Solutions to Physics: Principles with Applications, 5/E, Giancoli Chapter 9
|
|
- Georgiana Candace Poole
- 6 years ago
- Views:
Transcription
1
2
3
4 Solutions to Phsics: Principles with pplications, 5/E, Giancoli Chapter 9 CHPTER 9 6. We choose the coordinate sstem shown, with positive torques clockwise. We write τ = Iα about the point from the force diagram for the leg: τ = = 0; (15.0 kg)(9.80 m/s 2 )(0.350 m) (0.805 m), which gives = 63.9 N. ecause there is no acceleration of the hanging mass, we have =, or m = /g = (63.9 N)/(9.80 m/s 2 ) = Hip Joint 6.52 kg. CM 7. We choose the coordinate sstem shown, with positive torques clockwise. We write τ = Iα about the point from the force diagram for the beam and piano: τ = ( +! F N2 = 0, which gives F N2 = ( +! = ((300 kg)(9.80 m/s 2 ) +!(160 kg)(9.80 m/s 2 ) = N. We write F = ma from the force diagram for the beam and piano: F N1 + F N2 = 0, which gives F N1 = + F N2 = (300 kg)(9.80 m/s 2 ) +!(160 kg)(9.80 m/s 2 ) N = N. The forces on the supports are the reactions to these forces: N down, and N down. F N 1 FN2 8. We choose the coordinate sstem shown, with positive torques clockwise. We write τ = Iα about the point from the force diagram for the two beams: τ = ( +! F N2 = 0, which gives F N2 = ( +! = ([!(1000 kg)](9.80 m/s 2 ) +!(1000 kg)(9.80 m/s 2 ) = N. We write F = ma from the force diagram for the two beams: F N1 + F N2 = 0, which gives F N1 = + F N2 = (1000 kg)(9.80 m/s 2 ) +!(1000 kg)(9.80 m/s 2 ) N = N. F N 1 F N2 9. We must move the direction of the net force 10 to the right. We choose the coordinate sstem shown, with the -ais in the direction of the original net force. We write F from the force diagram for the tooth: F = cos 20 + cos 20 = F net sin 10 ; (2.0 N) cos 20 + cos 20 = F net sin 10 ; F = + sin 20 + sin 20 = F net cos 10 ; + (2.0 N) sin 20 + sin 20 = F net cos F net 20 Page 9 1
5 Solutions to Phsics: Principles with pplications, 5/E, Giancoli Chapter 9 We have two equations for the two unknowns: F net, and. When we eliminate F net, we get = 2.3 N. Page 9 2
6 Solutions to Phsics: Principles with pplications, 5/E, Giancoli Chapter We choose the coordinate sstem shown, with positive torques clockwise. We write τ = Iα about the support point from the force diagram for the board and people: τ = m 1 g( d) + m 2 gd = 0; m 1 g d d m 2 g (30 kg)(10 m d) + (70 kg)d = 0, which gives d = 3.0 m from the adult. F N 11. We choose the coordinate sstem shown, with positive torques clockwise. We write τ = Iα about the support point from the force diagram for the board and people: τ = m 1 g( d) (! d) + m 2 gd = 0; m 1 g d d m 2 g (30 kg)(10 m d) (15 kg)(5.0 m d) + (70 kg)d = 0, which gives d = 3.3 m from the adult. F N 12. From the force diagram for the mass we can write F cos = 0, or = cos 30. F = sin = 0, or sin 30 = = (200 kg)(9.80 m/s 2 ), which gives = N. FT1 Thus we have = cos 30 = ( N) cos 30 = N. 13. From the force diagram for the hanging light and junction we can write F cos 1 cos 2 = 0; FT1 cos 37 = cos 53 ; F sin 1 + sin 2 = 0; sin 37 + sin 53 = (30 kg)(9.80 m/s 2 ). 2 1 When we solve these two equations for the two unknowns,, and, we get = N, and = N. 14. From the force diagram for the seesaw and children we can write F = F N m 1 g m 2 g = 0, or F N = (m 1 + m 2 + M)g = (30 kg + 25 kg kg)(9.80 m/s 2 ) = N. d F N m 1 g m 2 g Page 9 3
7 Solutions to Phsics: Principles with pplications, 5/E, Giancoli Chapter We choose the coordinate sstem shown, with positive torques clockwise. We write τ = Iα about the support point from the force diagram for the cantilever: τ = F 2 d +! = 0; F 2 (20.0 m) + (1200 kg)(9.80 m/s 2 )(25.0 m) = 0, which gives F 2 = N. For the forces in the -direction we have F = F 1 + F 2 = 0, or F 1 = F 2 = (1200 kg)(9.80 m/s 2 ) ( N) = F 1 F 2 d N (down). 16. From the force diagram for the sheet we can write F = cos cos = 0, which gives. F sin + sin = 0; 2 sin = ; 2 sin 3.5 = (0.60 kg)(9.80 m/s 2 ), which gives = 48 N; The tension is so much greater than the weight because onl the vertical components balance the weight. 17. We choose the coordinate sstem shown, with positive torques clockwise. We write τ = Iα about the lower hinge from the force diagram for the door: τ = F (H 2)!w = 0; w F F F [2.30 m 2(0.40 m)] (13.0 kg)(9.80 m/s 2 )!(1.30 m), H F which gives F = 55.2 N. We write F = ma from the force diagram for the door: F = F + F = 0; 55 N + F = 0, which gives F = 55.2 N. The top hinge pulls awa from the door, and the bottom hinge pushes on the door. F = F + F = 0. ecause each hinge supports half the weight, we have F = F =!(3.0 kg)(9.80 m/s 2 ) = 63.7 N. F Thus we have top hinge: F = 55.2 N, F = 63.7 N; bottom hinge: F = 55.2 N, F = 63.7 N. 18. We choose the coordinate sstem shown, with positive torques clockwise. We write τ = Iα about the support point from the force diagram for the seesaw and bos: τ = + m 2 g! + m 3 g m 1 g! = 0; + (35 kg)!(3.6 m) + (25 kg) (50 kg)!(3.6 m) = 0, which gives = 1.1 m. The third bo should be 1.1 m from pivot on side of lighter bo. m 1 g F N m 3 g m 2 g Page 9 4
