Understanding Static Equilibrium Principles

Chapter 9 Static Equilibrium

Static Equilibrium Objects in the real world have multiple forces acting on them at all times without any motion in the objects Objects on tables, leaning objects, hanging objects When there is no motion, then forces in each direction cancel out so that ? = 0 in all directions When an object has a net force of zero, and is at rest, the object is said to be in static equilibrium There are two conditions that must be held in order for something to be in static equilibrium These conditions stem from Newton s 2nd Law for linear and rotational motion

First Condition of Static Equilibrium For an object to be at rest, the sum of the force acting on it must add up to zero ??= 0 ??= 0 ?2 ?1 ??

Example: Chandelier Calculate the tensions ?? and ?? in the two cords that are connected to the vertical cord supporting the 200 ?? chandelier. We need a free-body diagram. For equilibrium, we want to choose a spot where all 3 forces are acting ? = 60 ??: ???+ ???= 0 ??: ??? ??= 0 ?? ?? ??sin ? ?? = 0 ??cos ? + ??= 0 ?? ?? sin ? ??= ??= ??cos ? ??= 2260 ? ??= 1130 ?

Second Condition of Static Equilibrium If an object is to remain at rest, the net torque applied to it must be zero ? = 0 While the first condition ensures that the linear acceleration must be zero, our second condition for equilibrium ensures that the angular acceleration of an object must be zero Rotation occurs around pivot points, so we must add the torques that act about a pivot and set them equal to 0 to ensure no movement ?1 ?2

Example: Seesaw A board of mass ? = 4.0 ?? serves as a seesaw. Person A has a mass of 30 ?? and sits 2.5 ? from the pivot point, P. Person B has a mass of 25 ??. What distance from the pivot most person B be to balance the see saw? We still a free-body diagram. Now, we set up a system of equations. 2 come from ? = 0 and 1 comes from ? = 0. We do this for every static equilibrium problem! ? ?? ? ??? ??? ?? = 0 ?? 0 = 0 The forces act around a pivot so the creates a torque onto the system. We need to consider the torque from every force. ?1 ? ???? ????= 0 ?2 ?? ?? ?? Note that the Normal force and weight of the see saw act at the pivot, so they cannot produce a torque

Example: Seesaw A board of mass ? = 4.0 ?? serves as a seesaw. Person A has a mass of 30 ?? and sits 2.5 ? from the pivot point, P. Person B has a mass of 25 ??. What distance from the pivot most person B be to balance the see saw? We just need to solve each equation ?? ? ??? ??? ?? = 0 ? ???? ????= 0 ? ? = 578 ? ??=?? ???? ??=??? ????? ?1 Person B weighs less, so they need a greater lever arm to cancel the torque ?2 ??= 3.0 ? ?? ?? ??

Example: Ladder A 5.0 ? long ladder leans against a wall at a point 4.0 ? above a cement floor. The ladder is uniform and has mass ? = 12.0 ??. Assuming the wall is frictionless, but the floor is not, determine the forces exerted on the ladder by the floor and by the wall. To solve this, we need to set up the free body diagram and equilibrium conditions ?? ??= 0 ??= 0 ? = 0 ?? ?? ?? = 0 ?? ? ??= 0 5 ? ?? 4 ? ? = ?? ??= 118 ? ?? Notice we simply start with one condition, solve as far as we can then move onto the next. Now we only have the torque condition left ? 3 ?

Example: Ladder A 5.0 ? long ladder leans against a wall at a point 4.0 ? above a cement floor. The ladder is uniform and has mass ? = 12.0 ??. Assuming the wall is frictionless, but the floor is not, determine the forces exerted on the ladder by the floor and by the wall. ? = ?? ??= 118 ? ? = 0 For torque, we now need to pick a pivot point. This can by anywhere along the ladder, but we can choose a smart place to simplify the torques. Pick a spot where the most forces are acting ?? 5 ? Since forces at a pivot do not create torque, only ?? and ?? will create a torque ?? 4 ? ?? ? ????sin ?? ????sin ?? = 0 ? 3 ? What are the lever arms for each force?

Aside: Defining the Lever Arm Note the normal way we define torque is to do tip to tail and multiply the force and lever arm, and consider the angle between the two The length of the ladder is 5 ?, and we can use trig to find out the angle ??= sin 14/5 = 53.1 ?? The torque due to the wall is then, ?? ??= ???sin ?? = ??5.0 ? sin 53.1 = 4 ?? 5 ? ?? 4 ? ? ?? ? 3 ?

Aside: Defining the Lever Arm Note the normal way we define torque is to multiply the force and lever arm, and consider the angle between the two The length of the ladder is 5 ?, and we can use trig to find out the angle ??= sin 14/5 = 53.1 ?? The torque due to the wall is then, ?? ??= ???sin ?? = ??5.0 ? sin 53.1 = 4 ?? ? 5 ? The second way to do torque is to draw the lever arm perpendicular to the force. Since we know the wall force is exerted 4 ? above the ground, we immediately know the lever arm ?? 4 ? ?? ? ??= ???sin ? = ??4.0 ? sin 90 = 4 ?? 3 ? Both get the same result!

