contact pushing against each other.
(b) Find Δv which their change in velocity; this is v_final - 0. To find v_final, you need to find how far the cart moves over how much time after the carts are separated. You can use v = d/t only when the cart is not accelerating.
(c) Use the impulse equation to find the average force exerted while there is contact.

After you calculate the average force on each cart, verify that the two average forces are equal in magnitude and opposite in direction to within your experimental accuracy. Remember that velocity is a vector.

 

[Tasdel]

I suggest that you forget about the details of the acceleration because you don't know what they are and you cannot assume that the acceleration is constant. The kinematic equations do not apply here. All you can calculate is the average force that each person exerts on the other. You know the impulse equation F_avg.Δt = m Δv.
For each cart, you need to:
(a) Find the time Δt that they are in contact pushing against each other.
(b) Find Δv which their change in velocity; this is v_final - 0. To find v_final, you need to find how far the cart moves over how much time after the carts are separated. You can use v = d/t only when the cart is not accelerating.
(c) Use the impulse equation to find the average force exerted while there is contact.

After you calculate the average force on each cart, verify that the two average forces are equal in magnitude and opposite in direction to within your experimental accuracy. Remember that velocity is a vector.

Thank you so much. So I went ahead and calculated..
Connor = Vc=d/t
Vc=0.2m/(41frames/240)s
Vc=0.20m/0.17s
Vc=1.17m/s
FΔt=mΔv
F=((91.4kg)*(1.17m/s))/(0.78s time of contact)
F=137.1NHailey
Vh=d/t
Vh=-0.2m/(65frames/240)s
Vh=-0.2m/0.27s
Vh=-0.74m/s
FΔt=mΔv
F=((148.4)*(-0.74))/(0.78)
F=-140.8N


Impulse for Connor
137.1x0.78= 106.94Ns

Impulse for Hailey
-140.8Nx0.78= -109.5Ns

So they almost completely cancel out, so the average force should be 0 right? Or should it be -1.8N
Fave.=(137.1+(-140.8))/2
Fave=-1.8N

 

[haruspex]

the average force should be 0 right?

The sum of the average forces should be zero, since they must be equal and opposite. The discrepancy will mostly be from rounding the frame counts to whole numbers. E.g. 41 frames is ±½ a frame, so a 1% error margin.
I assume the question wants the average magnitude of force each experiences, so I suggest taking the average of the two absolute values.

 

[Tasdel]

The sum of the average forces should be zero, since they must be equal and opposite. The discrepancy will mostly be from rounding the frame counts to whole numbers. E.g. 41 frames is ±½ a frame, so a 1% error margin.
I assume the question wants the average magnitude of force each experiences, so I suggest taking the average of the two absolute values.

Thanks for the response haruspex.
The question was. Determine the average force acting on Both Hailey and Connor as they explode apart.
The way I'm interpreting it, is that what's the average force of the absolute values.. so even though one of them is in the negative direction I'm just going to add the sum off both values and divide by two to get the average. Sorry for the silly questions.

Maybe if it asked what the Net foce is it could be 0N...

This is what I think.
(137.1+148.4)/2
Fave=142.8N

 

[haruspex]

Thanks for the response haruspex.
The question was. Determine the average force acting on Both Hailey and Connor as they explode apart.
The way I'm interpreting it, is that what's the average force of the absolute values.. so even though one of them is in the negative direction I'm just going to add the sum off both values and divide by two to get the average. Sorry for the silly questions.

Maybe if it asked what the Net foce is it could be 0N...

This is what I think.
(137.1+148.4)/2
Fave=142.8N

Yes, that's what I was suggesting.

 

[Tasdel]

Yes, that's what I was suggesting.

Oh hehe then I interpreted what you said correctly thanks!

 

[kuruman]

I have two observations.
1. I don't think that "average force" in this context means adding the action reaction magnitudes and dividing by 2. The force as a function of time, $F(t)$, is an unknown. However the impulse-momentum theorem says that $\int{F(t)~dt}=m \Delta v$. The integral (area under the curve) can be replaced with the equal area of a rectangle of width $\Delta v$ and height $F_{avg.}$ in which case the impulse-momentum equation becomes $F_{avg.} \Delta t=m \Delta v$. In this case, $F_{avg.~Connor}=137.1~N$ and $F_{avg.~Hailey}=-142.8~N$. No further averaging is required.
2. The above numbers for the two averages should be reported to the appropriate number of significant figures. The relevant rule is, For quantities created from measured quantities by multiplication and division, the calculated result should have as many significant figures as the measured number with the least number of significant figures^*. Here, it looks like the number of sig. figs. is 2. If you change your average force calculations from 4 sig. figs. to 2, what do you get?

_______________________________
^*https://en.wikipedia.org/wiki/Significant_figures

 

[haruspex]

don't think that "average force" in this context means adding the action reaction magnitudes and dividing by 2

Agreed, but ideally the two numbers for that average would have come out the same. Since they did not, because of measurement errors, it seemed reasonable to take their average.

On reflection, though, consideration should be given to exactly where the errors are likely to have come from and the corresponding uncertainties.
If we assume that rounding the frame counts to whole numbers is the greatest source of error, the fractional error is greater for the smaller frame count. This says the averaging of the averages should be biased towardsthe value derived from the larger frame count, in roughly the ratio of the two counts: (137.1x41+148.4x65)/(41+65).