2/4100%+B. Weighing a rock(Q-4) Balance the meter stick with two weights and a rock.O cm100cmposition (cm) weight (gwt) lever arm (cm) torque (gwt cm)Rock97. 16+oр3 XXXXXXXXXXXXXX25.16tao3 99.U 1 0-05Weight 2 40. 701003 150.0l 005Weight 1(Q-5) Assume that the torques balance and use this fact to calculate the weight of the rock.Weale(Q-6) Measure the weight of the rock using an electronic balance.W measCalculate the percent discrepancy.WmeasWeale% discrepancy =x 100% =WmeasUp to this point, we have not considered the weight of the meter stick or the upward forceacting on the meter stick due to the fulcrum. Why can we get away with this? Rotational EquilibriumName:Section: 2AL-_Date performed: /-I-Lab station:Partners:A. Conditions for equilibrium(Q-1) Determine the location of the center of gravity of the meter stick by balancing themeter stick alone.Teg = (50.39L±_0.03)cm(Q-2) Now balance the meter stick with 3 weights attached. Use three different weights,each between 100 gwt and 500 gwt.O cm100 cmposition (cm)weight (gwt)lever arm (cm)torque (gwt cm)126.30 + 0.0399.91± 0.0525.05+ 0.03150.el + 0.053 87.27 ± 0. 03 249.16± 0.05土Sum of clockwise torques =Sum of counterclockwise torques =Use this space for calculations.(Q-3) Do the torques balance? Use the discrepancy test and explain your results.N.BNo need to convort the weight in New tons, keep gwt unit

Hello. Since you have posted multiple questions with multiple subparts (solved and unsolved) and not…

cm %3D (Q-2) Now balance the meter stick with 3 weights attached. Use three different weights, each between 100 gwt and 500 gwt. O cm 100 cm position (cm) 126.30 + 0.03 weight (gwt) 99.91+ 0.05 lever arm (cm) torque (gwt cm) 2 5.05 + 0,03 150.ol + 0.05 87.27 + 0.03 249.16+ 0.05 土土 Sum of clockwise torques Sum of counterclockwise torques

cm %3D (Q-2) Now balance the meter stick with 3 weights attached. Use three different weights, each between 100 gwt and 500 gwt. O cm 100 cm position (cm) 126.30 + 0.03 weight (gwt) 99.91+ 0.05 lever arm (cm) torque (gwt cm) 2 5.05 + 0,03 150.ol + 0.05 87.27 + 0.03 249.16+ 0.05 土土 Sum of clockwise torques Sum of counterclockwise torques

College Physics

11th Edition

ISBN:9781305952300

Author:Raymond A. Serway, Chris Vuille

Publisher:Raymond A. Serway, Chris Vuille

Chapter1: Units, Trigonometry. And Vectors

Section: Chapter Questions

Problem 1CQ: Estimate the order of magnitude of the length, in meters, of each of the following; (a) a mouse, (b)...

See similar textbooks

Related questions

Q: A wheel of inner radius r, = 13.4 cm and outer radius r, = 35.0 cm shown in the figure below is free…

Q: The (Figure 1) shows two thin beams joined at right angles. The vertical beam is 12.0 kg and 1.00 m…

A: Given : Mass of vertical beam (M1) = 12 kg Length of vertical beam (L1) = 1 m Mass of horizontal…

Q: 28. The fishing pole in Figure P10.28 makes an angle of W 20.0° with the horizontal. What is the…

A: Given: Force, F = 100 NRadius, r = 2 mFind: The torque exerted by the fish about an axis…

Q: A uniform 1200 N piece of medical apparatus that is 3.5 m long is suspended horizontally by two…

Q: When tightening a bolt, you push perpendicularly on a wrench with a force of 167 N at a distance of…

A: Given, Force, F=167N Distance, r=0.14m

Q: F = N-m 124 N at an angle 37.0° below the horizontal? The force is applied at a point L = 2.02 m…

A: Step 1:

Q: PROBLEM 1-A uniform beam (uniform rod) is being used to support a sign: Suddenly, theifmas…

A: The objective of the question is to find the angular acceleration of the beam at the moment the…

Q: The body in the figure is pivoted at O. Three forces act on it in the directions shown: FA = 15.0 N…

A: Magnitude of force FA = 15N Distance of point A from point O is rA = 5.80m Angle between vector…

Q: The masses m1 = 2.9 kg and m2 = 10.5 kg are placed on the beam shown below to balance. If the…

Q: 31 32

A: 31) the expression for the balancing moment with respect to the point where normal force is acting.