Aside: Defining the Lever Arm We can do the same for the gravitational force. For the first method, the force acts halfway up the ladder, so ? = 2.5 ?. We need trig to find out the angle between them ??= sin 13/5 = 36.9 The torque due to gravity is then, ?? = ?? 2.5 ? sin 36.9 ??= ???sin ?? = 1.5 ?? ?? 5 ? ?? 4 ? ?? ? ?? 3 ?

Aside: Defining the Lever Arm We can do the same for the gravitational force. For the first method, the force acts halfway up the ladder, so ? = 2.5 ?. We need trig to find out the angle between them ??= sin 13/5 = 36.9 The torque due to gravity is then, ?? = ?? 2.5 ? sin 36.9 ??= ???sin ?? = 1.5 ?? ?? The second way to do torque is to draw the lever arm perpendicular to the force. We know the force is halfway up the ladder, so this cuts the base in half to ? = 1.5 ? 5 ? ?? 4 ? ?? ??= ???sin ? = ?? 1.5 ? sin 90 = 1.5 ?? ?? Again, we get the same result! The method you use will depend on whether the problem give you angles or sides ? 3 ?

Example: Ladder A 5.0 ? long ladder leans against a wall at a point 4.0 ? above a cement floor. The ladder is uniform and has mass ? = 12.0 ??. Assuming the wall is frictionless, but the floor is not, determine the forces exerted on the ladder by the floor and by the wall. ? = ?? ??= 118 ? ? ???? ????= 0 ?? ?? 1.5 ? sin 90 ??4.0 ? sin 90 = 0 ?? ??=?? 1.5 ? 4.0 ? 5 ? ?? 4 ? ? = 44 ? ??= 44 ? ??= 118 ? ?? ? 3 ? ??

Example: Hanging Sign A shop sign weighing 21.9 ?? hangs from the end of a uniform 15.8 ?? beam that is 1.70 ? long. Find the tension in the supporting wire (at 35 and 1.35 ? from the pivot) and the horizontal ?? and vertical ?? forces exerted by the hinge on the beam at the wall We consider the torque acting on the beam, so we need a free- body diagram of all forces acting on it, as well as a pivot point. ?? ? Now we can list our equations for static equilibrium. Let s start with forces ? ?? ??? ??: ?cos ? + ??= 0 ??:?sin ? ??? ???+ ??= 0 ??? ??= ?sin ? + ??? + ??? ??= ?cos ?

Example: Hanging Sign A shop sign weighing 21.9 ?? hangs from the end of a uniform 15.8 ?? beam that is 1.70 ? long. Find the tension in the supporting wire (at 35 and 1.35 ? from the pivot) and the horizontal ?? and vertical ?? forces exerted by the hinge on the beam at the wall Now we can consider the torques ??= ?cos ? ?? ??= ?sin ? + ??? + ??? ? ?:???sin ? ????? ?????= 0 ? ?? ??? ???sin ? = ????? + ????? ??? ? =????? + ????? ??sin ? 0.85 ??? + 1.70 ??? 1.35 sin 35 ? =

Example: Hanging Sign A shop sign weighing 21.9 ?? hangs from the end of a uniform 15.8 ?? beam that is 1.70 ? long. Find the tension in the supporting wire (at 35 and 1.35 ? from the pivot) and the horizontal ?? and vertical ?? forces exerted by the hinge on the beam at the wall ??= ?cos ? 0.85 ??? + 1.70 ??? 1.35 sin 35 ? = ?? ??= ?sin ? + ??? + ??? ? ? ? = 641 ? ?? ??? ??? ??= 525 ? ??= 641 ? cos 35 ??= 641 ? sin 35 + ??? + ??? ??= 1.80 ?

Elasticity If a force is exerted on an object, the length of the object changes This is true in all objects, so we can approximate all objects as if they behaved like a spring ? = ? Since we do not notice the springiness of objects, their spring constant must be very large as a result After a certain amount of force is applied, we break the spring. The amount you can stretch or compress an object without breaking is called the elastic limit 0 When the force is within the elastic limit, it will always return to its original shape Once surpassed, the object will not return to its original length

Youngs Modulus If we apply a force on an object, it acts over the area. We call this the stress on the object and define it as, ? Stress =? ? The ratio of how much the object is stretched compared to its rest length is called the strain on the object and is defined as, Strain = ? =stress strain =1 ? ? 0 0 0 ? We use these to define the spring constant of the object that we call Young s Modulus We can use this equation to find out how much an objects length will change depending on the force applied

Example: Piano Wire A 1.60 ? long steel piano wire has a diameter 0.20 ??. How great is the tension of the wire if it stretches 0.25 ?? when tightened? To find tension, we must solve fore the force that is causing the wire to stretch =1 ? ? 0 ? = ? ? = ??2 ? ? 0 ? = (2.0 1011 ?/?2)0.0025 ? (3.14 10 6) 1.60 ? ? = 980 ?