8 Solutions to Phsics: Principles with pplications, 5/E, Giancoli Chapter We choose the coordinate sstem shown, with positive torques clockwise. For the torques about the point we have τ = F 1 d + = 0; F 1 (1.0 m) + (60 kg)(9.80 m/s 2 )(3.0 m) = 0, which gives F 1 = N (down). For the torques about the point we have τ = F 2 d + ( + d)= 0; F 2 (1.0 m) = (60 kg)(9.80 m/s 2 )(4.0 m m), which gives F 2 = N (up). F 1 F2 d 20. We choose the coordinate sstem shown, with positive torques clockwise. For the torques about the point we have τ = F 1 d + + [!( + d) d] = 0; F 1 (1.0 m) = (60 kg)(9.80 m/s 2 )(3.0 m) (35 kg)(9.80 m/s 2 )(1.0 m), which gives F 1 = N (down). For the torques about the point we have τ = F 2 d + +!( + d) = 0; F 2 (1.0 m) = (60 kg)(9.80 m/s 2 )(3.0 m) + (35 kg)(9.80 m/s 2 )(2.0 m), which gives F 2 = N (up). F 1 d F2 21. From Eample 7 16 we have = (20.4/100)(1.60 m) = m. From Table 7 1, we have = [( )/100](1.60 m) = m; M =![( )/100](60.0 kg) = kg. We choose the coordinate sstem shown, with positive torques clockwise. We write τ = Iα about the hip joint from the force diagram for the leg: τ = = 0; (10.35 kg)(9.80 m/s 2 )(0.326 m) (0.770 m) = 0. which gives = 42.9 N. ecause there is no acceleration of the hanging mass, we have =, or m = /g = (42.9 N)/(9.80 m/s 2 ) = 4.38 kg. Hip Joint CM 22. We choose the coordinate sstem shown, with positive torques clockwise. We write τ = Iα about the support point from the force diagram for the beam: τ = + F 1 (d 1 + d 2 + d 3 + d 4 ) F 3 (d 2 + d 3 + d 4 ) F 4 (d 3 + d 4 ) F 5 d 4!(d 1 + d 2 + d 3 + d 4 ) = 0; F 1 (10.0 m) (4000 N)(8.0 m) (3000 N)(4.0 m) (2000 N)(1.0 m) (250 kg)(9.80 m/s 2 )(5.0 m) = 0, which gives F 1 = N. We write τ = Iα about the support point from the force diagram for the beam: τ = F 2 (d 1 + d 2 + d 3 + d 4 ) + F 3 d 1 F 4 (d 1 + d 2 ) F 5 (d 1 + d 2 + d 3 )!(d 1 + d 2 + d 3 + d 4 ) = 0; F 2 (10.0 m) (4000 N)(2.0 m) (3000 N)(6.0 m) F 1 F 3 F4 F5 d 1 d 2 d 3 d 4 F 2 Page 9 5
9 Solutions to Phsics: Principles with pplications, 5/E, Giancoli Chapter 9 (2000 N)(9.0 m) (250 kg)(9.80 m/s 2 )(5.0 m) = 0, which gives F 2 = N. Page 9 6
10 Solutions to Phsics: Principles with pplications, 5/E, Giancoli Chapter We choose the coordinate sstem shown, with positive torques clockwise. We write τ = Iα about the point from the force diagram for the beam: τ = ( sin α) +! = 0; F W sin 50 + (30 kg)(9.80 m/s 2 )! = 0, which gives = N. Note that we find the torque produced b the tension b finding the torques produced b the components. We write F = ma from the force diagram for the beam: F = F W cos α = 0; F W ( N) cos 50 = 0, which gives F W = 123 N. F = F W + sin α = 0; F W + ( N) sin 50 (30 kg)(9.80 m/s 2 ) = 0, which gives F W = 147 N. For the magnitude of F W we have F W = (F W 2 + F W 2 ) 1/2 = [(123 N) 2 + (147 N) 2 ] 1/2 = N. We find the direction from tan = F W /F W = (147 N)/(123 N) = 1.19, which gives = 50. Thus the force at the wall is F W = N, 50 above the horizontal. α 24. ecause the backpack is at the midpoint of the rope, the angles are equal. The force eerted b the backpacker is the tension in the rope. From the force diagram for the backpack and junction we can write F cos cos = 0, or = = F; F sin + sin = 0, or 2F sin =. (a) We find the angle from tan = h/! = (1.5 m)/!(7.6 m) = 0.395, or = When we put this in the force equation, we get 2F sin 21.5 = (16 kg)(9.80 m/s2 ), which gives F = N. (b) We find the angle from tan = h/! = (0.15 m)/!(7.6 m) = , or = When we put this in the force equation, we get 2F sin 2.26 = (16 kg)(9.80 m/s2 ), which gives F = N. 25. We choose the coordinate sstem shown, with positive torques clockwise. For the torques about the CG we have τ CG = F N1 ( ) F N2 = 0; (35.1 kg)g(170 cm ) (31.6 kg)g = 0, which gives = 89.5 cm from the feet. F N1 CG F N2 Page 9 7
