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 force must equal zero n Net eternal torque must equal zero q Center of gravity q Solving static equilibrium problems
Static and Dynamic Equilibrium q Equilibrium implies the object is at rest (static) or its center of mass moves with a constant velocity (dynamic) q We will consider only with the case in which linear and angular velocities are equal to zero, called static equilibrium : v CM 0 and w 0 q Eamples n n n n n Book on table Hanging sign Ceiling fan off Ceiling fan on Ladder leaning against wall
Conditions for Equilibrium q The first condition of equilibrium is a statement of translational equilibrium q The net eternal force on the object must equal zero! F net!! å F ma et 0 q It states that the translational acceleration of the object s center of mass must be zero
Conditions for Equilibrium q If the object is modeled as a particle, then this is the only condition that must be satisfied!! F net å Fet 0 q For an etended object to be in equilibrium, a second condition must be satisfied q This second condition involves the rotational motion of the etended object
Conditions for Equilibrium q The second condition of equilibrium is a statement of rotational equilibrium q The net eternal torque on the object must equal zero! t net!! åt Ia 0 et q It states the angular acceleration of the object to be zero q This must be true for any ais of rotation
Conditions for Equilibrium qthe net force equals zero åf! 0 n If the object is modeled as a particle, then this is the only condition that must be satisfied qthe net torque equals zero å! t 0 n This is needed if the object cannot be modeled as a particle qthese conditions describe the rigid objects in the equilibrium analysis model
Static Equilibrium q (A) (B) (C) (D) Consider a light rod subject to the two forces of equal magnitude as shown in figure. Choose the correct statement with regard to this situation: The object is in force equilibrium but not torque equilibrium. The object is in torque equilibrium but not force equilibrium The object is in both force equilibrium and torque equilibrium The object is in neither force equilibrium nor torque equilibrium
Equilibrium Equations q For simplicity, We will restrict the applications to situations in which all the forces lie in the y plane. q Equation 1: q Equation 2:! F! t net net! å Fet 0 : Fnet 0 Fnet, y 0 Fnet, z! åt : t 0 t 0 t, et 0 net, net, y net, z q There are three resulting equations F å F 0 F t net, net, y net, z å åt F et, et, y et, z 0 0 0 0
q EX: A seesaw consisting of a uniform board of mass m pl and length L supports at rest a father and daughter with masses M and m, respectively. The support is under the center of gravity of the board, the father is a distance d from the center, and the daughter is a distance 2.00 m from the center. q A) Find the magnitude of the upward force n eerted by the support on the board. q B) Find where the father should sit to balance the system at rest.
A) Find the magnitude of the upward force n eerted by the support on the board. B) Find where the father should sit to balance the system at rest. F n t net, y net, z mgd n - mg - Mg - m g 0 mg mgd æ ç è t m M + Mg + m d ö d ø + t 2m M pl + t g < + t pl - Mg + 0 + 0 0 Mg f pl n 2.00 m F F t net, net, y net, z å F å F åt et, et, y et, z 0 0 0
Ais of Rotation q The net torque is about an ais through any point in the y plane q Does it matter which ais you choose for calculating torques? q NO. The choice of an ais is arbitrary q If an object is in translational equilibrium and the net torque is zero about one ais, then the net torque must be zero about any other ais q We should be smart to choose a rotation ais to simplify problems
B) Find where the father should sit to balance the system at rest. Rotation ais O t mgd net, z m M d mgd - Mg + 0 + 0 0 æ ç è t + t + t + t Mg ö d ø f 2m M pl n Rotation ais P t 0 - Mg( d - Mgd - Mg - m mgd net, z æ ç è t + t + t + t m M d Mg ö d ø f 2m M pl + ) - m pl pl n gd + nd 0 gd + ( Mg + mg + m pl g) d 0 P O F F t net, net, y net, z å F å F åt et, et, y et, z 0 0 0
Center of Gravity q The torque due to the gravitational force on an object of mass M is the force Mg acting at the center of gravity of the object q If g is uniform over the object, then the center of gravity of the object coincides with its center of mass q If the object is homogeneous and symmetrical, the center of gravity coincides with its geometric center
Where is the Center of Mass? q Assume m 1 1 kg, m 2 3 kg, and 1 1 m, 2 5 m, where is the center of mass of these two objects? m + m A) CM 1 m B) CM 2 m C) CM 3 m D) CM 4 m E) CM 5 m 1 1 CM m + 1 m 2 2 2
Center of Mass (CM) q An object can be divided into many small particles n Each particle will have a specific mass and specific coordinates q The coordinate of the center of mass will be CM å i å i m q Similar epressions can be found for the y coordinates i m i i
Center of Gravity (CG) q All the various gravitational forces acting on all the various mass elements are equivalent to a single gravitational force acting through a single point called the center of gravity (CG) Mg CG m g 1 1 CG 1 ( m1 + m2 + m3 +!) g + m g + m g +! 2 2 2 3 3 3 CG CG q If g g2 3 1 g! q then CG m11 + m22 + m m + m + m 1 2 3 3 +! +! 3 å mi å m i i
CG of a Ladder q A uniform ladder of length l rests against a smooth, vertical wall. When you calculate the torque due to the gravitational force, you have to find center of gravity of the ladder. The center of gravity should be located at A B D C E
Ladder Eample q 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 q at which the ladder does not slip.
