Static Equilibrium. Lana Sheridan. Dec 5, De Anza College

tatic Equilibrium Lana heridan De Anza College Dec 5, 2016

Last time simple harmonic motion

Overview Introducing static equilibrium center of gravity

tatic Equilibrium: ystem in Equilibrium Knowing that an object is in equilibrium can give a lot of information about the forces on the object. Previously: a system was in equilibrium if the net force was zero. equilibrium constant velocity or at rest Also, acceleration is zero.

tatic Equilibrium: Extended ystem in Equilibrium Now we consider extended rigid objects.

tatic Equilibrium: Extended ystem in Equilibrium Now we consider extended rigid objects. Forces can cause rotations (torques).

tatic Equilibrium: Extended ystem in Equilibrium Now we consider extended rigid objects. Forces can cause rotations (torques). Force Equilibrium F net = i F i = 0 Rotational Equilibrium τ net,o = i τ i = 0 about any axis O.

tatic Equilibrium: Extended ystem in Equilibrium Rigid Object in Equilibrium A rigid object is said to be in equilibrium if F net = i F i = 0 and τ net,o = i τ i,o = 0 about any axis O. equilibrium v = const. and ω = const. a = 0 and α = 0

tatic Equilibrium tatic Equilibrium is the special case that the object is also at rest: v CM = 0 ω = 0

Question situation: zero net force does not mean an absence of translational motion. Quick Quiz 12.1 1 Consider the object subject to the two forces of equal magnitude. Choose the correct statement for this situation. F d d CM F 2 F Figure 12.2 (Quick Quiz 12.1) Two forces of equal magnitude are applied at equal distances from the center of mass of a rigid object. These conditions describe the condition is a statement of tra acceleration of the object s ce tial reference frame. The seco it states that the angular acce case of static equilibrium, wh equilibrium is at rest relative t speed (that is, v CM 5 0 and v Q uick Quiz 12.1 Consider th in Figure 12.2. Choose the (a) The object is in force eq is in torque equilibrium bu force equilibrium and torqu equilibrium nor torque equ Q uick Quiz 12.2 Consider th Choose the correct stateme force equilibrium but not to librium but not force equili and torque equilibrium. (d) torque equilibrium. (A) The object is in force equilibrium but not torque equilibrium. (B) The object is in torque equilibrium but not force equilibrium. (C) The object is in both force equilibrium and torque equilibrium. (D) The object is in neither F 1 force equilibrium nor torque equilibrium. 1 erway & Jewett, page 364. The two vector expression in general, to six scalar equa and three from the second (c

Question mean an absence of translational motion. Quick Quiz 12.1 1 Consider the object subject to the two forces of equal magnitude. Choose the correct statement for this situation. F (B) The object is in torque equilibrium Fbut 2 not force equilibrium. F 1 d d CM F Figure 12.2 (Quick Quiz 12.1) Two forces of equal magnitude are applied at equal distances from the center of mass of a rigid object. condition is a statement of tra acceleration of the object s ce tial reference frame. The seco it states that the angular acce case of static equilibrium, wh equilibrium is at rest relative t speed (that is, v CM 5 0 and v Q uick Quiz 12.1 Consider th in Figure 12.2. Choose the (a) The object is in force eq is in torque equilibrium bu force equilibrium and torqu equilibrium nor torque equ Q uick Quiz 12.2 Consider th Choose the correct stateme force equilibrium but not to librium but not force equili and torque equilibrium. (d) (A) The object is in force equilibrium but not torque equilibrium. torque equilibrium. (C) The object is in both force equilibrium and torque equilibrium. (D) The object is in neither force equilibrium nor torque equilibrium. 1 erway & Jewett, page 364. The two vector expression in general, to six scalar equa and three from the second (c complex system involving seve

Rotational Equilibrium? A system is in rotational equilibrium if τ net = 0 for any possible axis. uppose the system is in force equilibrium. Then F net = 0. If F net = 0 and τ net,o = 0 for a particular axis O, then τ net,o = 0 for any axis O. Rotational Equilibrium from Force Equilibrium If an object is in force equilibrium and the net torque is zero about one axis, then the net torque must be zero about any other axis.

Rotational Equilibrium? Rotational Equilibrium from Force Equilibrium If an object is in force equilibrium and the net torque is zero about one axis, then the net torque must be zero about any other axis. uppose τ net,o = 0 and F net = 0. ION 12.2 More on the Center of Gravity 365, the F 2 F 1 τ net,o = i r i F i ) ional ough nslarque F 3 O r 1 r 1 r r O F4 Figure 12.4 Construction showing that if the net torque is zero about origin O, it is also zero about any = (r i r ) F i i = 0 (r i F i ) r F i i i = τ net,o = 0

