Log out Go to App
Go to App

Learning Materials

Features

Discover

Chapter 4: Problem 444

A mass of \(\mathrm{M} \mathrm{kg}\) is suspended by a weight-less string, the horizontal force that is required to displace it until the string makes an angle of \(60^{\circ}\) with the initial vertical direction is (A) \(\mathrm{Mg} / \sqrt{3}\) (B) \(\mathrm{Mg} \cdot \sqrt{2}\) (C) \(\mathrm{Mg} / \sqrt{2}\) (D) \(\mathrm{Mg} \cdot \sqrt{3}\)

Short Answer

Expert verified
The horizontal force required to displace the mass until the string makes an angle of 60 degrees with the initial vertical direction is \(F = \frac{Mg}{\sqrt{3}}\).

Step by step solution

01

Draw the forces diagram

First, draw a diagram with the mass M, the string, and the angles. The forces acting on the mass are tension (T) in the string (that has a vertical and a horizontal component), gravitational force (Mg), and the horizontal force F.
02

Apply equilibrium of forces

We will analyze the vertical and horizontal forces when the mass is in equilibrium. In the vertical direction, the vertical component of tension balances the gravitational force: \(T_v = Mg\) In the horizontal direction, the horizontal component of tension balances the horizontal force: \(T_h = F\) Since the string makes an angle of 60 degrees with the initial vertical direction, the angle between the string and the horizontal direction is 30 degrees (90 - 60 degrees). Thus, we can find the relation between the vertical and horizontal components of tension using trigonometry.
03

Relate tension components

We can find the relationship between the vertical and horizontal components of tension using the angle between the string and the horizontal direction: Using sine function, we have: \(T_h = T \cdot \sin(30^{\circ})\) Using cosine function, we have: \(T_v = T \cdot \cos(30^{\circ})\) Now, we know that \(T_v = Mg\), and \(T_h = F\). Therefore, \(Mg = T \cdot \cos(30^{\circ})\) and \(F = T \cdot \sin(30^{\circ})\)
04

Solve for F

Now, we need to find the value of F in terms of M and g. Divide the equations: \(F = T \cdot \sin(30^{\circ})\) and \(Mg = T \cdot \cos(30^{\circ})\) We obtain: \(F / Mg = \sin(30^{\circ}) / \cos(30^{\circ})\) We know that \(\sin(30^{\circ}) = 1/2\) and \(\cos(30^{\circ}) = \sqrt{3}/2\), so: \(F / Mg = (1/2) / (\sqrt{3}/2)\)
05

Simplify and find the answer

Finally, simplify the expression for F: \(F = Mg \cdot ((1/2) / (\sqrt{3}/2))\) Multiply both sides by 2: \(F = Mg / \sqrt{3}\) This corresponds to the choice (A) in the problem. Therefore, the horizontal force required to displace the mass until the string makes an angle of 60 degrees with the initial vertical direction is: \(F = \frac{Mg}{\sqrt{3}}\)

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.

Start your free trial

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

An ice-cream has a marked value of \(700 \mathrm{kcal}\). How many kilo-watt- hour of energy will it deliver to the body as it is digested $(\mathrm{J}=4.2 \mathrm{~J} / \mathrm{cal})$ (A) \(0.81 \mathrm{kwh}\) (B) \(0.90 \mathrm{kwh}\) (C) \(1.11 \mathrm{kwh}\) (D) \(0.71 \mathrm{kwh}\)

A spring of spring constant \(10^{3} \mathrm{~N} / \mathrm{m}\) is stretched initially \(4 \mathrm{~cm}\) from the unscratched position. How much the work required to stretched it further by another \(5 \mathrm{~cm}\) ? (A) \(6.5 \mathrm{NM}\) (B) \(2.5 \mathrm{NM}\) (C) \(3.25 \mathrm{NM}\) (D) \(6.75 \mathrm{NM}\)

A metal ball of mass \(2 \mathrm{~kg}\) moving with a velocity of $36 \mathrm{~km} / \mathrm{h}$ has a head on collision with a stationary ball of mass \(3 \mathrm{~kg}\). If after the collision, the two balls move together, the loss in kinetic energy due to collision is (A) \(40 \mathrm{~J}\) (B) \(60 \mathrm{~J}\) (C) \(100 \mathrm{~J}\) (D) \(140 \mathrm{~J}\)

A spring is compressed by \(1 \mathrm{~cm}\) by a force of \(4 \mathrm{~N}\). Find the potential energy of the spring when it is compressed by \(10 \mathrm{~cm}\) (A) \(2 \mathrm{~J}\) (B) \(0.2 \mathrm{~J}\) (C) \(20 \mathrm{~J}\) (D) \(200 \mathrm{~J}\)

An electric motor develops \(5 \mathrm{KW}\) of power. How much time will it take to lift a water of mass \(100 \mathrm{~kg}\) to a height of $20 \mathrm{~m}\( ? \)\left(\mathrm{g}=10 \mathrm{~m} / \mathrm{s}^{2}\right)$ (A) \(4 \mathrm{sec}\) (B) \(5 \mathrm{sec}\) (C) \(8 \mathrm{sec}\) (D) \(10 \mathrm{sec}\)
See all solutions

Recommended explanations on English Textbooks

Free Response Essay

Read Explanation

English Language Study

Read Explanation

English Grammar Summary

Read Explanation

Sociolinguistics

Read Explanation

Cues and Conventions

Read Explanation

Language and Social Groups

Read Explanation
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