11 Solutions to Phsics: Principles with pplications, 5/E, Giancoli Chapter We choose the coordinate sstem shown, with positive torques clockwise. We write τ = Iα about the point from the force diagram for the beam and sign: τ = ( sin ) + +! = 0; (sin 41.0 )(1.35 m) + (215 N)(1.70 m) + (135 N)!(1.70 m) = 0, which gives = 542 N. Note that we find the torque produced b the tension b finding the torques produced b the components. We write F = ma from the force diagram for the beam and sign: F = F hinge cos = 0; F hinge (542 N) cos 41.0 = 0, which gives F hinge = 409 N. F = F hinge + sin = 0; F hinge + (542 N) cos N 135 N = 0, which gives F hinge = 6 N (down). F hingev F hingeh 27. We choose the coordinate sstem shown, with positive torques clockwise. We write τ = Iα about the point from the force diagram for the pole and light: τ = H + cos +! cos = 0; (3.80 m) + (12.0 kg)(9.80 m/s 2 )(7.5 m) cos 37 + (8.0 kg)(9.80 m/s 2 )!(7.5 m) cos 37 = 0, which gives = N. We write F = ma from the force diagram for the pole and light: F = F H = 0; F H N = 0, which gives F H = N. F = F V = 0; F V (12.0 kg)(9.80 m/s 2 ) (8.0 kg)(9.80 m/s 2 ) = 0, which gives F V = N. H F V F H 28. We choose the coordinate sstem shown, with positive torques counterclockwise. We write τ = Iα about the point from the force diagram for the ladder: τ = (!) cos F N2 sin = 0, which gives F N2 = /2 tan. We write F = ma from the force diagram for the ladder: F fr F N2 = 0, which gives F fr = F N2 = /2 tan. We write F = ma from the force diagram for the ladder: F N1 = 0, which gives F N1 =. For the bottom not to slip, we must have F fr ² µf N1, or /2 tan ² µ, from which we get tan ³ 1/2µ. The minimum angle is min = tan 1 (1/2µ). F fr F N1 O F N2 Page 9 8
Physics 101 Lecture 12 Equilibrium
Physics 101 Lecture 12 Equilibrium Assist. Prof. Dr. Ali ÖVGÜN EMU Physics Department www.aovgun.com Static Equilibrium q Equilibrium and static equilibrium q Static equilibrium conditions n Net eternal
More informationTorque. Objectives. Assessment. Assessment. Equations. Physics terms 6/2/14
Objectives Calculate torque given the lever arm (perpendicular distance) and the force. Calculate torque in newton meters and in pound feet. Interpret positive and negative signs in the context of torque.
More informationSection 2: Static Equilibrium II- Balancing Torques
Section 2: Static Equilibrium II- Balancing Torques Last Section: If (ie. Forces up = Forces down and Forces left = Forces right), then the object will have no translatory motion. In other words, the object
More informationSolutions to Phsics: Principles with Applications, 5/E, Giancoli Chapter 4 CHAPTER 4 1. If we select the sled and child as the object, we appl Newton s second law to find the force: F = ma; F = (60.0 kg)(1.15
More informationStatic equilibrium. Objectives. Physics terms. Assessment. Brainstorm. Equations 6/3/14
Static equilibrium Objectives State the conditions of static equilibrium in terms of forces and torques. Draw a free-body diagram of a lever showing all forces. Use the condition of equilibrium to solve
More informationStatic Equilibrium and Torque
10.3 Static Equilibrium and Torque SECTION OUTCOMES Use vector analysis in two dimensions for systems involving static equilibrium and torques. Apply static torques to structures such as seesaws and bridges.
More informationUnit 4 Statics. Static Equilibrium Translational Forces Torque
Unit 4 Statics Static Equilibrium Translational Forces Torque 1 Dynamics vs Statics Dynamics: is the study of forces and motion. We study why objects move. Statics: is the study of forces and NO motion.
More informationChapter 12 Static Equilibrium
Chapter Static Equilibrium. Analysis Model: Rigid Body in Equilibrium. More on the Center of Gravity. Examples of Rigid Objects in Static Equilibrium CHAPTER : STATIC EQUILIBRIUM AND ELASTICITY.) The Conditions
More informationCenter of Mass. A baseball thrown into the air follows a smooth parabolic path. A baseball bat thrown into the air does not follow a smooth path.
Center of Mass A baseball thrown into the air follows a smooth parabolic path. A baseball bat thrown into the air does not follow a smooth path. The bat wobbles about a special point. This point stays
More informationP12 Torque Notes.notebook. March 26, Torques
Torques The size of a torque depends on two things: 1. The size of the force being applied (a larger force will have a greater effect) 2. The distance away from the pivot point (the further away from this
More informationAP Physics Multiple Choice Practice Torque
AP Physics Multiple Choice Practice Torque 1. A uniform meterstick of mass 0.20 kg is pivoted at the 40 cm mark. Where should one hang a mass of 0.50 kg to balance the stick? (A) 16 cm (B) 36 cm (C) 44
More informationStatics. Phys101 Lectures 19,20. Key points: The Conditions for static equilibrium Solving statics problems Stress and strain. Ref: 9-1,2,3,4,5.
