/*! This file is auto-generated */ .wp-block-button__link{color:#fff;background-color:#32373c;border-radius:9999px;box-shadow:none;text-decoration:none;padding:calc(.667em + 2px) calc(1.333em + 2px);font-size:1.125em}.wp-block-file__button{background:#32373c;color:#fff;text-decoration:none} Q4MCQ A heavy ball suspended by a cabl... [FREE SOLUTION] | 91Ó°ÊÓ

91Ó°ÊÓ

A heavy ball suspended by a cable is pulled to the side by a horizontal force \(\vec F\), as shown in Fig. 9–43. If angle \(\theta \) is small, the magnitude of the force F can be less than the weight of the ball because

(a) the force holds up only part of the ball’s weight.

(b) even though the ball is stationary, it is not really in equilibrium.

(c) \(\vec F\) is equal to only the x component of the tension in the cable.

(d) the original statement is not true. To move the ball, \(\vec F\) must be at least equal to the ball’s weight.

Short Answer

Expert verified

The correct option is (c).

Step by step solution

01

Concepts

At equilibrium, the net force is zero.For this problem, the horizontal and vertical forces balance each other for the equilibrium of the ball.

02

Explanation

Let m be the mass of the ball and T be the tension in the cable. The cable makes \(\theta \) angle with the vertical.

Condition for equilibrium:

For the horizontal forces, the magnitude of the force is equal to the x component of the tension force.

For the vertical force, the weight is equal to the y component of the tension force.

Hence, option (c) is correct.

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 91Ó°ÊÓ!

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


Question:A 50-story building is being planned. It is to be 180.0 m high with a base 46.0 m by 76.0 m. Its total mass will be about\({\bf{1}}{\bf{.8 \times 1}}{{\bf{0}}^{\bf{7}}}\;{\bf{kg}}\)and its weight therefore about\({\bf{1}}{\bf{.8 \times 1}}{{\bf{0}}^{\bf{8}}}\;{\bf{N}}\). Suppose a 200-km/h wind exerts a force of\({\bf{950}}\;{\bf{N/}}{{\bf{m}}^{\bf{2}}}\)over the 76.0-m-wide face (Fig. 9–80). Calculate the torque about the potential pivot point, the rear edge of the building (where acts in Fig. 9–80), and determine whether the building will topple. Assume the total force of the wind acts at the midpoint of the building’s face, and that the building is not anchored in bedrock. [Hint:\({\vec F_{\rm{E}}}\)in Fig. 9–80 represents the force that the Earth would exert on the building in the case where the building would just begin to tip.]



A 65.0-kg painter is on a uniform 25-kg scaffold supported from above by ropes (Fig. 9–84). There is a 4.0-kg pail of paint to one side, as shown. Can the painter walk safely to both ends of the scaffold? If not, which end(s) is dangerous, and how close to the end can he approach safely?

(II) The Leaning Tower of Pisa is 55 m tall and about 7.7 m in radius. The top is 4.5 m off center. Is the tower in stable equilibrium? If so, how much farther can it lean before it becomes unstable? Assume the tower is of uniform composition.

(II) A 110-kg horizontal beam is supported at each end. A 320-kg piano rests a quarter of the way from one end. What is the vertical force on each of the supports?

(II) If a compressive force of \({\bf{3}}{\bf{.3 \times 1}}{{\bf{0}}^{\bf{4}}}\;{\bf{N}}\) is exerted on the end of a 22 cm long bone of a cross-sectional area of \({\bf{3}}{\bf{.6}}\;{\bf{c}}{{\bf{m}}^{\bf{2}}}\), (a) will the bone break, and (b) if not, by how much does it shorten?

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.