Note: You will receive no credit for late submissions. To learn more, read your instructor's Grading Policy. An Eccentric Professor

Similar documents
Chapter 11 Equilibrium

Physics 201 Homework 8

Chapter 4. Forces and Newton s Laws of Motion. continued

Newton s Law of Motion

TEACHER ANSWER KEY November 12, Phys - Vectors

PHY121 #8 Midterm I

C B A T 3 T 2 T What is the magnitude of the force T 1? A) 37.5 N B) 75.0 N C) 113 N D) 157 N E) 192 N

Lecture 6. Weight. Tension. Normal Force. Static Friction. Cutnell+Johnson: , second half of section 4.7

Physics 1A Lecture 10C

AP Physics C Fall Final Web Review

6. Block and Tackle* Block and tackle

Midterm Solutions. mvr = ω f (I wheel + I bullet ) = ω f 2 MR2 + mr 2 ) ω f = v R. 1 + M 2m

PHY231 Section 2, Form A March 22, Which one of the following statements concerning kinetic energy is true?

Chapter 4: Newton s Laws: Explaining Motion

B) 286 m C) 325 m D) 367 m Answer: B

Physics 2A, Sec B00: Mechanics -- Winter 2011 Instructor: B. Grinstein Final Exam

Copyright 2011 Casa Software Ltd. Centre of Mass

HW Set II page 1 of 9 PHYSICS 1401 (1) homework solutions

Solution Derivations for Capa #11

A Determination of g, the Acceleration Due to Gravity, from Newton's Laws of Motion

PHYSICS 111 HOMEWORK SOLUTION, week 4, chapter 5, sec 1-7. February 13, 2013

PHY231 Section 1, Form B March 22, 2012

9. The kinetic energy of the moving object is (1) 5 J (3) 15 J (2) 10 J (4) 50 J

Conceptual Questions: Forces and Newton s Laws

5. Forces and Motion-I. Force is an interaction that causes the acceleration of a body. A vector quantity.

v v ax v a x a v a v = = = Since F = ma, it follows that a = F/m. The mass of the arrow is unchanged, and ( )

Serway_ISM_V1 1 Chapter 4

If you put the same book on a tilted surface the normal force will be less. The magnitude of the normal force will equal: N = W cos θ

LAB 6: GRAVITATIONAL AND PASSIVE FORCES

Two-Body System: Two Hanging Masses

1 of 7 9/5/2009 6:12 PM

LAB 6 - GRAVITATIONAL AND PASSIVE FORCES

Fundamental Mechanics: Supplementary Exercises

Chapter 5 Using Newton s Laws: Friction, Circular Motion, Drag Forces. Copyright 2009 Pearson Education, Inc.

AP Physics Circular Motion Practice Test B,B,B,A,D,D,C,B,D,B,E,E,E, m/s, 0.4 N, 1.5 m, 6.3m/s, m/s, 22.9 m/s

AP Physics Applying Forces

B Answer: neither of these. Mass A is accelerating, so the net force on A must be non-zero Likewise for mass B.

Practice Test SHM with Answers

F N A) 330 N 0.31 B) 310 N 0.33 C) 250 N 0.27 D) 290 N 0.30 E) 370 N 0.26

At the skate park on the ramp

Objective: Equilibrium Applications of Newton s Laws of Motion I

Steps to Solving Newtons Laws Problems.

Physics 11 Assignment KEY Dynamics Chapters 4 & 5

Chapter 18 Static Equilibrium

Physics: Principles and Applications, 6e Giancoli Chapter 4 Dynamics: Newton's Laws of Motion

CHAPTER 6 WORK AND ENERGY

Name Period WORKSHEET: KINETIC AND POTENTIAL ENERGY PROBLEMS. 1. Stored energy or energy due to position is known as energy.

