# Finding tension in tarzan swing

• vu10758
In summary, Tarzan is attempting to swing across a river on a vine. Starting from rest, the tension in the rope at the bottom of the swing can be determined using the law of conservation of energy and Newton's second law. The tension is a function of the angle theta and other necessary variables, and is found to be equal to mg(2cos(theta) - 1). While Tarzan may feel like his weight is equal to the tension, this is not the case as he is accelerating in a circular motion.
vu10758
Tarzan (mass m) is trying to swing across a river on a long vine (length L) Assume that he starts from rest. Determine the tension in the rope at the bottom of the arc as a function of the angle theta, the arc angle between his starting point and ending point and whatever other variables necessary. Hints: Draw Free body diagrams for Tarzan when the rope is vertical.

I drew the free body diagram. It has tension pointing up and gravity pointing down. However, I don't know what to do.

vu10758 said:
It has tension pointing up and gravity pointing down.
So far, so good. What does Newton's 2nd law tell you? Hint: What's the acceleration?

vu10758 said:
Tarzan (mass m) is trying to swing across a river on a long vine (length L) Assume that he starts from rest. Determine the tension in the rope at the bottom of the arc as a function of the angle theta, the arc angle between his starting point and ending point and whatever other variables necessary. Hints: Draw Free body diagrams for Tarzan when the rope is vertical.

I drew the free body diagram. It has tension pointing up and gravity pointing down. However, I don't know what to do.

You mean, tension pointing in the direction of the rope, right?
Use the fact that energy is conserved.

Btw, I'm sure I saw a few more 'Tarzan' problems in the last few days, so using the search option might be useful.

Using the law of conservation of energy

K_i + U_i + W_other = K_f + U_f

(1/2)m(v_i)^2 + mgLcos(theta) = (1/2)m(v_f)^2 + mgL

I let the height equals zero at the top

v_i is zero since it starts from rest

mgLcos(theta) = (1/2)m(v_f)^2 + mgL

the masses cancel

gLcos(theta) = (1/2)(v_f)^2 + gL
gLcos(theta) - gL = (1/2)(v_f)^2
2(gLcos(theta) - gL) = (v_f)^2

I know the the motion is circular so I know

F = m(v^2)/r

In this case, v is v_f and r is L

F = m(v^2)/L

I want to solve for the force of tension which is in the same direction as the centripetal force in this case

so

F = m(2gLcos(theta) - gL) / L
F = mL(2gcos(theta) - g)/L
F = m (2gcos(theta) - g)
F = mg (2cos(theta) - 1)

T = mg(2cos(theta) - 1)

Is this correct?

Just a few corrections:
vu10758 said:
Using the law of conservation of energy

K_i + U_i + W_other = K_f + U_f

(1/2)m(v_i)^2 + mgLcos(theta) = (1/2)m(v_f)^2 + mgL

I let the height equals zero at the top
If you let the height = 0 at the top (top of the vine, I presume), then the initial and final heights will be negative, since they are below the top:
(1/2)m(v_i)^2 - mgLcos(theta) = (1/2)m(v_f)^2 - mgL

The main thing is that Tarzan drops: his final height is lower than the initial.

v_i is zero since it starts from rest

mgLcos(theta) = (1/2)m(v_f)^2 + mgL

the masses cancel

gLcos(theta) = (1/2)(v_f)^2 + gL
gLcos(theta) - gL = (1/2)(v_f)^2
2(gLcos(theta) - gL) = (v_f)^2
Realize that the left hand side of this is negative--the correction above will fix that.

I know the the motion is circular so I know

F = m(v^2)/r

In this case, v is v_f and r is L

F = m(v^2)/L

I want to solve for the force of tension which is in the same direction as the centripetal force in this case

so

F = m(2gLcos(theta) - gL) / L
F = mL(2gcos(theta) - g)/L
F = m (2gcos(theta) - g)
F = mg (2cos(theta) - 1)

T = mg(2cos(theta) - 1)
Right approach, but realize that the tension is not the only force acting on Tarzan. (Use the net force in the above equations.)

how about making tarzan's weight equal and opposite to the tension in the rope?

billiards said:
how about making tarzan's weight equal and opposite to the tension in the rope?
That will not do. When Tarzan swings on the rope he is traveling on a circular path, which means he is accelerating. There must be a net force acting upwards when he is at the bottom of the swing. The tension must be greater than his weight.

Tarzan will feel like his weight is the tension in the rope, because he has to pull that hard to hold on. But what he feels is not his true weight.

Last edited:
billiards said:
how about making tarzan's weight equal and opposite to the tension in the rope?
Huh? You're supposed to solve for the tension, not just assume it to be something. (The tension doesn't equal Tarzan's weight--if it did, he'd be in equilibrium.)

## 1. How is tension calculated in a Tarzan swing?

The tension in a Tarzan swing can be calculated using the formula: T = (m * v^2) / L, where T is the tension, m is the mass of the swinger, v is the velocity at the lowest point of the swing, and L is the length of the swing.

## 2. What factors affect the tension in a Tarzan swing?

The tension in a Tarzan swing is affected by the mass of the person swinging, the velocity at the lowest point of the swing, and the length of the swing. Other factors such as air resistance and the angle of the swing can also affect tension.

## 3. How does the tension in a Tarzan swing change during the swing?

The tension in a Tarzan swing is constantly changing during the swing. It is highest at the top of the swing and lowest at the bottom. This is because at the top, the potential energy is at its maximum and the kinetic energy is at its minimum, and vice versa at the bottom.

## 4. How does the tension in a Tarzan swing relate to the force experienced by the swinger?

The tension in a Tarzan swing is directly related to the force experienced by the swinger. As tension increases, the force on the swinger also increases. This is because tension is a type of force, specifically a pulling force.

## 5. Can the tension in a Tarzan swing be too high?

Yes, the tension in a Tarzan swing can be too high. If the tension is too high, it can exceed the strength of the swing or the material it is attached to, leading to potential safety hazards. It is important to carefully calculate and adjust the tension to ensure the safety of the swinger.

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