Physics 111: Mechanics Lecture 12 Static Equilibrium Static in addition to Dynamic Equilibrium Conditions as long as Equilibrium Conditions as long as Equilibrium

Physics 111: Mechanics Lecture 12 Static Equilibrium Static in addition to Dynamic Equilibrium Conditions as long as Equilibrium Conditions as long as Equilibrium www.phwiki.com

Physics 111: Mechanics Lecture 12 Static Equilibrium Static in addition to Dynamic Equilibrium Conditions as long as Equilibrium Conditions as long as Equilibrium

Waetjen, Sage, Executive Producer has reference to this Academic Journal, PHwiki organized this Journal Physics 111: Mechanics Lecture 12 Dale Gary NJIT Physics Department Static Equilibrium Equilibrium in addition to static equilibrium Static equilibrium conditions Net external as long as ce must equal zero Net external torque must equal zero Center of gravity Solving static equilibrium problems Static in addition to Dynamic Equilibrium Equilibrium implies the object is at rest (static) or its center of mass moves with a constant velocity (dynamic) We will consider only with the case in which linear in addition to angular velocities are equal to zero, called “static equilibrium” : vCM = 0 in addition to w = 0 Examples Book on table Hanging sign Ceiling fan – off Ceiling fan – on Ladder leaning against wall

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Conditions as long as Equilibrium The first condition of equilibrium is a statement of translational equilibrium The net external as long as ce on the object must equal zero It states that the translational acceleration of the object’s center of mass must be zero Conditions as long as Equilibrium If the object is modeled as a particle, then this is the only condition that must be satisfied For an extended object to be in equilibrium, a second condition must be satisfied This second condition involves the rotational motion of the extended object Conditions as long as Equilibrium The second condition of equilibrium is a statement of rotational equilibrium The net external torque on the object must equal zero It states the angular acceleration of the object to be zero This must be true as long as any axis of rotation

Conditions as long as Equilibrium The net as long as ce equals zero If the object is modeled as a particle, then this is the only condition that must be satisfied The net torque equals zero This is needed if the object cannot be modeled as a particle These conditions describe the rigid objects in the equilibrium analysis model Static Equilibrium Consider a light rod subject to the two as long as ces of equal magnitude as shown in figure. Choose the correct statement with regard to this situation: The object is in as long as ce equilibrium but not torque equilibrium. The object is in torque equilibrium but not as long as ce equilibrium The object is in both as long as ce equilibrium in addition to torque equilibrium The object is in neither as long as ce equilibrium nor torque equilibrium Equilibrium Equations For simplicity, We will restrict the applications to situations in which all the as long as ces lie in the xy plane. Equation 1: Equation 2: There are three resulting equations

A seesaw consisting of a uni as long as m board of mass mpl in addition to length L supports at rest a father in addition to daughter with masses M in addition to m, respectively. The support is under the center of gravity of the board, the father is a distance d from the center, in addition to the daughter is a distance 2.00 m from the center. A) Find the magnitude of the upward as long as ce n exerted by the support on the board. B) Find where the father should sit to balance the system at rest. A) Find the magnitude of the upward as long as ce n exerted by the support on the board. B) Find where the father should sit to balance the system at rest. Axis of Rotation The net torque is about an axis through any point in the xy plane Does it matter which axis you choose as long as calculating torques NO. The choice of an axis is arbitrary If an object is in translational equilibrium in addition to the net torque is zero about one axis, then the net torque must be zero about any other axis We should be smart to choose a rotation axis to simplify problems

B) Find where the father should sit to balance the system at rest. O P Rotation axis O Rotation axis P Center of Gravity The torque due to the gravitational as long as ce on an object of mass M is the as long as ce Mg acting at the center of gravity of the object If g is uni as long as m over the object, then the center of gravity of the object coincides with its center of mass If the object is homogeneous in addition to symmetrical, the center of gravity coincides with its geometric center Where is the Center of Mass Assume m1 = 1 kg, m2 = 3 kg, in addition to x1 = 1 m, x2 = 5 m, where is the center of mass of these two objects A) xCM = 1 m B) xCM = 2 m C) xCM = 3 m D) xCM = 4 m E) xCM = 5 m

