# Finding Tension Between Garfield and John at the Olympics Parking Lot

• Lolagoeslala
In summary: Maybe a1 and a2).Sorry, for the first part it is supposed to be mg (typo).I have no idea how far x and y are because the sketch is just rough.a1= acceleration of john and the skateboarda2= acceleration of garfieldIf we are not given the distance how can we have an equation?Sorry, for the first part it is supposed to be mg (typo).I have no idea how far x and y are because the sketch is just rough.a1= acceleration of john and the skateboarda2= acceleration of garfieldIf we are not given the distance how can we have an equation?You don't need to be given the
Lolagoeslala

## Homework Statement

Garfield (a cat) and John (his human) did not go to the Olympics (not feline up to
it). However, they got in trouble. John sits on a frictionless skateboard on a
horizontal parking lot. A light strong rope tied to a solid post passes horizontally
to the skate board, passes around a pulley on the skateboard, returns horizontally
to pass above the post, around a second pulley and vertically down into a yawning
chasm. Garfield hangs on tightly to the bottom end of the rope. Garfield has a
mass of 20 kg (too much lasagna), John and the skateboard have a combined mass
of 60 kg. Calculate the tension in the rope before John hits the post or Garfield
bottoms out.

## The Attempt at a Solution

My FBD diagrams

http://s1176.beta.photobucket.com/user/LolaGoesLala/media/c_zps40e98e7e.png.html

Ok so this is what i came up with to find the tensin between the cat and the by and skateboard

m
Fnet = Fg - T
ma = md - T

M
Fnet = Fg + Fn + T
Fnet = T
Ma = T

Combining
a = (mg/m+M)

To find the tension
Ma = T
M(mg/m+M) = T
T = (M x m/M +m)g
T = ( 60000g x 20000g/60000g + 20000g)(9.8m/s^2)
T = 147000N

so that is the tension between john/skateboard and the cat
however how do i account for the nail and the tension between john/skateboard and the nail ?
Here is a rough sketch of what i think it looks like

http://s1176.beta.photobucket.com/user/LolaGoesLala/media/f_zpsafd5a9e3.png.html

Is there a mechanical advantage involved? How many rope segments are pulling on each mass? How might this affect your FBD?

Are the accelerations of the two objects the same? If not, how are they related?

gneill said:
Is there a mechanical advantage involved? How many rope segments are pulling on each mass? How might this affect your FBD?

Are the accelerations of the two objects the same? If not, how are they related?

Wait are those questions that you are asking me? or? Because all its asking for is the tension before john/skateboard hit the post

Lolagoeslala said:
Wait are those questions that you are asking me? or? Because all its asking for is the tension before john/skateboard hit the post

You need to address the questions in order to solve the problem correctly; their answers affect the equations you write to determine the tension.

Last edited:
gneill said:
You need to address the questions in order to solve the problem correctly; their answers affect the equations you write to determine the tension.

I looked at all the prospects and then found the equations

Lolagoeslala said:
I looked at all the prospects and then found the equations

I would suggest that you reconsider the net force acting on John and the related rates of the motions of John and Garfield. The pulley arrangement makes a difference.

gneill said:
I would suggest that you reconsider the net force acting on John and the related rates of the motions of John and Garfield. The pulley arrangement makes a difference.

So are you trying to say that for john and the skateboard i would need to include the tenion twice since there is another piece of tension connected to the nail/post.
So like this?

M
Fnet = Fg + Fn + T + T
Fnet = T+ T
Ma = T^2

Lolagoeslala said:
So are you trying to say that for john and the skateboard i would need to include the tenion twice since there is another piece of tension connected to the nail/post.
That's the idea, yes.
So like this?

M
Fnet = Fg + Fn + T + T
Fnet = T+ T
Ma = T^2

Well, you should keep X and Y components separate (Fg and Fn act in a different direction than the T's), but the sum of the forces acting in the X-direction will be T + T. That's 2T (not T squared).

Then consider how the accelerations of the two masses are related; the pulley arrangement will impose a relationship between them --- they are not equal.

gneill said:
That's the idea, yes.