Physics 125 Practice Exam #3 Chapters 6-7 Professor Siegel

General Physics Lab: Atwood s Machine

VELOCITY, ACCELERATION, FORCE

while the force of kinetic friction is fk = µ

Physics 111: Lecture 4: Chapter 4 - Forces and Newton s Laws of Motion. Physics is about forces and how the world around us reacts to these forces.

WORKSHEET: KINETIC AND POTENTIAL ENERGY PROBLEMS

Chapter 7: Momentum and Impulse

Ideal Cable. Linear Spring - 1. Cables, Springs and Pulleys

1 of 40 03/20/ :49 PM

Kinetic Friction. Experiment #13

Kinetic Friction. Experiment #13

HW Set VI page 1 of 9 PHYSICS 1401 (1) homework solutions

2 Newton s First Law of Motion Inertia

Angular acceleration α

Work, Energy & Momentum Homework Packet Worksheet 1: This is a lot of work!

Sample Questions for the AP Physics 1 Exam

MULTIPLE CHOICE. Choose the one alternative that best completes the statement or answers the question.

Chapter 7 Homework solutions

AP1 Dynamics. Answer: (D) foot applies 200 newton force to nose; nose applies an equal force to the foot. Basic application of Newton s 3rd Law.

AP Physics: Rotational Dynamics 2

A) N > W B) N = W C) N < W. speed v. Answer: N = W

Worksheet #1 Free Body or Force diagrams

ANSWER KEY. Work and Machines

AP Physics - Chapter 8 Practice Test

AP Physics C. Oscillations/SHM Review Packet

Acceleration due to Gravity

W02D2-2 Table Problem Newton s Laws of Motion: Solution

Lecture 7 Force and Motion. Practice with Free-body Diagrams and Newton s Laws

Experiment 4. Vector Addition: The Force Table

Chapter 07 Test A. Name: Class: Date: Multiple Choice Identify the choice that best completes the statement or answers the question.

Weight The weight of an object is defined as the gravitational force acting on the object. Unit: Newton (N)

Physics Exam 2 Chapter 5N-New

FRICTION, WORK, AND THE INCLINED PLANE

E X P E R I M E N T 8

The Big Idea. Key Concepts

Work, Energy and Power Practice Test 1

circular motion & gravitation physics 111N

Prelab Exercises: Hooke's Law and the Behavior of Springs

6. Vectors Scott Surgent (surgent@asu.edu)

PHYS 211 FINAL FALL 2004 Form A

Lecture 16. Newton s Second Law for Rotation. Moment of Inertia. Angular momentum. Cutnell+Johnson: 9.4, 9.6

Force on Moving Charges in a Magnetic Field

TORQUE AND FIRST-CLASS LEVERS

AP Physics 1 Midterm Exam Review

5.1 The First Law: The Law of Inertia

Review Chapters 2, 3, 4, 5

Awell-known lecture demonstration1

When showing forces on diagrams, it is important to show the directions in which they act as well as their magnitudes.

D Alembert s principle and applications

Torque and Rotation. Physics

Tennessee State University

Pulleys, Work, and Energy

Chapter 6 Work and Energy

Transcription:

