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

A 0.120-kg, 50.0-cm-long uniform bar has a small 0.055-kg mass glued to its left end and a small 0.110-kg mass glued to the other end. The two small masses can each be treated as point masses. You want to balance this system horizontally on a fulcrum placed just under its center of gravity. How far from the left end should the fulcrum be placed?

Short Answer

Expert verified
The fulcrum should be placed approximately 29.82 cm from the left end.

Step by step solution

01

Determine the Position of the Center of the Uniform Bar

The center of gravity (CG) for a uniform bar is located at its geometric center. Since the bar is 50.0 cm long, its center will be at 25.0 cm from either end. Let us denote this position as \( x_b = 25.0 \text{ cm} \).
02

Calculate the Moments about a Chosen Point

We will calculate the moments about the left end of the bar, which we will call point O. We need to consider the uniform bar and the two point masses at the ends. The formula for the moment of a mass \( m \) located at position \( x \) is given by \( m \cdot x \). Thus:1. For the 0.055-kg mass at the left: \( M_1 = 0.055 \times 0 = 0 \).2. For the 0.110-kg mass on the right: \( M_2 = 0.110 \times 50.0 = 5.5 \text{ kg} \cdot \text{cm} \).3. For the 0.120-kg bar: \( M_b = 0.120 \times 25.0 = 3.0 \text{ kg} \cdot \text{cm} \).
03

Calculate the Total Moment about Point O

To find the center of gravity, we need the balance of moments to equal zero, meaning the sum of all moments must be used:\[ \text{Total Moment} = M_1 + M_b + M_2 \]\[ = 0 + 3.0 + 5.5 = 8.5 \text{ kg} \cdot \text{cm} \].
04

Calculate the Total Mass of the System

Add up the masses of all components:1. Mass of 0.055 kg glue: \( m_1 = 0.055 \text{ kg} \).2. Mass of the bar: \( m_b = 0.120 \text{ kg} \).3. Mass of 0.110 kg glue: \( m_2 = 0.110 \text{ kg} \).\( \text{Total Mass} = m_1 + m_b + m_2 = 0.055 + 0.120 + 0.110 = 0.285 \text{ kg} \).
05

Calculate the Center of Gravity from the Left End

The center of gravity position \( x_{cg} \) can be calculated by dividing the total moment by the total mass:\[ x_{cg} = \frac{\text{Total Moment}}{\text{Total Mass}} = \frac{8.5}{0.285} \approx 29.82 \text{ cm} \].

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!

Key Concepts

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

Uniform Bar
A uniform bar is a straight object that has the same cross-sectional dimensions and material properties along its entire length. This means that the mass is distributed evenly across the bar. In the context of physics problems, a uniform bar is often used to simplify calculations.
The uniform distribution of mass allows us to pinpoint the center of gravity (CG) easily. We calculate it to be precisely at the geometric center of the bar. For example, if a bar is 50.0 cm long, its CG will be at 25.0 cm from either end.
This concept helps us to understand how the weight of the bar is balanced. Such an understanding is crucial when performing further calculations like finding moments and balancing the bar on a fulcrum.
Moment of Inertia
The moment of inertia is a measure of an object's resistance to rotation around an axis. It depends on how the object's mass is distributed with respect to the pivot point.
For a point mass, the moment of inertia can be calculated simply by multiplying the mass by the square of the distance from the pivot:
  • For a point mass, the formula is given as: \[ I = m imes r^2 \]where \( I \) is the moment of inertia, \( m \) is the mass, and \( r \) is the distance from the distance to the axis of rotation.
When dealing with multiple masses, like a bar with point masses on either end, the moment of inertia for each mass must be considered separately.
By understanding the moment of inertia, we can calculate the total effects of all masses on the system's rotation.
Point Masses
Point masses refer to small objects whose size can be ignored, and they are considered as having their entire mass concentrated at a single point. This concept is useful in simplifying various calculations like those involving the center of gravity or moments.
In a physics problem, each point mass affects the balance of the system. We calculate the contribution of each point mass using their masses and their distances from a reference point. For example:
  • A 0.055-kg point mass on the left contributes minimally since it rests very close to the reference point.
  • A 0.110-kg mass on the right contributes significantly more as it affects the balance further away from the pivot.
Simplifying a system with point masses allows for straightforward calculations on how they will affect the whole system.
Fulcrum Balance
Fulcrum balance involves positioning the pivot point such that the system remains in equilibrium. This concept is essential when you want to balance asymmetrical systems, such as a bar with different masses glued to its ends.
The balance point, or the fulcrum, is typically placed directly beneath the center of gravity. In achieving balance, the sum of clockwise moments must equal the sum of counterclockwise moments around the fulcrum.
  • Moments are calculated by multiplying the weight of each mass with its distance from the fulcrum.
  • Adjusting the fulcrum position can re-distribute these moments to maintain equilibrium.
With the correct understanding of fulcrum balance, we can predictably determine where the fulcrum should be placed so that the entire system stays stable.