Q: The body in the figure is pivoted at O, and two forces act on it as shown. If r₁-2.83 m, r₂-2.40 m.…

A: r1-2.83 m. r2-2.40m F1-5.08 N. F₂-5.99 N. and θ-79.0°, θ-49.0°, Take a torque that tends to cause…

Q: An 8.0-cm radius disk with a rotational inertia of 0.12 kg.m² is free to rotate on a horizontal…

A: 1:1. Relate Torque and Angular Acceleration:• Torque (τ): This is a rotational force that causes an…

Q: 1. Find the net torque on the wheel about the axis O shown below. Include the associated direction…

Q: Problem 6: Rotational Dynamics - Torque A 0u cm, uniform, 50 N shelf is supported horizontally by…

Q: PROBLEM 1-A uniform beam (uniform rod) is being used to support a sign: Suddenly, thefama horizontal…

A: The objective of this problem is to find the angular acceleration of the beam at the moment the…

Q: Find the net torque on the wheel in the figure below about the axle through O, taking a = 15.0 cm…

A: The net torque with reapect to the center O is -4.09Nm.Explanation:Please see the attached image for…

Q: A 20 kg wooden plank is supported by a fulcrum at the midpoint. A 79 kg man is positioned 1.5 m from…

A: Given data: Mass of wooden plank (m) = 20 kg Mass of man (m1) = 79 kg Distance of man from fulcrum…

Q: Problem 7 A potter's wheel -a thick stone disk or radius 0.6 m and mass 100 kg- is freely rotating…

A: Radius of wheel, R = 0.6 m Mass, m= 100 kg Initial angular velocity, ωi =1 rad/sec Final angular…

Q: PROBLEM 1-A uniform beam (uniform rod) is being used to support a sign: Suddenly, the fe horizontal…

Q: Hélp What is the rotational inertia of a solid iron disk of mass 40.0 kg, with a thickness of 5.00…

Q: Summation of Forces Forces are given; Force 1 = 12 N @ 110° Force 2 = 15 N @260° Force 3 = 5 1 - 6…

A: First we have to resolve the forces into Cartesian co-ordinates then simply adding x and y…

Q: A torque of 79.5 N m is applied to a solid, uniform disk of radius 66.2 cm. The disk accelerates at…

A: Write the expression for the moment of inertia of the disk. I=12mr2

Q: 56. A solid steel cylinder is standing (on one of its ends) vertically on the floor. The length of…

A: Write the formula for Young’s modulus.

Q: The wheel has a mass of 10 kg and total radius of gyration of .15m. A force of 35 N is pulled as…

A: Angular acceleration is defined as the time rate of change of angular displacement with time. It is…

Q: In the figure here, a cylinder having a mass of 4.5 kg can rotate about its central axis through…

Q: A 240-N solid sphere that is 0.200-m in radius rolls 6.00-m down a ramp that is inclined at 37.0°…

A: Given: Weight of the solid sphere = W = 240 N Radius of the sphere = r = 0.2 m Distance rolled down…

Q: 16. In unit-vector notation, what is the net torque about the origin on a flea located at…

A: Given that,flea located at coordinates, r= 0,-4.0m,5.0mcan be written as, r→=0i^-4.0j+5k^where…

Q: 4. Rank the torques in order from largest to smallest: A: mass mo, radius ro B: mass 2mo, radius ro…

A: Given Rank the torque in order from largest to smallest. A.. mass m0 radius r0 , force F0 B..…

Q: What is the torque by the trash can about the center of the seesaw, in N-m? Use g = 10 m/s². Your…

Concept explainers

Rotational Equilibrium And Rotational Dynamics

In physics, the state of balance between the forces and the dynamics of motion is called the equilibrium state. The balance between various forces acting on a system in a rotational motion is called rotational equilibrium or rotational dynamics.

Equilibrium of Forces

The tension created on one body during push or pull is known as force.

