Warning: foreach() argument must be of type array|object, bool given in /var/www/html/web/app/themes/studypress-core-theme/template-parts/header/mobile-offcanvas.php on line 20

Tarzan swings from a vine. When he is at the bottom of his swing, as shown in Figure 5.63, which is larger in magnitude: the force by the Earth on Tarzan, the force by the vine (a tension force) on Tarzan, or neither (same magnitude)? Explain how you know this.

Short Answer

Expert verified

The vine's tension force is greater than the gravitational force.

Step by step solution

01

Definition of the tension force

Tension force is the force produced when a load is applied in a directed away from one or more ends of a material, usually to the cross-section of the material. A tension force is also referred to as a "pulling" force.

02

Representation of the diagram

As the Tarzan swings, the net force acting on him at the bottom of the swing acts in the direction of the rate of change of momentum, which is upward.

Tarzan's veins exert a stronger upward force than their veins exert a downward force.

Tarzan must use this upward force to drive himself upward.

The free body diagram for the given situation is shown in the below figure,

here mgis the gravitation force acting on the person, Tis the tension force offered

by vine, and Fcis the centripetal force.

03

Applying the Newtons second law

Write the net force acting on the Tarzan, according to Newton's second law of motion.

Fc=T-mg

Reorganize the above equation to obtain T.

T=Fc+mg

Thus, the vine's tension force is greater than the gravitational force.

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 box of mass 40 kghangs motionless from two ropes, as shown in Figure. The angle is 38ยฐ. Choose the box as the system. The xaxis runs to the right, the yaxis runs up, and the zaxis is out of the page.

(a) Draw a free-body diagram for the box.

(b) Isdpโ†’/dtof the box zero or nonzero?

(c) What is the ycomponent of the gravitational force acting on the block? (A component can be positive or negative).

(d) What is theycomponent of the force on the block due to rope 2?

(e) What is the magnitude of localid="1657085603204" Fโ†’2?

(f) What is thexcomponent of the force on the block due to rope 2?

(g) What is the xcomponent of the force on the block due to rope 1?

A particle moving at nearly the speed of light (vโ‰ˆc) passes through a region where it is subjected to a magnetic force of constant magnitude that is always perpendicular to the momentum and has a magnitude of 2ร—10-10N. As a result, the particle moves along a circular arc with a radius of 8m. What is the magnitude of the momentum of this particle?

A child rides on a playground merry-go-round, from the center. The merry-go-round makes one complete revolution every 5 s. How large is the net force on the child? In what direction does the net force act?

Use a circular pendulum to determine . You can increase the accuracy of the time it takes to go around once by timingN revolutions and then dividing by N. This minimizes errors contributed by inaccuracies in starting and stopping the clock. It is wise to start counting from zero (0,1,2,3,4,5)rather than starting from (0,1,2,3,4,5)represents only four revolutions, not five). It also improves accuracy if you start and stop timing at a well-defined event, such as when the mass crosses in front of an easily visible mark. This was the method used by Newton to get an accurate value of g. Newton was not only a brilliant theorist but also an excellent experimentalist. For a circular pendulum, he built a large triangular wooden frame mounted on a vertical shaft, and he pushed this around and around while making sure that the string of the circular pendulum stayed parallel to the slanting side of the triangle.

A student said, "When the Moon goes around the Earth, there is an inward force due to the Moon and an outward force due to centrifugal force, so the net force on the Moon is zero." Give two or more physics reasons why this is wrong.

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