Explanation Given data: Refer to the Figure P8.22 in the textbook for the position of the board along with the box on the table. Mass of the box ( m Box ) is 2.0 kg. Mass of the board ( m Board ) is 6.0 kg. Formula used: Write the expression for static equilibrium condition of the torque as follows. ∑ τ = 0 (1) Here, ∑ τ is the sum of the torques acting on any object. Write the expression for torque in terms of force and distance as follows. τ = F d (2) Here, F is the force, and d is the distance. Write the expression for force due to the weight of the object as follows. w = m g (3) Here, m is the mass of the object, and g is the acceleration due to gravity, which is 9.8 m s 2 . From the given Figure, the length of the board is 50 cm. The board to be in stable state, the center of gravity of the board must be at its center. Therefore, the center of gravity of the board lies at 25 cm from its any end. From the analysis, the center of the gravity of the board lies at 5.0 cm from the end of the table. The given system to be in the stable state, the net torque must be zero. The clockwise torque in the system is due to the box and the counterclockwise torque in the system is due to the board. Compute the sum of the torque by modifying the expression in equation (1) as follows. τ counter-clock + ( − τ clock ) = 0 (4) Here, τ counter-clock is the counterclockwise torque in the system, which is due to the board, and τ clock is the clockwise torque in the system, which is due to the box. Write the expression for counterclockwise torque in the system by using the equation (2) and given figure as follows. τ counter-clock = w Board L end of table-center of gravity (5) Here, w Board is the force due to weight of the board, and L end of table-center of gravity is the distance between end of the table to the center of gravity of the board, which is 5.0 cm. Calculation of force due to weight of the board: Modify the expression in equation (3) to find the force due to weight of the board as follows. w Board = m Board g Substitute 6.0 kg for m Board , and 9.8 m s 2 for g , w Board = ( 6.0 kg ) ( 9.8 m s 2 ) = 58.8 kg ⋅ m s 2 = 58.8 N { ∵ 1 kg ⋅ m s 2 = 1 N } Calculation of counterclockwise torque in the system: Substitute 58.8 N for w Board , 5.0 cm for L end of table-center of gravity in equation (5), τ counter-clock = ( 58

Student Workbook for College Physics: A Strategic Approach Volume 1 (Chs. 1-16)

3rd Edition

ISBN: 9780321908865

Author: Randall D. Knight (Professor Emeritus), Brian Jones, Stuart Field

Publisher: PEARSON

See similar textbooks

Concept explainers

Center of Gravity of a system

The centre of gravity is a property that is based on the geometry of a rigid body or any particle. It is described as an equilibrium point or a hypothetical location where the total weight of the body is concentrated.

Question

Chapter 8, Problem 22P

To determine

To find: The distance from the center of the board to the end of the table before the board begins to tilt.

Expert Solution & Answer

Want to see the full answer?

Check out a sample textbook solution

See solution

Chapter 8, Problem 21P

Chapter 8, Problem 23P

Students have asked these similar questions

No chatgpt pls will upvote

Chapter 8 Solutions

Student Workbook for College Physics: A Strategic Approach Volume 1 (Chs. 1-16)

Ch. 8 - An object is acted upon by two (and only two)...Ch. 8 - Prob. 2CQ Ch. 8 - Could a ladder on a level floor lean against a...Ch. 8 - If you are using a rope to raise a tall mast,...Ch. 8 - Prob. 5CQ Ch. 8 - Prob. 6CQ Ch. 8 - Prob. 7CQ Ch. 8 - A spring exerts a 10 N force after being stretched...Ch. 8 - The left end of a spring is attached to a wall....Ch. 8 - A spring is attached to the floor and pulled...