Question

2/4
100%
+
B. Weighing a rock
(Q-4) Balance the meter stick with two weights and a rock.
O cm
100cm
position (cm) weight (gwt) lever arm (cm) torque (gwt cm)
Rock
97. 16+oр3 XXXXXX
XXXXXXXX
25.16tao3 99.U 1 0-05
Weight 2 40. 701003 150.0l 005
Weight 1
(Q-5) Assume that the torques balance and use this fact to calculate the weight of the rock.
Weale
(Q-6) Measure the weight of the rock using an electronic balance.
W meas
Calculate the percent discrepancy.
Wmeas
Weale
% discrepancy =
x 100% =
Wmeas
Up to this point, we have not considered the weight of the meter stick or the upward force
acting on the meter stick due to the fulcrum. Why can we get away with this?

Rotational Equilibrium
Name:
Section: 2AL-_
Date performed: /-I-
Lab station:
Partners:
A. Conditions for equilibrium
(Q-1) Determine the location of the center of gravity of the meter stick by balancing the
meter stick alone.
Teg = (50.39L±_0.03)cm
(Q-2) Now balance the meter stick with 3 weights attached. Use three different weights,
each between 100 gwt and 500 gwt.
O cm
100 cm
position (cm)
weight (gwt)
lever arm (cm)
torque (gwt cm)
126.30 + 0.03
99.91± 0.05
25.05
+ 0.03
150.el + 0.05
3 87.27 ± 0. 03 249.16± 0.05
土
Sum of clockwise torques =
Sum of counterclockwise torques =
Use this space for calculations.
(Q-3) Do the torques balance? Use the discrepancy test and explain your results.
N.B
No need to convort the weight in New tons, keep gwt unit

Expert Solution

This question has been solved!

Explore an expertly crafted, step-by-step solution for a thorough understanding of key concepts.

This is a popular solution!

SEE SOLUTION Check out a sample Q&A here

Step 1

Step 2

Step 3

Step 4

Trending now

This is a popular solution!

Step by step

Solved in 4 steps with 14 images

SEE SOLUTION Check out a sample Q&A here

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.

Similar questions

I just want to make sure that I am doing this question correctly and also the answer is 155N. Moment on left side = Moment on right side. 10kg*10m=100 6kg*4m=24 100+24=124 124/8m= 15.5N 15.5*10m/s 2 gravity = 155N
Answer the question in full details, thank you very much, Answer on the correct significant figures:, and the final answer is not 307.12 rad/s^2 or 310 rad/s^2: A uniform, solid 2.5 kg cylinder can rotate about an axis through its center at O. The forces applied are : F1 =45N, F2 =4.3N, F3 = 5.5N,and F4 = 4.8 N. Also, R1 = 10. cmand R3 = 4.2cm. Find the magnitude (in rad/s^2) and direction (+ denotes counterclockwise and — denotes clockwise) of theangular acceleration of the cylinder.
Can somebody help please ?
In which of the three cases would it be easiest to calculate the torque using METHOD 2? (You may select more than one case)
In unit-vector notation, what is the net torque about the origin on a flea located at coordinates (0, -7.23 m, 1.81 m) when forces F₁ = (4.92 N) and F₂ = (-6.97 N) act on the flea? Number 48.187 î 0 Ĵ 0 k Units N-m
Find the net torque on the wheel as shown about the axle through O, taking a = 10.0 cm and b = 25.0 cm.
Which option is it?
A thin cord is wrapped around a uniform grindstone of radius 0.30 m and mass of 25 kg supported by bearings that produce negligible friction torque. The cord exerts a steady tension force on the grindstone, causing it to accelerate from rest to 60 rad/s in 12 s. (lgrindstone = 5 What is the magnitude of the tension force of the cord? O 5.6 N 11.3 N O 18.7 N O 37.5 N
Please ASAP
A 4 kg cat sits on a 30 kg, 4 m long beam that is supported by a cable. She sits 0.5 m in from the right, and the cable is supporting the beam at a point 1 m in from the left. The beam is free to rotate at the hinge. What is the tension in the cable and horizontal and vertical forces exerted by the hinge? What direction do the vertical and horizontal forces point?
Fg 120000 N FA+FB 165000 %3D FA+120000= 165000 FA=45000 N Ex : What is the value of the force (F) required to balance the system in the figure? M = 5 kg and F makes and angle (0 =37°) below 3 m 2 m OM Answer: ET about center 0 ITMl+ITpl 0 %3D 71Page
Calculate the torque produced by the tension T = 20 N at point 0.