Equilibrium at a Point

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Sherilyn Chase
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the use of the free- bod diagram to isolate a sstem for analsis Understand the

1 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 Understand the reac3on provided b the connec3on to a rope Understand the reac3on provided b the connec3on to a spring 2 1

2 Tools Basic Trigonometr Pthagorean Theorem Algebra Scalar analsis of forces 3 Scalar Representa3on We noted that the magnitude of forces is alwas posi3ve. A nega3ve sign in front of a magnitude actuall represents the direc3on of the force and not its absolute magnitude. A posi3ve sign is in the direc3on of the label. A nega3ve sign is in the direc3on awa from the label. 4 2

3 Equilibrium The unifing concept to this course is that of sta3c equilibrium Sta3c equilibrium in this course means that there is no change in velocit with 3me 5 Equilibrium Stated mathema3call, sta3c equilibrium would be Δ v Δt = 0 6 3

4 Equilibrium If we epress this as an instantaneous change in velocit, the epression becomes dv dt = 0 7 Equilibrium From phsics, ou ma remember that instantaneous change in velocit is also known as accelera3on so d v dt = a = 0 8 4

5 Equilibrium 9 Now also from phsics, ou ma remember that accelera3on is produced b the ac3on of a force on a mass This is Newton s first law of mo3on dv dt = a = 0 F = ma Equilibrium 10 Mass is a scalar quan3t so there is no direc3on assigned to it. Force however will have the same direc3on as the accelera3on dv dt = a = 0 F = ma 5

6 Equilibrium 11 We can then remove some of the elements in our long epression to get to the heart of the maoer in this course dv dt = a F = 0 = 0 = F m Equilibrium This means that the net resultant force ac3ng on a bod must be equal to zero for the bod to be in equilibrium F =

7 Equilibrium If we have a number of forces ac3ng on a bod, then the vector sum of those forces (the resultant) must be equal to 0 F = 0 i 13 Equilibrium Since the resultant is equal to 0, the coefficient of each of the components of the resultant must be equal to 0 F i = 0 F j = 0 F z k =

8 Equilibrium 15 If we take the sign from the direc3on of the vector components 3mes the magnitude for each of the components we can then write F F F z = = = Free- Bod Diagram The second important concept that we introduce is the idea of a free- bod diagram The free- bod diagram is an isola3on of an element and the iden3fica3on of all forces which are ac3ng on that element B an element, we mean a part of a mechanical sstem 16 8

9 Free- Bod Diagram Forces can act on a sstem as either an applied force from some eternal source or Forces can act due to the connec3on of the sstem we have isolated to some other sstem in response to the applied forces these forces are known as reac3ons 17 Free- Bod Diagram In order to be able to draw a correct free- bod diagram (I will use FBD for the free- bod diagram) we have to understand what tpes of forces/reac3ons are generated through different tpes of connec3ons 18 9

10 Problem F Problem F3-4 Nothing is the sstem is accelerating. Since the block is at rest, nothing is moving at all. What forces are acting on the block? 20 10

11 Springs Springs are sstems that eert forces that are propor3onal to the condi3on of the spring Springs ma be in compression (pushed together) or in tension (stretched apart) 21 Springs Springs that are in compression tend to push on the objects that the are connected to Springs that are in tension (stretched) tend to pull on the objects that the are connected to Springs alwas generate a force that is along the ais of the spring 22 11

12 Springs 23 The force that the eert will be propor3onal to the amount the are compressed or etended The magnitude of the force will be given b F = k Δl ( ) k is the spring constant and has dimensions of force/length Δl is the amount b which the spring changes from its original unstretched length. If Δl is positive, the spring is stretched and it pulls on what it is connected to. If Δl is negative, the spring is compressed and it pushes on what it is connected to. Problem F3-4 We will assume that the spring is being stretched and draw a force pulling on the block along the line of the 24 12

13 Weight Weight alwas acts toward the center of the earth In tpical problems, that will be towards the booom of the page If the weight or mass of an element is not given, ou ma assume that it is negligible in the analsis that is being done 25 Problem F3-4 We can add a force representing the action of gravit on the block (the weight) acting straight down

14 Smooth Surface A smooth or fric3onless surface provides a force that is perpendicular to the surface. It onl stops something from moving through it. 27 Problem F3-4 Finall, we add a force representing the reaction of the smooth surface on the bo

15 Problem F3-4 We can consider the bo as a point and make all the force coincident at a point. 29 Problem F3-4 We can label each of the forces acting on the sstem to make the work and representation a bit easier

16 Problem F3-4 In order to make the math easier, we can set the -ais along the 0.4m side and the - ais parallel to the 0.3m side. 31 Problem F3-4 Now we can write each of the forces in Cartesian form. = 4 5 =

17 Problem F3-4 Now we can write each of the forces in Cartesian form. = 0 = 33 Problem F3-4 Now we can write each of the forces in Cartesian form. ( ) ( ) = cos 45 0 = sin

18 Problem F3-4 Summing the forces in the and directions we have. F = cos( 45 0 ) F = sin( 45 0 ) 35 Problem F3-4 Both of these sums must be equal to 0. F = cos( 45 0 ) = 0 F = 3 5 F + F 2 3 sin( 450 ) =

19 Problem F3-4 is equal to the weight of the block and weight is equal to mass times gravit. F = mgcos( 45 0 ) = 0 F = mgsin( 45 0 ) = 0 37 Problem F3-4 Substituting what we know F F = 4 5 F kg ( ) 9.81 m s 2 = 3 5 F + F 5kg 1 2 ( ) 9.81 m s 2 ( ) = 0 cos 45 0 ( ) = 0 sin

20 Problem F3-4 Using the first equation and isolating kg = 5 4 5kg = 43.35N ( ) 9.81 m s 2 ( ) 9.81 m s 2 ( ) = 0 cos 45 0 ( ) cos Problem F3-4 Using the epression for a = 43.35N = k 1 ( Δl) 43.35N = 200 N m Δl Δl = 0.22m ( ) 40 20

21 Problem F3-4 So the spring is increased b 0.22m. From the diagram the stretched length of the spring is 0.5m, so the original length of the spring was 0.28m. 41 = 43.35N = k 1 ( Δl) 43.35N = 200 N m Δl Δl = 0.22m ( ) Problem 3-27 The 10-lb weight A is supported b the cord AC and roller C, and b spring AB. If the spring has an unstretched length of 8 in, and the weight is in equilibrium when d = 4 in., determine the k of the 42 21

22 Homework Problem 3-2 Problem 3-6 Problem

Physics 111. Lecture 10 (Walker: 5.5-6) Free Body Diagram Solving 2-D Force Problems Weight & Gravity. February 18, Quiz Monday - Chaps.

Physics 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