Phys101 Lectures 19,20 Statics Key points: The Conditions for static equilibrium Solving statics problems Stress and strain Ref: 9-1,2,3,4,5. Page 1 The Conditions for Static Equilibrium An object in static
More informationPractice. Newton s 3 Laws of Motion. Recall. Forces a push or pull acting on an object; a vector quantity measured in Newtons (kg m/s²)
Practice A car starts from rest and travels upwards along a straight road inclined at an angle of 5 from the horizontal. The length of the road is 450 m and the mass of the car is 800 kg. The speed of
More informationStatic Equilibrium; Torque
Static Equilibrium; Torque The Conditions for Equilibrium An object with forces acting on it, but that is not moving, is said to be in equilibrium. The first condition for equilibrium is that the net force
More informationPhysics 207: Lecture 26. Announcements. Make-up labs are this week Final hwk assigned this week, final quiz next week.
Torque due to gravit Rotation Recap Phsics 07: ecture 6 Announceents Make-up labs are this week Final hwk assigned this week, final quiz net week Toda s Agenda Statics Car on a Hill Static Equilibriu Equations
More informationChapter 12 Static Equilibrium; Elasticity and Fracture
2009 Pearson Education, Inc. This work is protected by United States copyright laws and is provided solely for the use of instructors in teaching their courses and assessing student learning. Dissemination
More informationEquilibrium. For an object to remain in equilibrium, two conditions must be met. The object must have no net force: and no net torque:
Equilibrium For an object to remain in equilibrium, two conditions must be met. The object must have no net force: F v = 0 and no net torque: v τ = 0 Worksheet A uniform rod with a length L and a mass
More informationRecap I. Angular position: Angular displacement: s. Angular velocity: Angular Acceleration:
Recap I Angular position: Angular displacement: s Angular velocity: Angular Acceleration: Every point on a rotating rigid object has the same angular, but not the same linear motion! Recap II Circular
More informationPhysics 2210 Homework 18 Spring 2015
Physics 2210 Homework 18 Spring 2015 Charles Jui April 12, 2015 IE Sphere Incline Wording A solid sphere of uniform density starts from rest and rolls without slipping down an inclined plane with angle
More information1. If we select the sled and child as the object, we apply Newton s second law to find the force: F = ma; F = (60.0 kg)(1.15 m/s 2 ) = 69.0 N.
CHAPTER 4 1. If we select the sled and child as the object, we appl Newton s second law to find the force: = ma; = (60.0 kg)(1.15 m/s ) = 69.0 N.. If we select the bike and rider as the object, we appl
More informationCHAPTER 12 STATIC EQUILIBRIUM AND ELASTICITY. Conditions for static equilibrium Center of gravity (weight) Examples of static equilibrium
CHAPTER 12 STATIC EQUILIBRIUM AND ELASTICITY As previously defined, an object is in equilibrium when it is at rest or moving with constant velocity, i.e., with no net force acting on it. The following
More informationHATZIC SECONDARY SCHOOL
HATZIC SECONDARY SCHOOL PROVINCIAL EXAMINATION ASSIGNMENT STATIC EQUILIBRIUM MULTIPLE CHOICE / 33 OPEN ENDED / 80 TOTAL / 113 NAME: 1. State the condition for translational equilibrium. A. ΣF = 0 B. ΣF
More informationPhysics 111. Lecture 10 (Walker: 5.5-6) Free Body Diagram Solving 2-D Force Problems Weight & Gravity. February 18, Quiz Monday - Chaps.
Phsics 111 Lecture 10 (Walker: 5.5-6) Free Bod Diagram Solving -D Force Problems Weight & Gravit Februar 18, 009 Quiz Monda - Chaps. 4 & 5 Lecture 10 1/6 Third Law Review A small car is pushing a larger
More informationWe define angular displacement, θ, and angular velocity, ω. What's a radian?
We define angular displacement, θ, and angular velocity, ω Units: θ = rad ω = rad/s What's a radian? Radian is the ratio between the length of an arc and its radius note: counterclockwise is + clockwise
More informationPhysics 111. Lecture 22 (Walker: ) Torque Rotational Dynamics Static Equilibrium Oct. 28, 2009
Physics 111 Lecture 22 (Walker: 11.1-3) Torque Rotational Dynamics Static Equilibrium Oct. 28, 2009 Lecture 22 1/26 Torque (τ) We define a quantity called torque which is a measure of twisting effort.
More informationApplications of Statics, Including Problem-Solving Strategies
Applications of Statics, Including Problem-Solving Strategies Bởi: OpenStaxCollege Statics can be applied to a variety of situations, ranging from raising a drawbridge to bad posture and back strain. We
More informationAnswers to selected problems from Essential Physics, Chapter 10
Answers to selected problems from Essential Physics, Chapter 10 1. (a) The red ones have the same speed as one another. The blue ones also have the same speed as one another, with a value twice the speed
More informationTorque. 1. Torque - To make an object rotate, we must apply a force at a certain distance away from the axis => ex) opening a door
Torque Level : Conceptual Physics Teacher : Kim 1. Torque - To make an object rotate, we must apply a force at a certain distance away from the axis => ex) opening a door Torque is defined as τ = r F where
More informationChapter 9 TORQUE & Rotational Kinematics
Chapter 9 TORQUE & Rotational Kinematics This motionless person is in static equilibrium. The forces acting on him add up to zero. Both forces are vertical in this case. This car is in dynamic equilibrium
More informationis the study of and. We study objects. is the study of and. We study objects.