1/5 2009/11/14 上午 11:08 Manage this Assignment: Chapter 12 Due: 12:00am on Saturday, July 3, 2010 Note: You will receive no credit for late submissions. To learn more, read your instructor's Grading Policy An Eccentric Professor Description: Determine the normal force exerted on an object in static equilibrium in an odd setup. Then, find the amount of load that must be added to break equilibrium. Uses applets. Realizing that she often doesn't have her students' full attention during class, a professor devises an elaborate device on which to stand while she lectures. The pulley is placed so that the string makes a 45.0-degree angle with the beam. The beam is uniform, 5.00 meters long, and has weight. The professor stands 2.00 meters from the pivot point and has weight. Find the tension in the rope in terms of the professor's weight and the weight of the beam. Hint A.1 How to approach the problem Since the professor and her contraption are in static equilibrium, the net torque on the beam must be zero. You should pick a covenient point around which to calculate the net torque. Ideally, you should pick a point at which the force exerted on the beam is neither one that you are given nor one that you are seeking. The torque about some point due to any force exerted at that point is zero:. Therefore, you don't have to worry about unknown forces acting at the point about which you are calculating the net torque. In this problem, which of the following is the best point around which to calculate the torques? the pivot at the wall the point where the professor is standing the center of mass of the beam the end of the beam where the rope is attached the center of the pulley Once you have found all of the torques (both magnitude and sign), add them up and set the sum equal to zero. This is the equilibrium condition:. Then, solve this equation for. Hint A.2 Find the net torque What is the net torque on the beam around the pivot where the beam connects to the wall? Use for the tension, for the professor's weight, and for the beam's weight. Hint A.2.1 Find the magnitude of the normal force What is the magnitude of the normal force exerted by the professor on the beam? There are two key ideas needed to solve this. First, the normal force exerted by the professor on the beam has the same magnitude as and opposite direction from (i.e., it is the Newton's third law partner) the normal force exerted by the beam on the professor. Second, the professor is in static equilibrium, and thus the net force on her must be zero. Hint A.2.1.1 Forces on the professor There are three forces acting on the professor: gravity (her weight), tension, and the normal force exerted by the beam. Gravity is directed downward, and the other two forces are directed upward. Thus, the equilibrium condition can be solved to find the magnitude of the normal force in terms of given quantities. Express your answer in terms of some or all of the variables,, and.

2/5 2009/11/14 上午 11:08 Hint A.2.2 Center of mass of the beam The weight of the beam acts at its center of mass. Since the beam is said to be uniform, the center of mass will be in the geometric center, 2.5 meters from each end. Express your answer in terms of some or all of the variables,, and. Take torques that result in counterclockwise angular acceleration being positive. Now set the net torque equal to zero and solve for. This will tell you the tension necessary for equilibrium, which is the tension in the rope, since we know that the professor/beam system is in equilibrium. Express your answer in terms of and. Part B If the professor's mass is 60 kilograms and the mass of the beam is 20 kilograms, what is the magnitude of the normal force exerted by the beam on the professor? Use 10 meters per second squared for the magnitude of the acceleration due to gravity. Hint B.1 How to approach the problem In you found an expression for the magnitude of the normal force in the course of determining the net torque on the beam. Use that expression, together with the formula for tension that you found above, to determine the magnitude of the normal force in this situation. Express your answer in newtons to two significant figures. This applet shows all of the forces and how they change as the weight of the professor is varied. As an incentive to her students, the professor attaches a light basket to the center of mass of the beam and tells them that every student who gets an A on the next test gets to put a ball of mass 4 kilograms into the basket. There are enough students in the class that if a high percentage get A's, the weight of the basket will hoist the professor into the air. Part C How many students have to get A's on the test to hoist the professor up into the air? Hint C.1 How to approach the problem You could simply add the weights of the balls as another torque in the problem. However, since the basket is located at the center of mass of the beam (where the weight of the beam is applied), the weight in the basket can be considered as an increase in the beam's weight. Before you can calculate the weight in the basket that will cause the professor to be hoisted into the air, you have to understand why this should happen at all. Specifically, you need to figure out what condition will lead to her moving upward, instead of staying in equilibrium. Which of the statements below best describes the point at which equilibrium is lost and the professor begins moving upward? The tension exceeds the normal force. The tension exceeds the weight of the professor. The weight of the beam exceeds the weight of the professor. The weight of the beam exceeds the tension. The center of mass of the system moves to the right so that it is no longer supported by the pulley. The change in weight of the beam causes the tension to be directed at an angle greater than 45 degrees.