Ch. 8 - A typical mattress has a network of springs that...Ch. 8 - Take a spring and cut it in half to make two...Ch. 8 - A wire is stretched right to its breaking point by...Ch. 8 - Steel nails are rigid and unbending. Steel wool is...Ch. 8 - The rod in Figure Q8.15 pivots around an axle at...Ch. 8 - Two children hold opposite ends of a lightweight,...Ch. 8 - Prob. 17MCQ Ch. 8 - Prob. 18MCQ Ch. 8 - Prob. 19MCQ Ch. 8 - Prob. 20MCQ Ch. 8 - Prob. 21MCQ Ch. 8 - Prob. 22MCQ Ch. 8 - You have a heavy piece of equipment hanging from a...Ch. 8 - Prob. 24MCQ Ch. 8 - Prob. 1P Ch. 8 - Prob. 2P Ch. 8 - Prob. 3P Ch. 8 - Prob. 4P Ch. 8 - Youre carrying a 3.6-m-long, 25 kg pole to a...Ch. 8 - Prob. 6P Ch. 8 - How much torque must the pin exert to keep the rod...Ch. 8 - Prob. 8P Ch. 8 - Prob. 9P Ch. 8 - Consider the procedure for measuring a womans...Ch. 8 - Prob. 11P Ch. 8 - Prob. 12P Ch. 8 - Prob. 13P Ch. 8 - Prob. 14P Ch. 8 - Prob. 15P Ch. 8 - Prob. 16P Ch. 8 - Prob. 17P Ch. 8 - The stability of a vehicle is often rated by the...Ch. 8 - You want to slowly push a stiff board across a 20...Ch. 8 - Prob. 20P Ch. 8 - A car manufacturer claims that you can drive its...Ch. 8 - Prob. 22P Ch. 8 - One end of a spring is attached to a wall. A 25 N...Ch. 8 - An orthodontic spring, connected between the upper...Ch. 8 - Experiments using optical tweezers measure the...Ch. 8 - A spring has an unstretched length of 10 cm. It...Ch. 8 - One end of a 10-cm-long spring is attached to the...Ch. 8 - A spring stretches 5.0 cm when a 0.20 kg block is...Ch. 8 - A 1.2 kg block is hung from a vertical spring,...Ch. 8 - You need to make a spring scale to measure the...Ch. 8 - A force stretches a wire by 1.0 mm. a. A second...Ch. 8 - What hanging mass will stretch a 2.0-m-long,...Ch. 8 - How much force does it take to stretch a...Ch. 8 - An 80-cm-long, 1.0-mm-diameter steel guitar string...Ch. 8 - A student is testing a 1.0 m length of...Ch. 8 - A 1.2-m-long steel rod with a diameter of 0.50 cm...Ch. 8 - A mineshaft has an ore elevator hung from a single...Ch. 8 - The normal force of the ground on the foot can...Ch. 8 - A three-legged wooden bar stool made out of solid...Ch. 8 - Prob. 40P Ch. 8 - A glass optical fiber in a communications system...Ch. 8 - The Achilles tendon connects the muscles in your...Ch. 8 - Prob. 43GP Ch. 8 - Prob. 44GP Ch. 8 - Using the information in Figure 8.2, calculate the...Ch. 8 - A woman weighing 580 N does a pushup from her...Ch. 8 - Prob. 47GP Ch. 8 - Prob. 48GP Ch. 8 - Prob. 49GP Ch. 8 - The wheel of mass m in Figure P8.50 is pulled on...Ch. 8 - A 5.0 kg mass hanging from a spring scale is...Ch. 8 - Two identical, side-by-side springs with spring...Ch. 8 - Two springs have the same equilibrium length but...Ch. 8 - Figure P8.54 shows two springs attached to a block...Ch. 8 - A 60 kg student is standing atop a spring in an...Ch. 8 - A 25 kg child bounces on a pogo stick. The pogo...Ch. 8 - Figure P8.57 shows a lightweight plank supported...Ch. 8 - In the hammer throw, an athlete spins a heavy mass...Ch. 8 - There is a disk of cartilage between each pair of...Ch. 8 - In Example 8.1, the tension in the biceps tendon...Ch. 8 - Larger animals have sturdier bones than smaller...Ch. 8 - Orb spiders make silk with a typical diameter of...Ch. 8 - Prob. 63MSPP Ch. 8 - Prob. 64MSPP Ch. 8 - Prob. 65MSPP

Knowledge Booster

Learn more about

Need a deep-dive on the concept behind this application? Look no further. Learn more about this topic, physics and related others by exploring similar questions and additional content below.

The distance from the center of the board to the end of the table before the board begins to tilt.

Solution Summary: The author calculates the distance between the center of the board and the table before it begins to tilt.

Concept explainers

To find: The distance from the center of the board to the end of the table before the board begins to tilt.

Want to see the full answer?

Chapter 8 Solutions