/*! 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} Problem 23 To loosen the lid on a jar of ja... [FREE SOLUTION] | 91Ó°ÊÓ

91Ó°ÊÓ

To loosen the lid on a jar of jam \(8.9 \mathrm{cm}\) in diameter, a torque of \(8.5 \mathrm{N} \cdot \mathrm{m}\) must be applied to the circumference of the lid. If a jar wrench whose handle extends \(15 \mathrm{cm}\) from the center of the jar is attached to the lid, what is the minimum force required to open the jar?

Short Answer

Expert verified
The required force is approximately 56.67 N.

Step by step solution

01

Understand the Problem

We need to calculate the minimum force required to open a jar lid which requires a torque of 8.5 N·m, by using a jar wrench with a handle extending 15 cm from the center (which acts as the lever arm). Torque is the product of force and the lever arm radius.
02

Convert Units

Ensure all measurements are in the same unit system. The handle extends 15 cm from the center, which should be converted to meters. Therefore, 15 cm = 0.15 m.
03

Determine Given Values

We know the required torque \ \( \tau = 8.5 \ \mathrm{N} \cdot \mathrm{m} \) and the length of the lever arm \ \( r = 0.15 \ \mathrm{m} \). We need to find the force \ \( F \).
04

Apply Torque Formula

Torque is defined as \ \( \tau = r \cdot F \cdot \sin\theta \). Here \( \theta = 90° \) (since force applied perpendicularly maximizes torque), so \ \( \sin\theta = 1 \). The equation simplifies to \ \( F = \frac{\tau}{r} \).
05

Calculate the Force

Substitute the known values into the equation: \ \[ F = \frac{8.5 \ \mathrm{N} \cdot \mathrm{m}}{0.15 \ \mathrm{m}} \]. Calculate this to get the required force.
06

Solution Calculation

Perform the division to find the force: \ \[ F = \frac{8.5}{0.15} = 56.67 \ \mathrm{N} \]. Thus, the force required is approximately 56.67 N.

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!

Key Concepts

These are the key concepts you need to understand to accurately answer the question.

Force Calculation
To solve any problem involving torque, you need to understand how to calculate force. Torque is essentially a rotational force applied to an object. The relationship between torque (\( \tau \)) and force (\( F \)) is determined by the torque formula:\[ \tau = r \cdot F \cdot \sin\theta \].Here \( r \) is the lever arm length, and \( \theta \) is the angle between the force applied and the lever arm. In many practical applications, the force is applied perpendicularly, making \( \theta = 90^\circ \), which means \( \sin\theta = 1 \).
To calculate the force required to achieve a certain torque when the force is applied perpendicularly, use the simplified formula:\[ F = \frac{\tau}{r} \].By substituting the provided torque and lever arm into this equation, we can determine the force necessary to achieve the same effect.
Lever Arm
The lever arm is a crucial concept when discussing torque, as it is the distance from the pivot point (or fulcrum) to where the force is applied. In the exercise with the jar wrench, the lever arm is the distance from the center of the jar to the point where the force is applied on the handle of the wrench.
This distance magnifies the force applied, increasing the torque with less effort.The longer the lever arm, the more torque you generate with the same amount of force. Therefore, extending the handle of the jar wrench helps create the necessary torque to open the lid with less force. This is because torque is directly proportional to the lever arm,\[ \tau = r \cdot F \], meaning when you increase \( r \), you decrease \( F \) required for the same torque.
Understanding how lever arms work allows us to use tools effectively, such as using a longer wrench to loosen tight bolts or jar lids.
Unit Conversion
Unit conversion is often a necessary step in solving physics problems, particularly when certain measurements need to be consistent. In our case, we must convert centimeters to meters before applying the torque formula.For our exercise, the handle extending 15 centimeters from the center must be converted to meters, as the standard unit for torque calculations is meters. Conversion from centimeters to meters is straightforward, as 1 meter equals 100 centimeters, so:\[ 15 \ ext{cm} = \frac{15}{100} \ ext{m} = 0.15 \ ext{m} \].
Keeping units consistent not only simplifies calculations but also helps avoid errors. In physics, always ensure that the units match the standard before applying formulas. When force, distance, or any measurement is given in non-standard units, convert them first.

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 \(0.015-\mathrm{kg}\) record with a radius of \(15 \mathrm{cm}\) rotates with an angular speed of \(33 \frac{1}{3} \mathrm{rpm} .\) Find the angular momentum of the record.

\(\mathrm{A} 47.0-\mathrm{kg}\) uniform rod \(4.25 \mathrm{m}\) long is attached to a wall with a hinge at one end. The rod is held in a horizontal position by a wire attached to its other end. The wire makes an angle of \(30.0^{\circ}\) with the horizontal, and is bolted to the wall directly above the hinge. If the wire can withstand a maximum tension of \(1450 \mathrm{N}\) before breaking, how far from the wall can a \(68.0-\mathrm{kg}\) person sit without breaking the wire?

A uniform crate with a mass of \(16.2 \mathrm{kg}\) rests on a floor with a coefficient of static friction equal to 0.571 . The crate is a uniform cube with sides \(1.21 \mathrm{m}\) in length. (a) What horizontal force applied to the top of the crate will initiate tipping? (b) If the horizontal force is applied halfway to the top of the crate, it will begin to slip before it tips. Explain.

A uniform metal rod, with a mass of \(3.7 \mathrm{kg}\) and a length of \(1.2 \mathrm{m},\) is attached to a wall by a hinge at its base. A horizontal wire bolted to the wall \(0.51 \mathrm{m}\) above the base of the rod holds the rod at an angle of \(25^{\circ}\) above the horizontal. The wire is attached to the top of the rod. (a) Find the tension in the wire. Find (b) the horizontal and \((c)\) the vertical components of the force exerted on the rod by the hinge.

A torque of \(0.12 \mathrm{N} \cdot \mathrm{m}\) is applied to an egg beater. (a) If the egg beater starts at rest, what is its angular momentum after \(0.65 \mathrm{s} ?\) (b) If the moment of inertia of the egg beater is \(2.5 \times 10^{-3} \mathrm{kg} \cdot \mathrm{m}^{2},\) what is its angular speed after \(0.65 \mathrm{s} ?\)

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.