Conditions for Equilibrium. Static Equilibrium. Rotational Equilibrium. Torque

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1 Static Equilibrium Equilibrium implies the object is at rest (static) or its center of mass moves with a constant velocity (dynamic) Static equilibrium is a common situation in engineering Principles involved are of particular interest to civil engineers, architects, and mechanical engineers Conditions for Equilibrium The net force equals zero ΣF = 0 If the object is modeled as a particle, then this is the only condition that must be satisfied The net torque equals zero Στ = 0 This is needed if the object cannot be modeled as a particle Torque Rotational Equilibrium τ = r x F Use the right-hand rule to determine the direction of the torque The tendency of the force to cause a rotation about O depends on F and the moment arm d Need the angular acceleration of the object to be zero For rotation, Στ = Iα For rotational equilibrium, Στ = 0 This must be true for any axis of rotation 1

2 Equilibrium Summary There are two necessary conditions for equilibrium The resultant external force must equal zero: ΣF = 0 This is a statement of translational equilibrium The acceleration of the center of mass of the object must be zero when viewed from an inertial frame of reference The resultant external torque about any axis must be zero: Στ = 0 This is a statement of rotational equilibrium The angular acceleration must equal zero Equilibrium Equations We will restrict the applications to situations in which all the forces lie in the xy plane These are called coplanar forces since they lie in the same plane There are three resulting equations ΣF x = 0 ΣF y = 0 Στ z = 0 Axis of Rotation for Torque Equation The net torque is about an axis through any point in the xy plane The choice of an axis is arbitrary If an object is in translational equilibrium and the net torque is zero about one axis, then the net torque must be zero about any other axis Center of Gravity 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) 2

3 Center of Gravity 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 If g is uniform over the object, then the center of gravity of the object coincides with its center of mass If the object is homogeneous and symmetrical, the center of gravity coincides with its geometric center Problem-Solving Strategy Equilibrium Problems Draw a of the system Isolate the object being analyzed Draw a free-body Show and label all external forces acting on the object Indicate the locations of all the forces For systems with multiple objects, draw a separate free-body for each object Problem-Solving Strategy Equilibrium Problems Establish a convenient coordinate system Find the components of the forces along the two axes Apply the first condition for equilibrium (ΣF=0) Be careful of signs Problem-Solving Strategy Equilibrium Problems Choose a convenient axis for 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 force that acts along a line passing through the origin produces a zero torque 3

4 Problem-Solving Strategy Equilibrium Problems Apply the second condition for equilibrium (Στ = 0) The two conditions of equilibrium will give a system of equations Solve the equations simultaneously If the solution gives a negative for a force, it is in the opposite direction than what you drew in the free body Weighted Hand Example Model the forearm as a rigid bar The weight of the forearm is ignored There are no forces in the x-direction Apply the first condition for equilibrium (ΣF y = 0) Weighted Hand Example Apply the second condition for equilibrium using the joint O as the axis of rotation (Στ =0) Generate the equilibrium conditions from the freebody Solve for the unknown forces (F and R) Horizontal Beam Example The beam is uniform So the center of gravity is at the geometric center of the beam The person is standing on the beam What are the tension in the cable and the force exerted by the wall on the beam? 4

5 Horizontal Beam Example... Draw a free-body Use the pivot in the problem (at the wall) as the pivot This will generally be easiest Note there are three unknowns (T, R, θ) Horizontal Beam Example The forces can be resolved into components in the freebody Apply the two conditions of equilibrium to obtain three equations Solve for the unknowns Ladder Example Ladder Example The ladder is uniform So the weight of the ladder acts through its geometric center (its center of gravity) There is static friction between the ladder and the ground Draw a free-body for the ladder The frictional force is ƒ = µn Let O be the axis of rotation Apply the equations for the two conditions of equilibrium Solve the equations 5

6 Ladder Example, Extended Add a person of mass M at a distance d from the base of the ladder The higher the person climbs, the larger the angle at the base needs to be Eventually, the ladder may slip 6