# Me Tarzan, You Jane. Force Problem.

• tambourine
In summary, the problem involves Jane trying to climb a vine to reach Tarzan's treetop hut. The vine can safely support the combined weight of Tarzan, Jane, and Cheeta. Using the given information about their masses and the length of the vine, the minimum time for Jane to climb can be calculated using the equations of motion. However, the answers provided by other students may not be correct as energy is not conserved in this situation.
tambourine
holy physics! i can't wrap my head around this.

## Homework Statement

Jane wishes to quickly scale a slender vine to visit Tarzan in his treetop hut. The vine is known to safely support the combined weight of Tarzan, Jane, and Cheeta. Tarzan has twice the mass of Jane, who has twice the mass of Cheeta. If the vine is 60 m long, what minimum time should Jane allow for the climb?

## The Attempt at a Solution

this is what I've scrounged so far:

d= 60 m
V1 = 0

jane wants to quickly scale the vine - let's find a, then t

cheetah's weight is x, jane is 2x, tarzan is 4x. the vine can hold 7x, so jane can apply 3.5x her own weight.

Fnet = ma
Fg = mg = ma
3.5mg = ma [cancel masses]
3.5 (9.81) = a
a = 34.3

then plug a into one of the kinematics to find t. is this correct?

edit: i got t = 1.87 s

i guess what i am trying to say is that it seems too easy now compared to the amount of time i just stared at the question so maybe i have made some terrible mistake

Last edited:
tambourine said:
holy physics! i can't wrap my head around this.

Interesting introduction...what you did seems right considering that her net acceleration would be constant.

ooo tambourine you were so close. You forgot a basic detail in your last equation. I am guessing that you used Ke=1/2kg*v^2 and then t=d/v

ok, when you use the Kenectic energy formula, you go a velocity of 34.31m/s correct?
than you probably did:

t=34.31m/s*60m, well that's where you were slightly wrong. Remember that the velocity changes from 0m/s to 34.31m/s. That means that you had to find the average velocity witch is 17.16m/s

now you can do this:
t=17.16m/s*60m
t= 3.5s

You were so close :(. Beautiful work though

Hi tiale11,

tiale11 said:
ooo tambourine you were so close. You forgot a basic detail in your last equation. I am guessing that you used Ke=1/2kg*v^2 and then t=d/v

ok, when you use the Kenectic energy formula, you go a velocity of 34.31m/s correct?
than you probably did:

t=34.31m/s*60m, well that's where you were slightly wrong. Remember that the velocity changes from 0m/s to 34.31m/s. That means that you had to find the average velocity witch is 17.16m/s

now you can do this:
t=17.16m/s*60m
t= 3.5s

You were so close :(. Beautiful work though

I don't believe your velocity of 34.31m/s is correct. How did you get that number?

It looks like you just set the change in kinetic energy equal to the change in potential energy, but that is not right for this problem. Energy is not conserved here. Or did you do something else?

Last edited:
i am 100% sure that i am correct.
to find the velocity, I used Ke= 1/2*m*v^2 and soleved for v
2x*9.81m/s^2*60m=1/2*2x*v^2.
and yes, you can use the potential energy as the kenetic energy. Think of droping Jane with the same acceleation(replaces gravity) that she went up. At the top she will have no kenetic energy but have potential energy. As she hits the floor she will have TRANSFERED all of that potential energy to make it become kenetic energy. The same concept for bottom up. Again i am sure my answer is correct. You have to remember CONSERVATION of ENERGY!
How is the energy not conserved here?

I don't know if you notice but your time is half of mine. your error is because you did not use the average velocity to find the time.
Show me how you found your velocity and time.

tiale11 said:
i am 100% sure that i am correct.
to find the velocity, I used Ke= 1/2*m*v^2 and soleved for v
2x*9.81m/s^2*60m=1/2*2x*v^2.
and yes, you can use the potential energy as the kenetic energy. Think of droping Jane with the same acceleation(replaces gravity) that she went up. At the top she will have no kenetic energy but have potential energy. As she hits the floor she will have TRANSFERED all of that potential energy to make it become kenetic energy. The same concept for bottom up. Again i am sure my answer is correct. You have to remember CONSERVATION of ENERGY!
How is the energy not conserved here?

I don't know if you notice but your time is half of mine. your error is because you did not use the average velocity to find the time.
Show me how you found your velocity and time.

I did not tell you what my time I got for the problem. I think the answer the original poster got and you got are both incorrect.

Since she is pulling herself up, work is being done on her, and energy is not conserved here.

Hi tambourine,

tambourine said:
holy physics! i can't wrap my head around this.