1. straightforward; 2. intermediate; 3. challenging 3. challenging Example 1. full solution available in the tudent 1. full solution olutions available Manual/tudy in the tudent Guide olutions Manual/tudy Guide M Master It tutorial av WebAssign WebAssign W Watch It video solution W Watch Enhanced It WebAssign video solut Enhanced WebAssig #1, page 379 ection 12.1 Analysis Model: Rigid Object in Equilibrium 1. What are the necessary conditions for equilibrium of the object shown in Figure P12.1? Calculate torques about an ection 12.1 Analysis Model: Rigid Object in Equilibrium 1. What are the necessary conditions for axis equilibrium through point of O. the object shown in Figure P12.1? Calculate torques about an axis through point O. 2. Why is the following situation u R x impossible? A uniform beam of O mass m R b 5 3.00 kg and length y, 5 1.00 m supports blocks F g u with masses m 1 5 5.00 kg and R x Figure P12.1 m 2 5 15.0 kg at two positions O as shown in Figure P12.2. The beam rests on two triangular blocks, with point P a distance d 5 0.300 F g m to the right of the center of gravity of the beam. The position of 2. Why is the following situation impossible? A uniform beam of mass m b 5 3.00 kg and length, 5 1.00 m supports blocks with masses m 5 5.00 kg and R y F y F y F x F x 4. M 5.

Example Conditions for equilibrium?

Example Conditions for equilibrium? Fx = 0 F x = R x

Example Conditions for equilibrium? Fx = 0 F x = R x Fy = 0 F y + R y = F g

Example Conditions for equilibrium? Fx = 0 F x = R x Fy = 0 F y + R y = F g Torques about axis through O?

Example Conditions for equilibrium? Fx = 0 F x = R x Fy = 0 F y + R y = F g Torques about axis through O? τ i = τ Fy τ Fx τ Mg i = F y l sin(90 θ) F x l sin θ Mg l sin(90 + θ) 2 = F y l cos θ F x l sin θ Mg l 2 cos θ = 0

Center of Gravity vs. Center of Mass Center of gravity the point in an extended object at which a single gravitational force acting is equivalent to the combination of all the individual gravitational forces acting on each mass element in the object.

al F y 0 ore on the Center of Gravity Center of Gravity vs. Center of Mass iron e deal cross with a rigid object, one of the forces we must consider is the gravrce acting on it, and we must know the point of application of this force. ed in ection 9.5, associated with every object is a special point called its ravity. The combination of the various gravitational forces acting on all fectly mass elements of the object is equivalent to a single gravitational force ugh this point. Therefore, to compute the torque due to the gravitaon an object of mass M, we need only consider the force M g acting at ke find on this an special average point? equilibrium As mentioned orienta- in ection 9.5, if we assume g is center of gravity. we er the object, the center of gravity of the object coincides with its cen-. Chapter To see why, 26) consider an object of arbitrary shape lying in the xy plane d in Figure 12.4. uppose the object is divided into a large number of masses m 1, m 2, m 3,... having coordinates (x 1, y 1 ), (x 2, y 2 ), (x 3, y 3 ),.... In.29, we defined the x coordinate of the center of mass of such an object Each particle of the object has a specific mass and specific e gravis force. y a m i x CM 5 m 1x 1 1 m 2 x 2 1 m 3 x 3 1 c a m i x i coordinates. m 1 1 m 2 1 m 3 1 c 5 i i alled milar equation its to define the y g on all (x 1, y 1 ) (x coordinate of the center of mass, replacwith its y counterpart. 2, y 2 ) ow examine the situation from another al force m point of view by considering the m 2 al force exerted on each particle 1 as shown in Figure 12.5. Each particle gravitacting eight mg at multiplied by its moment CM arm. For example, the magnitude of a torque about an axis through (x the 3, yorigin 3 ) equal in magnitude to the due to the force m 1 g m 1 is m 1 g 1 x 1, where 3 g 1 is the value of the gravitational x n at the me position of Othe particle of mass m 1. We wish to locate the center the point g is at which application of the single gravitational force M gcg its m 1 ceny plane Figure 12.4 An object can 1 m 2 1 m 3 1??? is the total mass of the object and g CG is the accelto gravity at the location of the center of gravity) has the same effect on be Each particle of the object has a specific mass and specific coordinates. Center of gravity the point in an extended object at which a (x single gravitational force acting is equivalent to 1, y 1 ) the (x 2, y 2 ) x m combination of m 2 O 1 all the individual gravitational forces acting on each(x 3 mass, y 3 ) element CM m in the object. 3 x If the gravitational field is uniform (the same Figure at 12.4 all An points) object can be then divided into many small particles. the center of gravity is the same point as the These center particles can of be used mass. to O y locate the center of mass. O y (x 1, y 1 ) m 1 g 1 CG (x 2, y 2 ) m 2 g 2 (x 3, y 3 ) m 3 g3 F g MgCG Figure 12.5 By dividing an object into many particles, we can find its center of gravity. x

Center of Gravity nowns in terms of the known In particular, the torque caused by a single force F g through the center of gravity must be the same as the net torque of all the forces m i g on all the masses m i in the system. t with your diagram. If you selected a lution for a force, do not be alarmed; s the opposite of what you guessed. object and confirm that each set s on the object and confirm that the When an object is supported below its center of mass, there should be no net torque due to gravity. AM M and length, supm f and m d, respeclled the fulcrum) is ther is a distance d /2 from the center. d n 2 rce n exerted by m f g M g m d g board and consider Figure 12.7 (Example 12.1) A balanced system.