Well, you should keep X and Y components separate (Fg and Fn act in a different direction than the T's), but the sum of the forces acting in the X-direction will be T + T. That's 2T (not T squared).

Then consider how the accelerations of the two masses are related; the pulley arrangement will impose a relationship between them --- they are not equal.

if i do that it does not work, please see my process

m
Fnet = Fg - T
ma = md - T

M
Fnet = Fg + Fn + T + T
Fnet = T + T
Ma = 2T

Combining
a = (mg + T/m+M)

To find the tension
Ma = 2T
M(mg+T/m+M) = 2T
Mmg + MT = 2T (m+M)
Mmg + MT = 2Tm + 2TM
Mmg = 2Tm + TM
Mmg = T(Tm +M)

Like this is confusing me?

Lolagoeslala said:
if i do that it does not work, please see my process

m
Fnet = Fg - T
ma = md - T
What is the 'd' in 'md'? Shouldn't that be mg?
M
Fnet = Fg + Fn + T + T
Fnet = T + T
Ma = 2T
Okay, BUT! The a's for the two masses are NOT THE SAME. They are related, but they are not the same value. The pulley arrangement makes them unequal. Check: If the cat descends by some distance 'y', by how much distance 'x' will the skateboard move?

Write an equation relating the two accelerations (And label them differently. Maybe a1 and a2).
Combining
a = (mg + T/m+M)
I don't see how you can write the above if they don't share the same acceleration. You need to sort out the relationship between the two accelerations before you can meaningfully combine the equations.

Last edited:
gneill said:
Lolagoeslala said:
if i do that it does not work, please see my process

m
Fnet = Fg - T
ma = md - T
[\quote]
What is the 'd' in 'md'? Shouldn't that be mg?

Okay, BUT! The a's for the two masses are NOT THE SAME. They are related, but they are not the same value. The pulley arrangement makes them unequal. Check: If the cat descends by some distance 'y', by how much distance 'x' will the skateboard move?

Write an equation relating the two accelerations (And label them differently. Maybe a1 and a2).

I don't see how you can write the above if they don't share the same acceleration. You need to sort out the relationship between the two accelerations before you can meaningfully combine the equations.

But if the cat moves down then john and its skateboard moves ahead so wouldn't the acceleration be the same.. we did questions similare to this but the acceleration is the same.

But then how would i do it in terms of a1 and a2?

ma1 = mg -T
Ma2 = 2T

ma1 + Ma2 = mg - T + 2T

Lolagoeslala said:
gneill said:
But if the cat moves down then john and its skateboard moves ahead so wouldn't the acceleration be the same.. we did questions similare to this but the acceleration is the same.
Maybe similar, but not the same... At some point you should have done problems involving pulley systems that involve mechanical advantage. The movements of the various parts of the system are related but not equal. For a given distance that the cat moves, the skateboard will move a different (but related) distance. You need to work out what the relationship between the movements is. The same relationship will apply to distance, velocity, and acceleration.

(Hint: Consider that the total length of the rope does not change. This should allow you to work out how far the skateboard moves for a given movement of the cat).
But then how would i do it in terms of a1 and a2?

ma1 = mg -T
Ma2 = 2T

ma1 + Ma2 = mg - T + 2T

You need to determine an equation relating the accelerations. That will give you three equations in three unknowns (a1, a2, T).

gneill said:
Lolagoeslala said:
Maybe similar, but not the same... At some point you should have done problems involving pulley systems that involve mechanical advantage. The movements of the various parts of the system are related but not equal. For a given distance that the cat moves, the skateboard will move a different (but related) distance. You need to work out what the relationship between the movements is. The same relationship will apply to distance, velocity, and acceleration.

(Hint: Consider that the total length of the rope does not change. This should allow you to work out how far the skateboard moves for a given movement of the cat).