Center of Mass (CM) An object can be divided into many small particles Each particle will have a specific mass in addition to specific coordinates The x coordinate of the center of mass will be Similar expressions can be found as long as the y coordinates Center of Gravity (CG) All the various gravitational as long as ces acting on all the various mass elements are equivalent to a single gravitational as long as ce acting through a single point called the center of gravity (CG) If then CG of a Ladder A uni as long as m ladder of length l rests against a smooth, vertical wall. When you calculate the torque due to the gravitational as long as ce, you have to find center of gravity of the ladder. The center of gravity should be located at C A B D E

Ladder Example A uni as long as m ladder of length l rests against a smooth, vertical wall. The mass of the ladder is m, in addition to the coefficient of static friction between the ladder in addition to the ground is s = 0.40. Find the minimum angle at which the ladder does not slip. Problem-Solving Strategy 1 Draw sketch, decide what is in or out the system Draw a free body diagram (FBD) Show in addition to label all external as long as ces acting on the object Indicate the locations of all the as long as ces Establish a convenient coordinate system Find the components of the as long as ces along the two axes Apply the first condition as long as equilibrium Be careful of signs Which free-body diagram is correct A uni as long as m ladder of length l rests against a smooth, vertical wall. The mass of the ladder is m, in addition to the coefficient of static friction between the ladder in addition to the ground is s = 0.40. gravity: blue, friction: orange, normal: green A B C D

A uni as long as m ladder of length l rests against a smooth, vertical wall. The mass of the ladder is m, in addition to the coefficient of static friction between the ladder in addition to the ground is s = 0.40. Find the minimum angle at which the ladder does not slip. Problem-Solving Strategy 2 Choose a convenient axis as long as calculating the net torque on the object Remember the choice of the axis is arbitrary Choose an origin that simplifies the calculations as much as possible A as long as ce that acts along a line passing through the origin produces a zero torque Be careful of sign with respect to rotational axis positive if as long as ce tends to rotate object in CCW negative if as long as ce tends to rotate object in CW zero if as long as ce is on the rotational axis Apply the second condition as long as equilibrium Choose an origin O that simplifies the calculations as much as possible A uni as long as m ladder of length l rests against a smooth, vertical wall. The mass of the ladder is m, in addition to the coefficient of static friction between the ladder in addition to the ground is s = 0.40. Find the minimum angle. O O O O A) B) C) D)

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A uni as long as m ladder of length l rests against a smooth, vertical wall. The mass of the ladder is m, in addition to the coefficient of static friction between the ladder in addition to the ground is s = 0.40. Find the minimum angle at which the ladder does not slip. Problem-Solving Strategy 3 The two conditions of equilibrium will give a system of equations Solve the equations simultaneously Make sure your results are consistent with your free body diagram If the solution gives a negative as long as a as long as ce, it is in the opposite direction to what you drew in the free body diagram Check your results to confirm Horizontal Beam Example A uni as long as m horizontal beam with a length of l = 8.00 m in addition to a weight of Wb = 200 N is attached to a wall by a pin connection. Its far end is supported by a cable that makes an angle of = 53 with the beam. A person of weight Wp = 600 N st in addition to s a distance d = 2.00 m from the wall. Find the tension in the cable as well as the magnitude in addition to direction of the as long as ce exerted by the wall on the beam.

Horizontal Beam Example The beam is uni as long as m So the center of gravity is at the geometric center of the beam The person is st in addition to ing on the beam What are the tension in the cable in addition to the as long as ce exerted by the wall on the beam Horizontal Beam Example, 2 Analyze Draw a free body diagram Use the pivot in the problem (at the wall) as the pivot This will generally be easiest Note there are three unknowns (T, R, q) The as long as ces can be resolved into components in the free body diagram Apply the two conditions of equilibrium to obtain three equations Solve as long as the unknowns Horizontal Beam Example, 3

Horizontal Beam Example, 3

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