Static Equilibrium Translational Forces Torque Unit 4 Statics Dynamics vs Statics is the study of and. We study objects. is the study of and. We study objects. Recall Newton s First Law All objects remain
More informationPHYSICS - CLUTCH CH 13: ROTATIONAL EQUILIBRIUM.
!! www.clutchprep.com EXAMPLE: POSITION OF SECOND KID ON SEESAW EXAMPLE: A 4 m-long seesaw 50 kg in mass and of uniform mass distribution is pivoted on a fulcrum at its middle, as shown. Two kids sit on
More informationTorque rotational force which causes a change in rotational motion. This force is defined by linear force multiplied by a radius.
Warm up A remote-controlled car's wheel accelerates at 22.4 rad/s 2. If the wheel begins with an angular speed of 10.8 rad/s, what is the wheel's angular speed after exactly three full turns? AP Physics
More informationThen, by Newton s third law: The knots are also in equilibrium. Newton s law applied to the left knot is. The y-equation gives T1 m1 g sin 1.
Chapter 7 Solutions 7.7. Model: The two hanging blocks, which can be modeled as particles, together with the two knots where rope meets with rope and rope meets with rope form a sstem. All the four objects
More informationChapter 9 Rotational Dynamics
Chapter 9 ROTATIONAL DYNAMICS PREVIEW A force acting at a perpendicular distance from a rotation point, such as pushing a doorknob and causing the door to rotate on its hinges, produces a torque. If the
More informationAnnouncements Oct 16, 2014
Announcements Oct 16, 2014 1. Prayer 2. While waiting, see how many of these blanks you can fill out: Centripetal Accel.: Causes change in It points but not Magnitude: a c = How to use with N2: Always
More informationSection 2: Static Equilibrium II- Balancing Torques
Section 2: Static Equilibrium II- Balancing Torques Last Section: If (ie. Forces up = Forces down and Forces left = Forces right), then the object will have no translatory motion. In other words, the object
More informationEquilibrium Notes 1 Translational Equilibrium
Equilibrium Notes 1 Translational Equilibrium Ex. A 20.0 kg object is suspended by a rope as shown. What is the net force acting on it? Ex. Ok that was easy, now that same 20.0 kg object is lifted at a
More informationPHYSICS - CLUTCH 1E CH 12: TORQUE & ROTATIONAL DYNAMICS.
!! www.clutchprep.com INTRO TO TORQUE TORQUE is a twist that a Force gives an object around an axis of rotation. - For example, when you push on a door, it rotates around its hinges. - When a Force acts
More informationPHYSICS - CLUTCH CH 12: TORQUE & ROTATIONAL DYNAMICS.
!! www.clutchprep.com TORQUE & ACCELERATION (ROTATIONAL DYNAMICS) When a Force causes rotation, it produces a Torque. Think of TORQUE as the equivalent of FORCE! FORCE (F) TORQUE (τ) - Causes linear acceleration
More informationRotational Kinetic Energy
Lecture 17, Chapter 10: Rotational Energy and Angular Momentum 1 Rotational Kinetic Energy Consider a rigid body rotating with an angular velocity ω about an axis. Clearly every point in the rigid body
More informationEquilibrium Notes 1 Translational Equilibrium
Equilibrium Notes 1 Translational Equilibrium A 20.0 kg object is suspended by a rope as shown. What is the net force acting on it? Ok that was easy, now that same 20.0 kg object is lifted at a velocity
More informationLECTURE 15 TORQUE AND CENTER OF GRAVITY
LECTURE 15 TORQUE AND CENTER OF GRAVITY 7.3 Torque Net torque 7.4 Gravitational torque and the center of gravity Calculating the position of the center of gravity What is unusual about this door? Learning
More informationConsider two students pushing with equal force on opposite sides of a desk. Looking top-down on the desk:
1 Bodies in Equilibrium Recall Newton's First Law: if there is no unbalanced force on a body (i.e. if F Net = 0), the body is in equilibrium. That is, if a body is in equilibrium, then all the forces on
More informationFr h mg rh h. h 2( m)( m) ( (0.800 kg)(9.80 m/s )
5. We consider the wheel as it leaves the lower floor. The floor no longer exerts a force on the wheel, and the only forces acting are the force F applied horizontally at the axle, the force of gravity
More informationChapter 9- Static Equilibrium
Chapter 9- Static Equilibrium Changes in Office-hours The following changes will take place until the end of the semester Office-hours: - Monday, 12:00-13:00h - Wednesday, 14:00-15:00h - Friday, 13:00-14:00h
More informationEquilibrium and Torque
Equilibrium and Torque Equilibrium An object is in Equilibrium when: 1. There is no net force acting on the object 2. There is no net Torque In other words, the object is NOT experiencing linear acceleration
More informationDefinition. is a measure of how much a force acting on an object causes that object to rotate, symbol is, (Greek letter tau)
Torque Definition is a measure of how much a force acting on an object causes that object to rotate, symbol is, (Greek letter tau) = r F = rfsin, r = distance from pivot to force, F is the applied force
More information5.3 Rigid Bodies in Three-Dimensional Force Systems
5.3 Rigid odies in Three-imensional Force Sstems 5.3 Rigid odies in Three-imensional Force Sstems Eample 1, page 1 of 5 1. For the rigid frame shown, determine the reactions at the knife-edge supports,,.
