Tarzan wants to swing on a vine across a river. He is standing on a ledge above the water’s edge, and the river is 5.00 m wide. The vine is attached to a tree branch that is 8.00 m directly above the opposite edge of the river. Initially the vine makes a 60.0° angle with the vertical as he is holding it. He swings across starting from rest, but unfortunately the vine breaks when the vine is 20.0° from the vertical. (a) Assuming Tarzan weighs 900.0 N, what was the tension in the vine just before it broke? (b) Does he land safely on the other side of the river?
The Correct Answer and Explanation is :
To solve the given problem, let’s first break down the information:
- Tarzan’s weight (W) = 900.0 N
- The vine makes a 60.0° angle with the vertical initially.
- The vine breaks when the angle with the vertical is 20.0°.
- The vine is 8.00 m above the opposite edge of the river.
Part (a): Tension in the vine just before it breaks
The forces acting on Tarzan when the vine is at a 20.0° angle with the vertical include:
- His weight (W) acting vertically downward (900.0 N).
- The tension (T) in the vine, which has both vertical and horizontal components.
To find the tension, we analyze the forces acting on Tarzan while he is swinging. The vertical component of the tension (T_vertical) balances the gravitational force (W), and the horizontal component (T_horizontal) provides the centripetal force that keeps Tarzan moving in a circular arc.
Vertical force balance:
[ T \cos(\theta) = W ]
where ( \theta = 20.0° ), so
[ T = \frac{W}{\cos(20.0°)} ]
Substitute ( W = 900.0 \, \text{N} ):
[ T = \frac{900.0}{\cos(20.0°)} ]
[ T = \frac{900.0}{0.9397} ]
[ T \approx 957.0 \, \text{N} ]
So, the tension in the vine just before it broke is approximately 957.0 N.
Part (b): Does Tarzan land safely on the other side of the river?
Now, we need to determine whether Tarzan will land safely on the other side of the river, which is 5.00 m wide.
Tarzan’s horizontal velocity just before the vine breaks can be determined from the centripetal force:
[ T_{\text{horizontal}} = T \sin(\theta) ]
The horizontal component of the velocity will be:
[ v = \sqrt{g L \sin(\theta)} ]
where ( g ) is acceleration due to gravity (9.81 m/s²) and ( L ) is the length of the vine.
Using the above, we can calculate the time Tarzan will take to swing horizontally across the river and if his trajectory will land him safely.