Tarzan swinging on a rope -- find the acceleration

In summary,Tarzan swings from a cliff at the end of a 20.0m vine that hangs from a high tree limb. Immediately after Tarzan steps off the cliff, the tension in the vine is 719N and makes an angle of 29.0deg with the vertical. Tarzan's weight is 822N.The magnitude of Tarzan's acceleration immediately after he steps off the cliff is 0.3m/s^2.
  • #1
isukatphysics69
453
8

Homework Statement


Tarzan swings from a cliff at the end of a 20.0m vine that hangs from a high tree limb. Immediately after Tarzan steps off the cliff, the tension in the vine is 719N and makes an angle of 29.0deg with the vertical. Tarzan's weight is 822N.

What is the magnitude of Tarzan's acceleration immediately after he steps off the cliff?

Homework Equations


netforce = m*a

The Attempt at a Solution


netforce =m*a
netforcex = 719cos71
netforcey = 719sin71-822
netforce= sqrt((719cos71)^2+(719sin71-822)^2))
a = 273.88/822
a=0.3m/s^2not sure what i am doing wrong here! >=[
 
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  • #2
isukatphysics69 said:
a = 273.88/822
Weight is not the same as mass.

In general, always write out and check your units in all steps. It will help you identify mistakes such as, in this case, getting a dimensionless acceleration.
 
  • #3
Orodruin said:
Weight is not the same as mass.

In general, always write out and check your units in all steps. It will help you identify mistakes such as, in this case, getting a dimensionless acceleration.
i see i made a stupid mistake there i meant 273/83.87 now i am getting 3.2 and its still wrong
 
  • #4
isukatphysics69 said:
netforce= sqrt((719cos71)^2+(719sin71-822)^2))
What is the result of this computation?
 
  • #5
Orodruin said:
What is the result of this computation?
273.8758934
 
  • #6
isukatphysics69 said:
273.8758934
I suggest you redo that math.

Edit: Also, again, units! The units here should be Newton. Otherwise it is not a force.
 
  • #7
Orodruin said:
I suggest you redo that math.

Edit: Also, again, units! The units here should be Newton. Otherwise it is not a force.
I put it into the calculator and got the same thing, i have the calculator on degree mode do you have yours on radian?
 
  • #8
isukatphysics69 said:
I put it into the calculator and got the same thing, i have the calculator on degree mode do you have yours on radian?
What is 90-29?
 
  • #9
Orodruin said:
What is 90-29?
omg i am so STUPID! >=[
 
  • #10
So the correct solution is what? There is an easier way of solving this problem, but full solutions are not allowed until the OP shows that they have solved the problem so I cannot tell you until you show your full correct solution.
 
  • #11
Orodruin said:
So the correct solution is what? There is an easier way of solving this problem, but full solutions are not allowed until the OP shows that they have solved the problem so I cannot tell you until you show your full correct solution.
netforce= sqrt((719cos61)^2+(719sin61-822)^2)) = 18.21643467

Edit: this is wrong also, there is something fundamental i am doing wrong here. i think i am supposed to be using v^2/R
 
  • #12
isukatphysics69 said:
netforce= sqrt((719cos71)^2+(719sin71-822)^2)) = 18.21643467

Edit: this is wrong also, there is something fundamental i am doing wrong here. i think i am supposed to be using v^2/R
First of all, that is still wrong and it still has 71 in it so it is unclear how you changed it. Second, just as he steps off the ledge, his velocity is zero and all acceleration is tangential. The radial acceleration (which is v^2/r) is zero.
 
  • #13
Orodruin said:
First of all, that is still wrong and it still has 71 in it so it is unclear how you changed it. Second, just as he steps off the ledge, his velocity is zero and all acceleration is tangential. The radial acceleration (which is v^2/r) is zero.
Sorry i am so tired i will try again and come back in 10 minutes i am screwing up big time on this problem and in LIFE! BRB
 
  • #14
I have no idea what i am doing wrong here whose going to be online tomorrow?? my school is closed again because of the storm and there's no tutoring and my stupid self needs help! >=[ i will get my phd in physics one day tarzan not going to stop me!
 
  • #15
Your original idea is workable. You are just not putting numbers into your calculator correctly. Double check everything.
 
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  • #16
I FIGURED IT OUT ITS 4.75 M/S^2!
 
  • #17
I FIGURED IT OUT ITS 4.75 M/S^2!
 
  • #18
Now i am trying to find the angle of acceleration with the vine... i did tan^-1(-193/348) and got -29 degrees but marked wrong what am i doing wrong here

FIGURED IT OUT I WAS DOING ANGLE WITH X AXIS NOT VINE NOW I UNDERSTAND ITS -90 DEGREES I AM SO STUPID
 
Last edited:
  • #19
What is -193 and 348? Consequently, what is the angle that you are computing?
 
  • #20
Orodruin said:
What is -193 and 348? Consequently, what is the angle that you are computing?
I edit post bro see above now I'm working on another problem thank you tho
 
  • #21
So here is the easier way of doing things.

Completely ignore the given forces, they are essentially useless for the solution to the problem. Since the vine has a fixed length, the motion must be along a circle and the vine takes up any force in the radial direction. This means that any force component in the radial direction will be canceled by the tension in the vine. Only the force component in the tangential direction remains and it is equal to the mass multiplied by the projection of the gravitational acceleration on the tangential direction. Since the acceleration is force/mass, the acceleration will just be the projection of the gravitational acceleration onto the tangential direction, i.e., ##g\sin(29^\circ) = 4.75~{\rm m/s}^2##.

Since the vine is radial and the motion tangential, the angle between them must be a right angle.
 
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Related to Tarzan swinging on a rope -- find the acceleration

1. What is the formula for calculating acceleration?

The formula for acceleration is a = (vf - vi) / t, where a is acceleration, vf is final velocity, vi is initial velocity, and t is time.

2. How is acceleration related to Tarzan swinging on a rope?

Acceleration is the change in velocity over time. In the case of Tarzan swinging on a rope, his velocity changes as he swings back and forth, so acceleration is a key factor in his movement.

3. Can you measure Tarzan's acceleration while he is swinging on a rope?

Yes, Tarzan's acceleration can be measured using a device called an accelerometer. This device can track changes in velocity and calculate acceleration.

4. Is Tarzan's acceleration constant while swinging on a rope?

No, Tarzan's acceleration is not constant while swinging on a rope. It changes as he moves back and forth, and can also be affected by external factors such as wind resistance and the weight of his body.

5. How does Tarzan's acceleration affect his speed while swinging on a rope?

Tarzan's acceleration directly affects his speed while swinging on a rope. The greater the acceleration, the faster he will swing. When his acceleration is zero, he will reach the top of his swing and momentarily stop before changing direction.

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