More informationPhys 1401: General Physics I
1. (0 Points) What course is this? a. PHYS 1401 b. PHYS 1402 c. PHYS 2425 d. PHYS 2426 2. (0 Points) Which exam is this? a. Exam 1 b. Exam 2 c. Final Exam 3. (0 Points) What version of the exam is this?
More informationPhysics 8 Wednesday, October 28, 2015
Physics 8 Wednesday, October 8, 015 HW7 (due this Friday will be quite easy in comparison with HW6, to make up for your having a lot to read this week. For today, you read Chapter 3 (analyzes cables, trusses,
More informationPhysics 101: Lecture 15 Torque, F=ma for rotation, and Equilibrium
Physics 101: Lecture 15 Torque, F=ma for rotation, and Equilibrium Strike (Day 10) Prelectures, checkpoints, lectures continue with no change. Take-home quizzes this week. See Elaine Schulte s email. HW
More informationUpthrust and Archimedes Principle
1 Upthrust and Archimedes Principle Objects immersed in fluids, experience a force which tends to push them towards the surface of the liquid. This force is called upthrust and it depends on the density
More information40 N cos 35 =49N. T 1 = 28 T y
16 (a) Analyzing vertical forces where string 1 and string 2 meet, we find T 1 = 40 N cos 35 =49N (b) Looking at the horizontal forces at that point leads to T 2 = T 1 sin 35 = (49 N) sin 35 =28N (c) We
More informationPhysics 2514 Lecture 13
Physics 2514 Lecture 13 P. Gutierrez Department of Physics & Astronomy University of Oklahoma Physics 2514 p. 1/18 Goals We will discuss some examples that involve equilibrium. We then move on to a discussion
More informationTorque. Physics 6A. Prepared by Vince Zaccone For Campus Learning Assistance Services at UCSB
Physics 6A Torque is what causes angular acceleration (just like a force causes linear acceleration) Torque is what causes angular acceleration (just like a force causes linear acceleration) For a torque
More informationApplications of Statics, Including Problem-Solving Strategies
OpenStax-CNX module: m42173 1 Applications of Statics, Including Problem-Solving Strategies OpenStax College This work is produced by OpenStax-CNX and licensed under the Creative Commons Attribution License
More informationDynamic equilibrium: object moves with constant velocity in a straight line. = 0, a x = i
Dynamic equilibrium: object moves with constant velocity in a straight line. We note that F net a s are both vector quantities, so in terms of their components, (F net ) x = i (F i ) x = 0, a x = i (a
More informationYour Comments. That s the plan
Your Comments I love physics as much as the next gal, but I was wondering. Why don't we get class off the day after an evening exam? What if the ladder has friction with the wall? Things were complicated
More informationChapter 8. Centripetal Force and The Law of Gravity
Chapter 8 Centripetal Force and The Law of Gravity Centripetal Acceleration An object traveling in a circle, even though it moves with a constant speed, will have an acceleration The centripetal acceleration
More informationLECTURE 22 EQUILIBRIUM. Instructor: Kazumi Tolich
LECTURE 22 EQUILIBRIUM Instructor: Kazumi Tolich Lecture 22 2 Reading chapter 11-3 to 11-4 Static equilibrium Center of mass and balance Static equilibrium 3 If a rigid object is in equilibrium (constant
More informationAP Physics 1- Torque, Rotational Inertia, and Angular Momentum Practice Problems FACT: The center of mass of a system of objects obeys Newton s second law- F = Ma cm. Usually the location of the center
More informationPhysics Technology Update James S. Walker Fourth Edition
Phsics Technolog Update James S. Walker Fourth Edition Pearson Education Lim ited Edinburgh Gat e Harlow Esse CM20 2JE England and Associated Com panies throughout the world Visit us on t he World Wide
More informationTEST REPORT. Question file: P Copyright:
Date: February-12-16 Time: 2:00:28 PM TEST REPORT Question file: P12-2006 Copyright: Test Date: 21/10/2010 Test Name: EquilibriumPractice Test Form: 0 Test Version: 0 Test Points: 138.00 Test File: EquilibriumPractice
More informationChapter 8. Rotational Equilibrium and Rotational Dynamics
Chapter 8 Rotational Equilibrium and Rotational Dynamics Force vs. Torque Forces cause accelerations Torques cause angular accelerations Force and torque are related Torque The door is free to rotate about
More informationReview questions. Before the collision, 70 kg ball is stationary. Afterward, the 30 kg ball is stationary and 70 kg ball is moving to the right.