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

The bulk modulus for bone is 15 GPa. (a) If a diver-in-training is put into a pressurized suit, by how much would the pressure have to be raised (in atmospheres) above atmospheric pressure to compress her bones by 0.10% of their original volume? (b) Given that the pressure in the ocean increases by 1.0 \(\times\) 10\(^4\) Pa for every meter of depth below the surface, how deep would this diver have to go for her bones to compress by 0.10%? Does it seem that bone compression is a problem she needs to be concerned with when diving?

An angler hangs a 4.50-kg fish from a vertical steel wire 1.50 m long and 5.00 \(\times\) 10\(^{-3}\) cm\(^2\) in cross-sectional area. The upper end of the wire is securely fastened to a support. (a) Calculate the amount the wire is stretched by the hanging fish. The angler now applies a varying force \(\overrightarrow{F}\) at the lower end of the wire, pulling it very slowly downward by 0.500 mm from its equilibrium position. For this downward motion, calculate (b) the work done by gravity; (c) the work done by the force \(\overrightarrow{F}\), (d) the work done by the force the wire exerts on the fish; and (e) the change in the elastic potential energy (the potential energy associated with the tensile stress in the wire). Compare the answers in parts (d) and (e). \(\textbf{Torques and Tug-of-War.}\) In a study of the biomechanics of the tug-of-war, a 2.0-m-tall, 80.0-kg competitor in the middle of the line is considered to be a rigid body leaning back at an angle of 30.0\(^\circ\) to the vertical. The competitor is pulling on a rope that is held horizontal a distance of 1.5 m from his feet (as measured along the line of the body). At the moment shown in the figure, the man is stationary and the tension in the rope in front of him is \({T_1} =\) 1160 N. Since there is friction between the rope and his hands, the tension in the rope behind him, \({T_2}\) is not equal to \({T_1}\). His center of mass is halfway between his feet and the top of his head. The coefficient of static friction between his feet and the ground is 0.65.

A uniform ladder 5.0 m long rests against a frictionless, vertical wall with its lower end 3.0 m from the wall. The ladder weighs 160 N. The coefficient of static friction between the foot of the ladder and the ground is 0.40. A man weighing 740 N climbs slowly up the ladder. Start by drawing a free-body diagram of the ladder. (a) What is the maximum friction force that the ground can exert on the ladder at its lower end? (b) What is the actual friction force when the man has climbed 1.0 m along the ladder? (c) How far along the ladder can the man climb before the ladder starts to slip?

A uniform metal bar that is 8.00 m long and has mass 30.0 kg is attached at one end to the side of a building by a frictionless hinge. The bar is held at an angle of 64.0\(^\circ\) above the horizontal by a thin, light cable that runs from the end of the bar opposite the hinge to a point on the wall that is above the hinge. The cable makes an angle of 37.0\(^\circ\) with the bar. Your mass is 65.0 kg. You grab the bar near the hinge and hang beneath it, with your hands close together and your feet off the ground. To impress your friends, you intend to shift your hands slowly toward the top end of the bar. (a) If the cable breaks when its tension exceeds 455 N, how far from the upper end of the bar are you when the cable breaks? (b) Just before the cable breaks, what are the magnitude and direction of the resultant force that the hinge exerts on the bar?

Two people are carrying a uniform wooden board that is 3.00 m long and weighs 160 N. If one person applies an upward force equal to 60 N at one end, at what point does the other person lift? Begin with a free-body diagram of the board.

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