Chapter 9- Static Equilibrium

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1 Chapter 9- Static Equilibrium

2 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

3 Old assignments and midterm exams (solutions have been posted on the web) can be picked up in my office (LB-212)

4 All marks, including assignments, have been posted on the web. Please, verify that all your marks have been entered in the list.

5 Chapter 9 The Conditions for Equilibrium Solving Statics Problems

6 Recalling Last Lectures

8 Angular Momentum and Its Conservation We have defined several angular quantities in analogy to linear motion. Type Linear Rotational Relation Displacement x θ x = rθ Velocity v ω v = rω Acceleration a α a = rα Dynamics F (force) τ (torque) τ = Fd Inertia m (mass) I (moment of inertia) Kinetic Energy Newton s 2 nd Law Equations of motion

9 Angular Momentum and Its Conservation 1) Angular momentum (8-26) L is called angular momentum.

10 Angular Momentum and Its Conservation 2) Work in angular momentum (8-27) Eq represents the work done by the torque τ when rotating an object through an angle θ. Power P (rate with it the work is done): (8-28)

11 Angular Momentum and Its Conservation 3) Impulse in angular momentum Impulse in rotational motion as (8-29)

12 Angular Momentum and Its Conservation 4) Generalized for of Newton s 2 nd Law General form of the rotational equivalent of Newton s 2 nd law: (8-30) Changes in the momentum of inertia ( L = Iω) are taken into account in the formalism of Newton s 2 nd law.

13 Angular Momentum and Its Conservation 5) Conservation of Angular Momentum The total angular momentum is conserved if there is NO external net torque acting on the system. (8-31)

14 Angular Momentum and Its Conservation Since NO external torque is applied on her, her total angular momentum should be conserved: Therefore, if her momentum of inertia decreases, her angular velocity has to increase to maintain the total angular momentum constant.

15 Today

16 Problem 8-62 (textbook): A 4.2-m-diameter merry-go-round is rotating freely with an angular velocity of 0.80 rad/s. Its total moment of inertia is 1760 Kg.m 2. Four people standing on the ground, each of mass 65 kg, suddenly step onto the edge of the merry-go-round. What is the angular velocity of the merry-go-round now? What if the people were on it initially and then jumped off in a radial direction (relative to the merry-go-round)?

17 . Problem 8-62 (textbook): The angular momentum of the merry-go-round and people combination will be conserved because there are no external torques on the combination. This situation is a totally inelastic collision, in which the final angular velocity is the same for both the merry-go-round and the people. Subscript 1 represents before the collision, and subscript 2 represents after the collision. The people have no initial angular momentum. L = L I ω = I ω I I 1 m-g-r m-g-r ω = ω = ω = ω I I + I I + 4M R 2 m-g-r people m-g-r person kgm = ( 0.80 rad s) 0.48 rad s 2 2 = 1760 kgm kg 2.1 m ( )( ) If the people jump off the merry-go-round radially, then they exert no torque on the merry-go-round, and thus cannot change the angular momentum of the merry-goround. The merry-go-round would continue to rotate at.0.80 rad/s. I

18 The Conditions for Equilibrium So far, we have worried about motion (velocity) and change of motion (net acceleration). However, there are many situations where we want to make sure objects remain in their original configuration of motion: at rest or constant velocity. In this lecture, we will deal with that situation where the original state of motion of an object is such that it is at rest. As we have seen, for an object to remain in its state of rest NO net force should act on it in other words, it does not have any acceleration. If external or internal net forces acts on an object, it can deform or fracture it. We want to find a state the equilibrium such that the object is not subjected to fractures or deformations

$Determining these forces allow engineers to build structures that can significantly minimize effects due to deformation or fracture.$

19 The Conditions for Equilibrium Static Static is the area of physics concerned about the calculation of the forces acting on structures that are in equilibrium. Determining these forces allow engineers to build structures that can significantly minimize effects due to deformation or fracture. No matter what, there is always at least one force acting on a object: gravity. If, for example, a book is at rest on a table, the normal force has to balance the gravitational force in order for the book to be found in equilibrium.

20 The Conditions for Equilibrium It is clear that if an object is to stay in its state of equilibrium NO net forces NOR net torque should act on the object In other words NO changes in translational or rotational motion must occur. The chandelier should not move in the vertical, horizontal or circular direction for equilibrium to be met.

21 The Conditions for Equilibrium First condition of equilibrium No change in translational motion The first condition for equilibrium is that the forces along each coordinate axis add to zero. (9-1) In the figure, the equilibrium condition implies that: 120 o

22 The Conditions for Equilibrium Second condition of equilibrium No change in rotational motion The second condition of equilibrium is that there be no torque around any axis; the choice of axis is arbitrary and we will assume that its direction is perpendicular to the xy plane of rotation. (9-2) In the figure, the sign supported by a beam of length l mounted by a hinge on a wall is in equilibrium. Chosen the hinge as the center of rotation, and counterclockwise as positive: l/2 l

23 The Conditions for Equilibrium Recalling an important note: We consider that the mass of an object with uniform mass distribution (or uniform object for short) is such that it can be assumed to be located at the center of the object. This point is know as either center of mass or center of gravity of the object. In the figure, assuming an uniform beam, its center of mass will be located at the middle of the beam (and this is why the force of gravity is shown to be acting on this point). l/2 l For an uniform sphere, its center of mass is located in its center.