Review questions Before the collision, 70 kg ball is stationary. Afterward, the 30 kg ball is stationary and 70 kg ball is moving to the right. 30 kg 70 kg v (a) Is this collision elastic? (b) Find the
More information= o + t = ot + ½ t 2 = o + 2
Chapters 8-9 Rotational Kinematics and Dynamics Rotational motion Rotational motion refers to the motion of an object or system that spins about an axis. The axis of rotation is the line about which the
More informationPhysics 2210 Fall smartphysics Rotational Statics 11/18/2015
Physics 2210 Fall 2015 smartphysics 17-18 Rotational Statics 11/18/2015 τ TT = L T 1 sin 150 = 1 T 2 1L Poll 11-18-01 τ TT = L 2 T 2 sin 150 = 1 4 T 2L 150 150 τ gg = L 2 MM sin +90 = 1 2 MMM +90 MM τ
More informationInclined Planes Worksheet Answers
Physics 12 Name: Inclined Planes Worksheet Answers 1. An 18.0 kg box is released on a 33.0 o incline and accelerates at 0.300 m/s 2. What is the coefficient of friction? m 18.0kg 33.0? a y 0 a 0.300m /
More informationt = g = 10 m/s 2 = 2 s T = 2π g
Annotated Answers to the 1984 AP Physics C Mechanics Multiple Choice 1. D. Torque is the rotational analogue of force; F net = ma corresponds to τ net = Iα. 2. C. The horizontal speed does not affect the
More informationStatic Equilibrium and Elasticity
Static Equilibrium and Elasticit CHPTER OUTLINE. The Conditions for Equilibrium. More on the Center of Gravit.3 Eamples of Rigid Objects in Static Equilibrium.4 Elastic Properties of Solids NSWERS TO QUESTIONS
More informationPhysics 111. Applying Newton s Laws. Lecture 9 (Walker: 5.4-5) Newton s Third Law Free Body Diagram Solving 2-D Force Problems Weight & Gravity
Phsics 111 Lecture 9 (Walker: 5.4-5) Newton s Third Law ree Bod Diagram Solving -D orce Problems Weight & Gravit Sept. 1, 009 Quiz Wednesda - Chaps. 3 & 4 Lecture 9 1/6 Newton s Third Law of Motion orces
More informationProblem Set #1 Chapter 21 10, 22, 24, 43, 47, 63; Chapter 22 7, 10, 36. Chapter 21 Problems
Problem Set #1 Chapter 1 10,, 4, 43, 47, 63; Chapter 7, 10, 36 Chapter 1 Problems 10. (a) T T m g m g (b) Before the charge is added, the cork balls are hanging verticall, so the tension is T 1 mg (0.10
More informationTorque and Static Equilibrium
Torque and Static Equilibrium Rigid Bodies Rigid body: An extended object in which the distance between any two points in the object is constant in time. Examples: sphere, disk Effect of external forces
More information1. An object is fired with an initial velocity of 23 m/s [R30 U]. What are the initial components of its velocity?
Physics 304 Unit 1 - Total Review 1. An object is fired with an initial velocity of 3 m/s [R30U]. What are the initial components of its velocity?. An object rolls off the top of a horizontal table. a)
More informationEquilibrium. Lecture 8 Physics 106 Spring Equilibrium. Equilibrium. Equilibrium. Balance of Forces: Balance of Forces: Balance of Torques:
Lecture 8 Physics 106 Spring 2006 http://web.njit.edu/~sirenko/ 3/8/2006 Andrei Sirenko, JIT 1 3/8/2006 Andrei Sirenko, JIT 2 3/8/2006 Andrei Sirenko, JIT 3 3/8/2006 Andrei Sirenko, JIT 4 3/8/2006 Andrei
More informationThe box is pushed by a force of magnitude 100 N which acts at an angle of 30 with the floor, as shown in the diagram above.
1. A small box is pushed along a floor. The floor is modelled as a rough horizontal plane and the 1 box is modelled as a particle. The coefficient of friction between the box and the floor is. 2 The box
More informationare (0 cm, 10 cm), (10 cm, 10 cm), and (10 cm, 0 cm), respectively. Solve: The coordinates of the center of mass are = = = (200 g g g)
Rotational Motion Problems Solutions.. Model: A spinning skater, whose arms are outstretched, is a rigid rotating body. Solve: The speed v rω, where r 40 / 0.70 m. Also, 80 rpm (80) π/60 rad/s 6 π rad/s.
More informationChapter 11 Rotational Dynamics and Static Equilibrium. Copyright 2010 Pearson Education, Inc.
