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Static Equilibrium AP Physics C
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Static Equilibrium For a system of particles to be in static equilibrium there are two conditions: it must be in translational equilibrium and it must be in rotational equilibrium. Translational equilibrium means the system is not changing its location. For a rigid body (i.e., a solid as opposed to a liquid or gas), translational equilibrium means that at least one point in the body is stationary. Rotational equilibrium means the system is not rotating about any axis.
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Static Equilibrium To be in translational equilibrium, the sum of the forces on the body must be zero, just as for a single particle. To be in rotational equilibrium, the turning effect of all the forces must also be zero. We call the turning effect of a force a torque (𝜏). For a rigid body to be in static equilibrium we must have both: 𝐹 =0 and 𝜏=0
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Torque The door is free to rotate about an axis through O
There are three factors that determine the effectiveness of the force in opening the door: The magnitude of the force The position of the application of the force The angle at which the force is applied (direction of the force)
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𝜏 = 𝑟 × 𝐹 (cross product of two vectors)
Torque 𝜏 = 𝑟 × 𝐹 (cross product of two vectors) t is the torque F is the force symbol is the Greek tau r is the length of the position vector Runs from the point where torque is to be evaluated and the point of application of the force SI unit is N.m
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Direction of Torque Torque is a vector quantity
The direction is perpendicular to the plane determined by the position vector and the force If the turning tendency of the force is counterclockwise, the torque will be positive If the turning tendency is clockwise, the torque will be negative
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Right Hand Rule Point the fingers in the direction of the position vector Curl the fingers toward the force vector The thumb points in the direction of the torque
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Multiple Torques When two or more torques are acting on an object, the torques are added As vectors If the net torque is zero, the object’s rate of rotation doesn’t change
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Check Point The figure shows an overhead view of a meter stick that can pivot about the dot at the position marked 20 (for 20cm). All five horizontal forces on the stick have the same magnitude. Rank those forces according to the magnitude of the torque that they produce, greatest first. F1: τ1 = r1 F1 sin φ1 = (20)Fsin(90o) = 20F (CCW) F2: τ2 = r2 F2 sin φ2 = (0)Fsin(60o) = 0 (no direction) F3: τ3 = r3 F3 sin φ3 = (20)Fsin(90o) = 20F (CW) F4: τ4 = r4 F4 sin φ4 = (20)Fsin(60o) = 17.3F (CW) F5: τ5 = r5 F5 sin φ5 = (80)Fsin(0o) = 0 (no direction)
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Equilibrium Example The woman, mass m, sits on the left end of the see-saw The man, mass M, sits where the see-saw will be balanced Solve for the unknown distance, x
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Equilibrium Example What is the normal force? m = 55 kg, M = 75 kg and the mass of the plank is 12 kg
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Axis of Rotation If the object is in equilibrium, it does not matter where you put the axis of rotation for calculating the net torque The location of the axis of rotation is completely arbitrary Often the nature of the problem will suggest a convenient location for the axis When solving a problem, you must specify an axis of rotation Once you have chosen an axis, you must maintain that choice consistently throughout the problem
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Notes About Equilibrium
A zero net torque does not mean the absence of rotational motion An object that rotates at uniform angular velocity can be under the influence of a zero net torque This is analogous to the translational situation where a zero net force does not mean the object is not in motion
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Solving Equilibrium Problems
Draw a diagram of the system Include coordinates and choose a rotation axis Isolate the object being analyzed and draw a free body diagram showing all the external forces acting on the object For systems containing more than one object, draw a separate free body diagram for each object
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Problem Solving, cont. Apply the Second Condition of Equilibrium
This will yield a single equation, often with one unknown which can be solved immediately Apply the First Condition of Equilibrium This will give you two more equations Solve the resulting simultaneous equations for all of the unknowns Solving by substitution is generally easiest
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Example of a Free Body Diagram (Forearm)
Isolate the object to be analyzed Draw the free body diagram for that object Include all the external forces acting on the object
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Example of a Free Body Diagram (Beam)
The free body diagram includes the directions of the forces The weights act through the centers of gravity of their objects
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Example A uniform horizontal beam with a length of l = 8.00 m and 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 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 exerted by the wall on the beam.
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Example, cont
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Example, cont
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Example of a Free Body Diagram (Ladder)
The free body diagram shows the normal force and the force of static friction acting on the ladder at the ground The last diagram shows the lever arms for the forces
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Example mg 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 s = Find the minimum angle at which the ladder does not slip.
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Example, cont.
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Check Point The figure gives an overhead view of a uniform rod in static equilibrium. (a) Can you find the magnitudes of unknown forces and by balancing the forces? (b) If you wish to find the magnitude of force by using a balance of torques equation, where should you place a rotation axis to eliminate from the equation? (c) The magnitude of turns out to be 65 N. What then is the magnitude of F ?
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Sample Problem A uniform beam, of length L and mass m 1.8 kg, is at rest on two scales. A uniform block, with mass M 2.7 kg, is at rest on the beam, with its center a distance L/4 from the beam’s left end. What do the scales read?
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Answer
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Example Figure 12-6a shows a safe (mass M 430 kg) hanging by a rope (negligible mass) from a boom (a 1.9 m and b 2.5 m) that consists of a uniform hinged beam (m 85 kg) and horizontal cable (negligible mass). (a) What is the tension Tc in the cable? In other words, what is the magnitude of the force on the beam from the cable?
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Answer
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