24 Solving Statics Problems 1. Choose one object at a time, and make a free-body diagram showing all the forces on it and where they act. 2. Choose a coordinate system and resolve forces into components. 3. Write equilibrium equations for the forces. 4. Choose any axis perpendicular to the plane of the forces and write the torque equilibrium equation. A clever choice here can simplify the problem enormously. 5. Make sure you correctly identify the components of the forces that can contribute to rotational motion. Forces that have zero lever arm do not contribute to rotational motion. 6. Solve.

25 Solving Statics Problems If you do not know the direction a force, just assume a generic one. For example, assume that F B points up and F A is unknown in the figure below. You initially assume that F A points up and then try to find the conditions of equilibrium for this problem. You also assume that the axis of rotation passes through the point of application of F A and that counterclockwise is positive. F A in your solution comes negative. It just means that it s in the opposite direction from the one you chose. This is trivial to fix, so don t worry about getting all the signs of the forces right before you start solving.

26 Solving Statics Problems If there is a cable or cord in the problem, it can support forces only along its length. Forces perpendicular to that would cause it to bend.

27 Problem 9-1 (textbook): Three forces are applied to a tree sapling, as shown in Fig. 9 41, to stabilize it. If and, find in magnitude and direction.

Problem 9-1 : If the tree is not accelerating, then the net force in all directions is 0. F F = F + F cos110+ F = 0 x C x A B F F C A B C x ( ) = F F cos110= 310 N 425 N cos110= 164.

28 Problem 9-1 : If the tree is not accelerating, then the net force in all directions is 0. F F = F + F cos110+ F = 0 x C x A B F F C A B C x ( ) = F F cos110= 310 N 425 N cos110= N F = F sin + F = 0 y y B B C y ( ) = F sin 110= 425 N sin 110= N ( ) ( ) = F + F = + = C C x C y 1 C y 1 o o o o o C x N N N N F N θ = tan = tan = 67.6, φ = = F N And so is 430 N, at an angle of 112 o clockwise from.

29 Problem 9-18 (textbook): Calculate (a) the tension F T in the wire that supports the 27-kg beam shown in Fig. 9 52, and (b) the force F r W exerted by the wall on the beam (give magnitude and direction).

30 Problem 9-18 : (a) The beam is in equilibrium, and so the net force and net torque on the beam must be zero. From the free-body diagram for the beam, calculate the net torque (counterclockwise positive) about the wall support point to find F r, and calculate the net force in both the x and y directions to find T the components of F r W. F o τ = FL sin 40 mgl 2= 0 T 2 ( 27 kg)( 9.80 m s ) mg = = = 2 sin 40 2 sin 40 T o o N N ( ) F = F F cos 40 = 0 F = F cos 40 = N cos 40 = N o o o x Wx T Wx T F F F y o = + sin 40 mg = 0 Wy T F mg F ( )( ) ( ) o 2 o = sin 40 = 27 kg 9.80 m s N sin 40 = N Wy T L

31 Problem 9-18 : (b) ( ) F = F F cos 40 = 0 F = F cos 40 = N cos 40 = N o o o x Wx T Wx T F = F + F mg = y Wy F mg F F o sin 40 0 T ( )( ) ( ) o 2 o = sin 40 = 27 kg 9.80 m s N sin 40 = N Wy T ( ) ( ) = F + F = + = W Wx Wx F θ = tan = tan = 40 F Wy 1 o Wx N N N N

32 Problem 9-27 (textbook): Consider a ladder with a painter climbing up it (Fig. 9 61). If the mass of the ladder is 12.0 kg, the mass of the painter is 55.0 kg, and the ladder begins to slip at its base when her feet are 70% of the way up the length of the ladder, what is the coefficient of static friction between the ladder and the floor? Assume the wall is frictionless.

33 Problem 9-27: The ladder is in equilibrium, so the net torque and net force must be zero. By stating that the ladder is on the verge of slipping, the static frictional force at the ground,, is at its maximum value and so: F Cx F Cx = µ F s Cy Since the person is standing 70% of the way up the ladder, the height of the ladder is: L = d 0.7= 2.8 m 0.7= 4.0 m y y The width of the ladder is: L = d 0.7= 2.1 m 0.7= 3.0 m x x

34 Problem 9-27 Torques are taken about the point of contact of the ladder with the ground, and counterclockwise torques are taken as positive. The three conditions of equilibrium are as follows: F Gy F = F F = 0 F = F x Cx W Cx W F = F Mg mg = y 0 Gy 2 ( M m) g ( )( ) = + = 67.0 kg 9.80 m s = N 1 ( ) τ = F L mg L Mgd = W y 2 x x Solving the torque equation gives: F W 0 ( )( ) + ( )( ) ( 2 ) 1 1 ml + Md 12.0 kg 3.0 m 55.0 kg 2.1 m 2 x x 2 = g = 9.80 m s = N L 4.0 m y FG x N µ = = = 0.50 s F N The coefficient of friction then is found to be G y

Static 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