Chapter 11 Rotational Dynamics and Static Equilibrium Units of Chapter 11 Torque Torque and Angular Acceleration Zero Torque and Static Equilibrium Center of Mass and Balance Dynamic Applications of Torque
More informationTorque and Rotational Equilibrium
Torque and Rotational Equilibrium Theory Torque is the rotational analog of force. If you want something to move (translate), you apply a force; if you want something to rotate, you apply a torque. Torque
More informationPhysics 8 Wednesday, October 25, 2017
Physics 8 Wednesday, October 25, 2017 HW07 due Friday. It is mainly rotation, plus a couple of basic torque questions. And there are only 8 problems this week. For today, you read (in Perusall) Onouye/Kane
More informationTorque. Introduction. Torque. PHY torque - J. Hedberg
Torque PHY 207 - torque - J. Hedberg - 2017 1. Introduction 2. Torque 1. Lever arm changes 3. Net Torques 4. Moment of Rotational Inertia 1. Moment of Inertia for Arbitrary Shapes 2. Parallel Axis Theorem
More informationChapter 8 continued. Rotational Dynamics
Chapter 8 continued Rotational Dynamics 8.6 The Action of Forces and Torques on Rigid Objects Chapter 8 developed the concepts of angular motion. θ : angles and radian measure for angular variables ω :
More informationPhysics 101 Lecture 11 Torque
Physics 101 Lecture 11 Torque Dr. Ali ÖVGÜN EMU Physics Department www.aovgun.com Force vs. Torque q Forces cause accelerations q What cause angular accelerations? q A door is free to rotate about an axis
More informationPhysics 211 Week 10. Statics: Walking the Plank (Solution)
Statics: Walking the Plank (Solution) A uniform horizontal beam 8 m long is attached by a frictionless pivot to a wall. A cable making an angle of 37 o, attached to the beam 5 m from the pivot point, supports
More informationApplication of Forces. Chapter Eight. Torque. Force vs. Torque. Torque, cont. Direction of Torque 4/7/2015
Raymond A. Serway Chris Vuille Chapter Eight Rotational Equilibrium and Rotational Dynamics Application of Forces The point of application of a force is important This was ignored in treating objects as
More informationGravitational potential energy
Gravitational potential energ m1 Consider a rigid bod of arbitrar shape. We want to obtain a value for its gravitational potential energ. O r1 1 x The gravitational potential energ of an assembl of N point-like
More informationAnnouncements Oct 17, 2013
Announcements Oct 17, 2013 1. No announcements! Colton - Lecture 14 - pg 1 Real satellites: http://science.nasa.gov/realtime/jtrack/3d/jtrack3d.html International space station, 340.5 km above surface
More informationChapter 18 Static Equilibrium
Chapter 8 Static Equilibrium Chapter 8 Static Equilibrium... 8. Introduction Static Equilibrium... 8. Lever Law... 3 Example 8. Lever Law... 5 8.3 Generalized Lever Law... 6 8.4 Worked Examples... 8 Example
More informationRotational Dynamics continued
Chapter 9 Rotational Dynamics continued 9.1 The Action of Forces and Torques on Rigid Objects Chapter 8 developed the concepts of angular motion. θ : angles and radian measure for angular variables ω :
More informationEquilibrium and Torque
Equilibrium and Torque Equilibrium An object is in Equilibrium when: 1. There is no net force acting on the object 2. There is no net Torque (we ll get to this later) In other words, the object is NOT
More informationStatic Equilibrium and Elasticity. Luis Anchordoqui
Static Equilibrium and Elasticity The Conditions for Equilibrium An object with forces acting on it, but that is not moving, is said to be in equilibrium. The Conditions for Equilibrium (cont d) The first
More informationΣF = 0 and Στ = 0 In 2-d: ΣF X = 0 and ΣF Y = 0 Goal: Write expression for Στ and ΣF
Thur Sept 24 Assign 5 Friday Exam Mon Oct 5 Morton 235 7:15-9:15 PM Email if conflict Today: Rotation and Torques Static Equilibrium Sign convention for torques: (-) CW torque (+) CCW torque Equilibrium
More information4.0 m s 2. 2 A submarine descends vertically at constant velocity. The three forces acting on the submarine are viscous drag, upthrust and weight.
1 1 wooden block of mass 0.60 kg is on a rough horizontal surface. force of 12 N is applied to the block and it accelerates at 4.0 m s 2. wooden block 4.0 m s 2 12 N hat is the magnitude of the frictional
More informationTorque and Rotational Equilibrium
Torque and Rotational Equilibrium Name Section Theory Torque is the rotational analog of force. If you want something to move (translate), you apply a force; if you want something to rotate, you apply
More informationAPPLIED MECHANICS I Resultant of Concurrent Forces Consider a body acted upon by co-planar forces as shown in Fig 1.1(a).
PPLIED MECHNICS I 1. Introduction to Mechanics Mechanics is a science that describes and predicts the conditions of rest or motion of bodies under the action of forces. It is divided into three parts 1.
More informationEquilibrium at a Point
Equilibrium at a Point Never slap a man who's chewing tobacco. - Will Rogers Objec3ves Understand the concept of sta3c equilibrium Understand the use of the free- bod diagram to isolate a sstem for analsis
More informationRIN: Monday, May 16, Problem Points Score Total 100
RENSSELER POLYTEHNI INSTITUTE TROY, NY FINL EXM INTRODUTION TO ENGINEERING NLYSIS ENGR-00) NME: Solution Section: RIN: Monda, Ma 6, 06 Problem Points Score 0 0 0 0 5 0 6 0 Total 00 N.B.: You will be graded
More informationProblem Set x Classical Mechanics, Fall 2016 Massachusetts Institute of Technology. 1. Moment of Inertia: Disc and Washer
8.01x Classical Mechanics, Fall 2016 Massachusetts Institute of Technology Problem Set 10 1. Moment of Inertia: Disc and Washer (a) A thin uniform disc of mass M and radius R is mounted on an axis passing
More informationLecture 8. Torque. and Equilibrium. Pre-reading: KJF 8.1 and 8.2
Lecture 8 Torque and Equilibrium Pre-reading: KJF 8.1 and 8.2 Archimedes Lever Rule At equilibrium (and with forces 90 to lever): r 1 F 1 = r 2 F 2 2 General Lever Rule For general angles r 1 F 1 sin θ
More informationModels and Anthropometry
Learning Objectives Models and Anthropometry Readings: some of Chapter 8 [in text] some of Chapter 11 [in text] By the end of this lecture, you should be able to: Describe common anthropometric measurements
More information