3/5 2009/11/14 上午 11:08 The tension exceeding the weight violates the equilibrium condition on the professor:. If the tension is greater than the professor's weight, then this condition cannot possibly be met, and so she must be moving (that is, no longer in equilibrium). While you worked, giving attention explicitly to the equilibrium condition of the beam, the value you used for the normal force was based on the equilibrium condition for the professor, which of course has to be met for the whole system to be in equilbrium. Now that you know the condition that breaks the equilibrium, you can just set up the equation for that condition and solve it. In this case, set, using the expression for that you found above. Solve for the weight of the beam that satisfies this condition. For any greater than that value, the professor will be lifted into the air, since the equilibrium conditions for the beam and for the professor can no longer be met. Express your answer as an integer. students Problem 12.38 Description: The raised span of the drawbridge shown in the figure has its 11000-kg mass distributed uniformly over its 14-m length. (a) Find the force exerted by the drawbridge hinge as F_vecF_x*i_unit+F_y*j_unit. Orts i_unit and j_unit are directed right... The raised span of the drawbridge shown in the figure has its 11000- mass distributed uniformly over its 14- length. Find the force exerted by the drawbridge hinge as. Orts and are directed right and upward respectively. Problem 12.33 Description: The boom in the crane of the figure is free to pivot about point P and is supported by the cable that joins halfway along its 18-m total length. The cable passes over a pulley and is anchored at the back of the crane. The boom has mass m1 distributed... The boom in the crane of the figure is free to pivot about point and is supported by the cable that joins halfway along its 18- total length. The cable passes over a pulley and is anchored at the back of the crane. The boom has mass 1800 distributed uniformly along its length, and the mass hanging from the end of the boom is 2500. The boom makes a 50 angle with the horizontal.

4/5 2009/11/14 上午 11:08 What is the tension in the cable that supports the boom? Problem 12.45 Description: A 5.0-m-long ladder has mass m and is leaning against a frictionless wall, making a 66 degree(s) angle with the horizontal. (a) If the coefficient of friction between the ladder and ground is 0.42, what is the mass of the heaviest person who can... A 5.0- -long ladder has mass 12.0 and is leaning against a frictionless wall, making a 66 angle with the horizontal. If the coefficient of friction between the ladder and ground is 0.42, what is the mass of the heaviest person who can safely ascend to the top of the ladder? (The center of mass of the ladder is at its center.) Problem 12.40 Description: A crane in a marble quarry is mounted on the rock walls of the quarry and is supporting a m2 slab of marble as shown in the figure. The center of mass of the m1 boom is located one-third of the way from the pivot end of its 15-m length, as shown. (a)... A crane in a marble quarry is mounted on the rock walls of the quarry and is supporting a 2200 slab of marble as shown in the figure. The center of mass of the 850 boom is located one-third of the way from the pivot end of its 15- length, as shown. Find the tension in the horizontal cable that supports the boom. Problem 12.39

5/5 2009/11/14 上午 11:08 Description: Climbers attempting to cross a stream place a m log against a vertical, frictionless ice cliff on the opposite side (see the figure). The log is inclined at 27 degree(s), and its center of gravity is one-third of the way along its 6.3-m length. (a)... Climbers attempting to cross a stream place a 335 log against a vertical, frictionless ice cliff on the opposite side (see the figure). The log is inclined at 27, and its center of gravity is one-third of the way along its 6.3- length. If the coefficient of friction between the left end of the log and the ground is 0.92, what is the maximum mass for a climber and pack to cross without the log slipping? Problem 12.32 Description: A uniform m1 ladder is leaning against a frictionless vertical wall, with which it makes a 15 degree(s) angle. The coefficient of friction between ladder and ground is 0.26. (a) Can a m2 person climb to the top of the ladder without it slipping? (b)... A uniform 6.8 ladder is leaning against a frictionless vertical wall, with which it makes a 15 angle. The coefficient of friction between ladder and ground is 0.26. Can a 65 person climb to the top of the ladder without it slipping? yes no Part B If so, how massive a person would make the ladder slip? Score Summary: Your score on this assignment is 0%. You received 0 out of a possible total of 53 points.