Problem-Solving Strategy 1 q Draw sketch, decide what is in or out the system q Draw a free body diagram (FBD) q Show and label all eternal forces acting on the object q Indicate the locations of all the forces q Establish a convenient coordinate system q Find the components of the forces along the two aes q Apply the first condition for equilibrium q Be careful of signs F F net, net, y å F å F et, et, y 0 0
Which free-body diagram is correct? q 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. gravity: blue, friction: orange, normal: green A B C D
q 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 q at which the ladder does not slip. å å F F y P f n mg P f f - P 0 n - mg 0,ma µ n µ mg s s mg
Problem-Solving Strategy 2 q Choose a convenient ais for calculating the net torque on the object n Remember the choice of the ais is arbitrary q Choose an origin that simplifies the calculations as much as possible n A force that acts along a line passing through the origin produces a zero torque q Be careful of sign with respect to rotational ais n n n positive if force tends to rotate object in CCW negative if force tends to rotate object in CW zero if force is on the rotational ais q Apply the second condition for equilibrium t net, z åt et, z 0
Choose an origin O that simplifies the calculations as much as possible? q 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. A) B) C) O D) O O mg mg mg mg O
q 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 q at which the ladder does not slip. åt 0 + 0 + sinq cosq q min O t min min n tan + t Pl sinq tanq -1 f + t min min 1 ( ) 2µ s g + t P l - mg cosq min 2 mg mg 2P 2µ mg tan -1 s 1 [ ] 2(0.4) 0 1 2µ! 51 s mg
Problem-Solving Strategy 3 q The two conditions of equilibrium will give a system of equations q Solve the equations simultaneously q Make sure your results are consistent with your free body diagram q If the solution gives a negative for a force, it is in the opposite direction to what you drew in the free body diagram q Check your results to confirm F F t net, net, y net, z å F å F åt et, et, y et, z 0 0 0
Horizontal Beam Eample q A uniform horizontal beam with a length of l 8.00 m and a weight of W b 200 N is attached to a wall by a pin connection. Its far end is supported by a cable that makes an angle of f 53 with the beam. A person of weight W p 600 N stands a distance d 2.00 m from the wall. Find the tension in the cable as well as the magnitude and direction of the force eerted by the wall on the beam.
Horizontal Beam Eample q The beam is uniform n So the center of gravity is at the geometric center of the beam q The person is standing on the beam q What are the tension in the cable and the force eerted by the wall on the beam?
Horizontal Beam Eample, 2 q Analyze n Draw a free body diagram n Use the pivot in the problem (at the wall) as the pivot n This will generally be easiest n Note there are three unknowns (T, R, q)
Horizontal Beam Eample, 3 q The forces can be resolved into components in the free body diagram q Apply the two conditions of equilibrium to obtain three equations q Solve for the unknowns
Horizontal Beam Eample, 3 0 sin sin 0 cos cos - - + å - å b p y W W T R F T R F f q f q N m m N m N l l W W d T l W W d l T b p b p z 313 )sin53 (8 ) )(4 (200 ) )(2 (600 sin ) 2 ( 0 ) 2 ( ) )( sin ( + + - - å! f f t N N T R T T W W T T W W R R b p b p 581 cos 71.7 )cos53 (313 cos cos 71.7 sin sin tan sin sin tan cos sin 1 ø ö ç ç è æ - + - + -!!! q f f f q f f q q q
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