Tarzan swings through the jungle on a massless vine. At the lowest point of his swing, is the tension in the vine greater than, less than, or equal to the gravitational force on Tarzan? Explain.

Short Answer

Expert verified

There must be a net inward force (toward center of circle) so tension must be greater than the gravitational force acting on Tarzan.

Step by step solution

01

Given information

The tension acting on the vine will create a centripetal force acting downward which is said to be as centripetal acceleration. Tarzan swings with the help of centripetal acceleration. The schematic diagram is shown in the figure.

02

Explanation

If the forces are equal at lowest point of his swing, he will continue to move in a straight line. If gravity is greater than tension, he will accelerate downward toward ground. So, tension must be greater than the gravitational force acting on Tarzan.

Unlock Step-by-Step Solutions & Ace Your Exams!

  • Full Textbook Solutions

    Get detailed explanations and key concepts

  • Unlimited Al creation

    Al flashcards, explanations, exams and more...

  • Ads-free access

    To over 500 millions flashcards

  • Money-back guarantee

    We refund you if you fail your exam.

Over 30 million students worldwide already upgrade their learning with Vaia!

One App. One Place for Learning.

All the tools & learning materials you need for study success - in one app.

Get started for free

Most popular questions from this chapter

A car runs out of gas while driving down a hill. It rolls through the valley and starts up the other side. At the very bottom of the valley, which of the free-body diagrams in FIGURE Q8.2 is correct? The car is moving to the right, and drag and rolling friction are

negligible.

The physics of circular motion sets an upper limit to the speed of human walking. (If you need to go faster, your gait changes from a walk to a run.) If you take a few steps and watch
what’s happening, you’ll see that your body pivots in circular motion over your forward foot as you bring your rear foot forward for the next step. As you do so, the normal force of the
ground on your foot decreases and your body tries to “lift off” from the ground.
a. A person’s center of mass is very near the hips, at the top of the legs. Model a person as a particle of mass m at the top of a leg of length L. Find an expression for the person’s maximum walking speed vmax.
b. Evaluate your expression for the maximum walking speed of a 70 kg person with a typical leg length of 70 cm. Give your answer in both m/s and mph, then comment, based on your
experience, as to whether this is a reasonable result. A “normal” walking speed is about 3 mph.

A 500 g steel block rotates on a steel table while attached to a 1.2-m-long hollow tube as shown in FIGURE CP8.70. Compressed air fed through the tube and ejected from a nozzle on the back of the block exerts a thrust force of 4.0 N perpendicular to the tube.

The maximum tension the tube can withstand without breaking is 50 N. If the block starts from rest, how many revolutions does it make before the tube breaks?

Space scientists have a large test chamber from which all the air can be evacuated and in which they can create a horizontal uniform electric field. The electric field exerts a constant horizontal force on a charged object. A 15gcharged projectile is launched with a speed of 6.0m/s at an angle35° above the horizontal. It lands 2.9m in front of the launcher. What is the magnitude of the electric force on the projectile?

Ramon and Sally are observing a toy car speed up as it goes around a circular track. Ramon says, “The car’s speeding up, so there must be a net force parallel to the track.” “I don’t think so,” replies Sally. “It’s moving in a circle, and that requires centripetal acceleration. The net force has to point to the center of the circle.” Do you agree with Ramon, Sally, or neither? Explain.

See all solutions

Recommended explanations on Physics Textbooks

View all explanations

What do you think about this solution?

We value your feedback to improve our textbook solutions.

Study anywhere. Anytime. Across all devices.

Sign-up for free