## Homework Statement

Jane wishes to quickly scale a slender vine to visit Tarzan in his treetop hut. The vine is known to safely support the combined weight of Tarzan, Jane, and Cheeta. Tarzan has twice the mass of Jane, who has twice the mass of Cheeta. If the vine is 60 m long, what minimum time should Jane allow for the climb?

## The Attempt at a Solution

this is what I've scrounged so far:

d= 60 m
V1 = 0

jane wants to quickly scale the vine - let's find a, then t

cheetah's weight is x, jane is 2x, tarzan is 4x. the vine can hold 7x, so jane can apply 3.5x her own weight.

Fnet = ma
Fg = mg = ma
3.5mg = ma [cancel masses]

I believe that here is your error. When you consider the force diagram acting on Jane, there are two forces acting on here--the tension and the weight force. But here you only have the tension acting on her. Once you put the weight force in the equation you should get the right answer.

In other words, you found the maximum tension was 3.5 times here weight; but part of that tension has to counteract the weight force. Only what's left over actually accelerates her upwards.

thanks for the responses!

tiale11, i don't quite understand what you did. how did you solve using the kinetic energy formula if mass wasn't given?

i got t=1.87 by plugging d=60 m, V1=0, and a=34.3 into d=V1t + 1/2at^2

alphysicist, this is the concept I'm having trouble with. would the tension and her weight both point in the same direction, downwards? and if so, what is propelling her up? she's applying tension on the vine, but that force points down. when she is at rest, hanging on the vine, would the tension force equal her weight? since without her, there is no tension.

can you please help me understand this? heck, anyone, feel free to jump in! we could talk in circles together. it would be fun.

tambourine said:
thanks for the responses!

tiale11, i don't quite understand what you did. how did you solve using the kinetic energy formula if mass wasn't given?

tiale11's approach was based on the belief that energy was conserved, which is not true here.

i got t=1.87 by plugging d=60 m, V1=0, and a=34.3 into d=V1t + 1/2at^2

alphysicist, this is the concept I'm having trouble with. would the tension and her weight both point in the same direction, downwards? and if so, what is propelling her up? she's applying tension on the vine, but that force points down. when she is at rest, hanging on the vine, would the tension force equal her weight? since without her, there is no tension.

can you please help me understand this? heck, anyone, feel free to jump in! we could talk in circles together. it would be fun.

In the force equation (from her force diagram), we need the force that the rope puts on her. If she pulls down on the rope, which way does the rope pull on her? That is the way I would think about it. So which way is the tension? And what equation would you get?

so.. the rope pulls up at her, equal and simultaneous-

Fnet = ma

if down is positive:

Fg - Ft = ma
3mg - Ft = ma ?

tambourine said:
so.. the rope pulls up at her, equal and simultaneous-

Fnet = ma

if down is positive:

Fg - Ft = ma
3mg - Ft = ma ?

No, I don't think that is right. The rope pulls up, and the weight pulls down, so:

T - W = m a

where T is tension and W is the weight. So what is W? And what is T? Remember that you had the tension in your first post. What do you get for a?

3mg - mg = ma
so 2mg = ma ?

tambourine said:
3mg - mg = ma
so 2mg = ma ?

In your original post, you found that the maximum tension allowed was 3.5 times her weight, which I think is the correct value.

oh, i can't see numbers straight anymore. meant 3.5. thank you alphysicist :D

Sure, glad to help! And I think these long hours are starting to get to me, too.

dudes, i am so sorry. I thought Jane did not climb but hung on onto the vine which acted like a spring :(. lol Now i see why you folks said that energy was not conserved. I was on a totaly different topic.
Ok so now let me get this right. the question asks to find the time that Jane will climb the 60m vine. Is it like a pulley problem with 2 directions? please explain so i can have a crack at it

tiale11 said:
dudes, i am so sorry. I thought Jane did not climb but hung on onto the vine which acted like a spring :(. lol Now i see why you folks said that energy was not conserved. I was on a totaly different topic.
Ok so now let me get this right. the question asks to find the time that Jane will climb the 60m vine. Is it like a pulley problem with 2 directions? please explain so i can have a crack at it

No, there's only one direction involved. The vine would be attached to the tree at the top.

The idea is that the problem gives you a way to find the upper limit of tension that the vine can withstand before breaking. The tension is the only upward force acting on Jane, so that allows you to find the maximum upward acceleration.

It's different because the normal thing is for ropes to be tied to the object, but that is not necessary; there just has to be some way for the rope and object to apply forces to each other.

Also, this doesn't help in finding the answer, but if you want to think more about the problem, you could say that it's not really the tension force that pulls her up; it's really the frictional force between the vine and her hand. You would then need to draw a force diagram for the vine, which would show that the tension force and frictional force are equal in magnitude.

ok so, the vine is like a rope, it is not moving at all. Jane does all the work. She does not hold on to a certain spot on the vine and it pulls her up. is that correct?

tiale11 said:
ok so, the vine is like a rope, it is not moving at all. Jane does all the work. She does not hold on to a certain spot on the vine and it pulls her up. is that correct?