Center of Gravity In the diagram the forces F g were in the y direction. The x-coordinate of the center of gravity satisfies: x CG M g CG = i x i m i g i where M = i m i is the total mass.

Center of Gravity In the diagram the forces F g were in the y direction. The x-coordinate of the center of gravity satisfies: x CG M g CG = i x i m i g i where M = i m i is the total mass. If g CG = g i i then the g i s cancel and this simplifies to: x CG = 1 m i x i M i

Question Quick Quiz 12.3 2 A meterstick of uniform density is hung from a string tied at the 25-cm mark. A 0.50-kg object is hung from the zero end of the meterstick, and the meterstick is balanced horizontally. What is the mass of the meterstick? (A) 0.25 kg (B) 0.50 kg (C) 1.0 kg (D) 2.0 kg 2 erway & Jewett, page 366.

Example 12.1 unknowns in terms of the known istent with your diagram. If you selected a r solution for a force, do not be alarmed; rce is the opposite of what you guessed. the object and confirm that each set ques on the object and confirm that the A seesaw 3 consisting of a uniform board of mass M and length l, supports at rest a father and daughter with masses m f and m d, respectively. The support (called the fulcrum) is under the center of gravity of the board, the father is a distance d from the center, and the daughter is a distance l/2 from the center. ted AM ass M and length, supasses m f and m d, respect (called the fulcrum) is e father is a distance d ce,/2 from the center. d n 2 rd force n exerted by m f g M g m d g Figure 12.7 (Example 12.1) A balanced system. the board and consider ughter as support forces applied on the directly board. to the board. The daughter would cause a support, whereas 3 the father would cause a counterclockwise rotation. erway & Jewett, page 367. Determine the magnitude of the upward force n exerted by the

eesaw - Example 12.1 Magnitude of the upward force?

eesaw - Example 12.1 Magnitude of the upward force? All the forces act in the vertical (y) direction. eesaw is at rest. Just equate downward forces with upward one!

eesaw - Example 12.1 Magnitude of the upward force? All the forces act in the vertical (y) direction. eesaw is at rest. Just equate downward forces with upward one! n = (M + m f + m d )g

eesaw - Example 12.1 unknowns in terms of the known istent with your diagram. If you selected a r solution for a force, do not be alarmed; rce is the opposite of what you guessed. the object and confirm that each set ques on the object and confirm that the A seesaw consisting of a uniform board of mass M and length l, supports at rest a father and daughter with masses m f and m d, respectively. The support (called the fulcrum) is under the center of gravity of the board, the father is a distance d from the center, and the daughter is a distance l/2 from the center. ted AM ass M and length, supasses m f and m d, respect (called the fulcrum) is e father is a distance d ce,/2 from the center. d n 2 rd force n exerted by m f g M g m d g Figure 12.7 (Example 12.1) A balanced system. the board and consider ughter as rest. forces applied directly to the board. The daughter would cause a support, whereas the father would cause a counterclockwise rotation. Determine where the father should sit to balance the system at

eesaw - Example 12.1 Where should the father sit to balance the system?

eesaw - Example 12.1 Where should the father sit to balance the system? Equate torques!

eesaw - Example 12.1 Where should the father sit to balance the system? Equate torques! Analyzing torques about the fulcrum point: τ net = 0 m f gd m d g l 2 = 0 m f gd = m d g l 2 d = m dl 2m f

Example 12.3 - lipping Ladder ticity A uniform ladder of length l, rests against a smooth, vertical wall. The mass of the ladder is m, and the coefficient of static friction between the ladder and the ground is µ s = 0.40. Find the minimum angle θ min at which the ladder does not slip., vertical wall (Fig. nt of static friction ind the minimum P n mbed. Do you want er and the surface dder stay up? imul surface. Does the model it as a rigid a u O b f s u mg Figure 12.9 (Example 12.3) (a) A uniform ladder at rest, leaning against a smooth wall. The ground is rough. (b) The forces on the ladder.

ummary tatic equilibrium center of gravity static equilibrium practice 4th Collected Homework! Due Thursday. Final Exam Tuesday, Dec 12, 9:15-11:15 am, 35 (classroom). (Uncollected) Homework erway & Jewett, Ch 11, onward from page 359. Questions: ection Qs 45, 53, 55 Read 12.1 12.3 of Chapter 12. Ch 12, onward from page 400. Questions: ection Qs 3, 11, 15, 19, 23, 25