You need to determine an equation relating the accelerations. That will give you three equations in three unknowns (a1, a2, T).

so for the cat this is my equation:
fnet = fg - T1
m1a1 = m1g - T1

and for the boy
fnet = T1 + T2
m2a2 = T1 + T2

right?

or..
i could...

make a big huge tension
so itll be like this

http://s1176.beta.photobucket.com/user/LolaGoesLala/media/ddgfffd_zpsbe230706.png.html

Where the equation could become:
cat
Fnet = Fg - T
m1a1 = m1g - T

and for the boy
Fnet = T2
m2a2 = T2

The rope has only one tension for its entire length, since the pulleys are massless and frictionless.

Take a close look at the forces acting on the skateboard pulley. Can you assign a tension to each of the "connections" there? In other words, how does your T2 relate to T?

Also, you still haven't stated the relationship that exists between the motion of the cat and the motion of the skateboard; how far does the skateboard move when the cat moves by some amount?

gneill said:
The rope has only one tension for its entire length, since the pulleys are massless and frictionless.

Take a close look at the forces acting on the skateboard pulley. Can you assign a tension to each of the "connections" there? In other words, how does your T2 relate to T?

Also, you still haven't stated the relationship that exists between the motion of the cat and the motion of the skateboard; how far does the skateboard move when the cat moves by some amount?

Well i have two ways i can do this. Since you said that the "rope has only one tension for its entire length, since the pulleys are massless and frictionless." I am going to assign the T2 to the little rope that connects the skateboard and the pulley. Well if the cat moves down 2m. the skateboard will move ahead 1m right? its a 1:2 ratio

Lolagoeslala said:
Well i have two ways i can do this. Since you said that the "rope has only one tension for its entire length, since the pulleys are massless and frictionless." I am going to assign the T2 to the little rope that connects the skateboard and the pulley.
Yes, but what is the relationship between T2 and T? Isolate that pulley and draw the FBD for it.
Well if the cat moves down 2m. the skateboard will move ahead 1m right? its a 1:2 ratio
Correct. That should tell you the ratio of a2 to a1 also (and v2 to v1 if it's of interest).

gneill said:
Yes, but what is the relationship between T2 and T? Isolate that pulley and draw the FBD for it.

Correct. That should tell you the ratio of a2 to a1 also (and v2 to v1 if it's of interest).

well for the T2 and T1

the ratio i came up with was T2 = 2T1

like this:
http://s1176.beta.photobucket.com/user/LolaGoesLala/media/dfdsdsd_zps2ce0a7cd.png.html

so since T2 = 2T1 and a1 = 2a2

(1) m1a1 = m1g - T
(2) m2a2 = T2

Times the first equation by two

2m1a1 = 2m1g - 2T
m2a2 = 2T

2m1(2a2) = 2m1g
m2a2

4m1a2 + m2a2 = 2m1g
a2 = 2m1g / (4m1 +m2)

when solving
a2 = 2(20kg)(9.8m/s^2) / 4(20kg)+(60kg)
a2 = 2.8 m/s^2

since a1 = 2a2
a1 = 5.6 m/s^2

and using the first equation
m1a1 = m1g - T
T = m1g - m1a1
T = (20kg)(9.8m/s^20 - (20kg)(5.6m/s^2)
T = 84 N

That looks good! Well done.

gneill said:
That looks good! Well done.
THANK YOU :D
it would have not been done without your help. :)

## 1. What caused the tension between Garfield and John at the Olympics parking lot?

The tension between Garfield and John at the Olympics parking lot was caused by a disagreement over a parking spot. Garfield wanted to park in a spot that John was already waiting for, leading to an argument between the two.

## 2. Did the tension between Garfield and John escalate into a physical altercation?

No, the tension between Garfield and John did not escalate into a physical altercation. They were able to resolve their disagreement peacefully.

## 3. How did Garfield and John resolve their tension at the Olympics parking lot?

After their initial argument, Garfield and John were able to calm down and communicate with each other. They were able to come to a compromise and find a different parking spot that worked for both of them.

It is unclear if this was the first time Garfield and John had a disagreement. However, it is likely that they have had disagreements before since they have known each other for some time.

## 5. Did the tension between Garfield and John affect their performance at the Olympics?

It is unknown if the tension between Garfield and John affected their performance at the Olympics. However, it is likely that they were able to put their disagreement behind them and focus on their respective events.

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