Right, that's how I visualized it; she's climbing up, hand over hand. (The problem would be the same, though, if she was just holding on and someone was pulling the vine up--assuming you can ignore the mass of the vine.)

o0o ok. now i am going to solve this using what alphysicist said about another person pulling Jane up a weigthless vine.
I am now going to assume that Tarzan, Jane and cheeta are the maximum weight on the vine that the other "person" can pull up. ****I am doing this like a pulley problem****
the tension of all of them will equal T=7x*a. Now, the acceleration that will be produced on Jane alone on the vine will be : a=(7x*a-2x*a)/2x=24.525m/s^2. Now, using d=vi*t+1/2*a*t^2 i solved for t and calculated it to be 2.212s. Well considering what i assumed for this problem, this should be right

tiale11 said:
o0o ok. now i am going to solve this using what alphysicist said about another person pulling Jane up a weigthless vine.
I am now going to assume that Tarzan, Jane and cheeta are the maximum weight on the vine that the other "person" can pull up. ****I am doing this like a pulley problem****
the tension of all of them will equal T=7x*a. Now, the acceleration that will be produced on Jane alone on the vine will be : a=(7x*a-2x*a)/2x=24.525m/s^2. Now, using d=vi*t+1/2*a*t^2 i solved for t and calculated it to be 2.212s. Well considering what i assumed for this problem, this should be right

man, finally! :). Well next time ill make sure i understand what the question is asking first before answering
thanks for clearing stuff out

Assuming 2.1s and 25 m/s^2 she ends at about 200 km/h, how does she stop?

Borek said:
Assuming 2.1s and 25 m/s^2 she ends at about 200 km/h, how does she stop?

There's a big branch that stops her, so all she needs is something for her head. That's where all of those egg drop experiments come in. Perhaps the old "insert her head in pantyhose and staple the ends to the box"?

Rofl!

here is the answer, which is same with the answer on ur textbook

solutions:
since it told u that the the vine could safetly support combined weight of Tarzan, Jane and Cheeta

therefore, the maximum tension force is the combined weight.

Assuming the mass of Cheeta is m, then Jane is 2m, therefore Tarzan is 4m.

so the maximum tension equals to T=m+2m+4m= 7m

According to Newton's second law, F(net)= mass times acceleration

and F(net)= 7mg-2mg=5mg
so 5mg= 2ma
therefore, the acceleration is 5mg/ 2m=2.5g (g=9.8m/s^2)

since the length of the vine is 60m, therefore the displacement is 60m

because: d= v(initial)t+(1/2)a(t)^2,the initial velocity equals zero.

therefore, t=2.2s

heyy, I just had this problem on a homework sheet so i was looking around for some help when i came across this forum. thanks for everyone's help, AND the explanations, they really help.

just so you guys know, the official answer, as this question was from a Sir Isaac Newton UW Physics Exam in '75, is t=2.2s so whoever got that answer must be right.

again, thanks for the help!

## 1. What is the "Me Tarzan, You Jane" force problem?

The "Me Tarzan, You Jane" force problem is a concept in physics that explores the forces involved in swinging from a vine, similar to the way Tarzan swings in the jungle. It is used to demonstrate the relationship between the force of gravity, the tension in the vine, and the centripetal force needed to maintain circular motion.

## 2. How is the "Me Tarzan, You Jane" force problem solved?

The "Me Tarzan, You Jane" force problem is solved by using the laws of physics, specifically Newton's laws of motion and the concept of centripetal force. This allows us to calculate the necessary forces involved in swinging from a vine, such as the tension in the vine and the required speed to maintain circular motion.

## 3. What are the key components of the "Me Tarzan, You Jane" force problem?

The key components of the "Me Tarzan, You Jane" force problem are gravity, tension, and centripetal force. Gravity is the force that pulls the swinging object towards the ground, tension is the force exerted by the vine on the object, and centripetal force is the force that keeps the object moving in a circular path.

## 4. How does the "Me Tarzan, You Jane" force problem relate to real-life scenarios?

The "Me Tarzan, You Jane" force problem can be applied to real-life scenarios, such as amusement park rides or sports activities like swinging on a rope. It helps us understand the forces involved in these activities and how they affect the motion of the object or person.

## 5. Why is it important to understand the "Me Tarzan, You Jane" force problem?

Understanding the "Me Tarzan, You Jane" force problem helps us understand the fundamental principles of physics and how they apply to real-life situations. It also allows us to calculate and predict the forces involved in swinging from a vine, which can be useful in designing safe and efficient structures or activities that involve circular motion.

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