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PHYSICS for Scientists and Engineers: Foundations and Connections Deborah Katz

What are the conditions for equilibrium?

How do static objects respond when forces are applied?Key QuestionsCengage

In this chapter, we take a closer look at the conditions that are necessary for an object to be at rest.1 neither A nor B. either A or B. both A and B.

A rigid body is in equilibrium if

Copyright 2008 Pearson Education, Inc., publishing as Pearson Addison-Wesley.2Static EquilibriumThe condition for a rigid body to be in static equilibrium is that there is no net force and no net torque p = 0 & L = 0.An important branch of engineering called statics analyzes buildings, dams, bridges, and other structures in total static equilibrium.

No matter which pivot point you choose, an object that is not rotating is not rotating about that point.

For a rigid body in total equilibrium, there is no net torque about any point. This is the basis of a problem-solving strategy.Copyright 2008 Pearson Education, Inc., publishing as Pearson Addison-Wesley.Types of Static Equilibrium

Static equilibrium is classified into three types: Stable, Unstable, and Neutral.To distinguish between these types, we must imagine what would happen if the object in equilibrium were displaced slightly and then released from its new resting position.If after being released, the forces acting on the object return it to its equilibrium position, the object is in stable static equilibrium.If, after being released, the object moves farther away from its equilibrium position, the object ins in unstable static equilibrium.4 For an extended object to be in static equilibrium, the net force and the net torque must be zero.

Conditions for Static EquilibriumFor a point mass object, the condition for static equilibrium is simply net force = 0.However, for an object with a size, the net force = 0 condition is not enough for it to be in static equilibrium.For a sizable object, the condition for static equilibrium is net force = 0 and net torque = 0.5

Which of these objects is in static equilibrium?

Static Equilibrium: Fnet, y = 0Rotation Equilibrium: net = 0If rotation axis at A: dAB*nB dAw*w = 0

If rotation axis at B: -dAB*nA + dBw*w = 0For an object in static equilibrium,the net torques at any axis-point ofrotation must be zero.peg

dAB = distance from point A to point BdAW = distance from point A to the c.o.g. of the hammerdBW = distance from point B to the c.o.g. of the hammer

Copyright 2008 Pearson Education, Inc., publishing as Pearson Addison-Wesley.Cross Product Revisited

Cross Product Revisited

A Day at the PlaygroundA granddaughter of mass mD and her grandfather of mass mF are on an adjustable seesaw of mass mS. The seesaw is a board of length l with notches on its underside so that the board can be moved relative to the pivot. For the seesaw to be in equilibrium when the granddaughter and grandfather sit on opposite ends, the pivot must be placed closer to the grandfather than to the granddaughter. Find an expression for the placement of the pivot relative to the seesaws center of mass.

Fnet = n1 w + n2 = 0Using point 1 as the pivot: tnet = -d1 w + (d1 + d2) n2 = 0

n2 = (d1 w) / (d1 + d2) = (1.5 m)(100 N)/(2 m) = 75 NGo back to Fnet: n1 = w n2 = 100 N 75 N = 25 NFor the purpose of calculating torque, the entire weight of an objectcan be considered as acting at a single point, the center of gravity. Forces on a Board on Sawhorses A board weighing 100 N sits across two sawhorses, as shown in the figure. What are the magnitudes of the normal forces of the sawhorses acting on the board?

Fnet = n1 w + n2 = 0 n1 = w n2This time we put the pivot point at the center of gravity: The torque about the center of gravity is: tnet = -d1 n1 + d2 n2 = 0Substituting n1 into the torque equation: -d1 (w - n2) + d2 n2 = 0

n2 = d1 w / (d1 + d2) = (1.5 m)(100 N)/(2.0 m) = 75 N Forces on a Board on Sawhorses A board weighing 100 N sits across two sawhorses, as shown in the figure. What are the magnitudes of the normal forces of the sawhorses acting on the board?

Finding the Center of Gravity of the Human Body A woman weighing 600 N lies on a 2.5-m-long, 60 N reaction board with her feet over the pivot. The scale on the right reads 250 N. What is the distance d from the womans feet to her center of gravity?

A 2-m-long board weighing 50 N extends out over the edge of a table, with 40% of the boards length off the table. How far beyond the table edge can a 25 N cat walk before the board begins to tilt?Tipping Overa) case 1 will tipb) case 2 will tipc) both will tipd) neither will tipConsider the two configurations of books shown below. Which of the following is true?1/21/41/21/412

19An object is in equilibrium if

Fnet = 0. net = 0.either A or B.both A and B.

20What does the scale read?

500 N1000 N2000 N4000 N

500 N21Dont Let the Rake Fall DownA rake is placed against a smooth wall so that it makes an angle q with the floor of the garage. Assume friction between the wall and the rake is negligible, but the rakes tines are made from a hard rubber that rests on the concrete floor. The coefficient of static friction mS for rubber on dry concrete is 1.0. Find the minimum value of q (other than 0) so that the rake does not slide. The rakes center of mass is one-fourth of the way up from the tines as shown.

What is the tension T in the ropeif the weight of board is 100 N.

Horizontal Beam ExampleConceptualizeThe 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?CategorizeThe system is at rest, categorize as a rigid object in equilibrium.

27Horizontal Beam Example, 2AnalyzeDraw a force 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).

Horizontal Beam Example, 3Analyze, cont.The forces can be resolved into components.Apply the two conditions of equilibrium to obtain three equations.Solve for the unknowns.FinalizeThe positive value for indicates the direction of R was correct in the diagram.

What is the tension T in the ropeif the weight of board, Wb is 100 N,the weight of the person, Wp is 75 N, l is 40 m, and d is 10 m.

What is the tension T in the rope if the weight of board, Wb is 100 N,the weight of the person, Wp is 75 N, l is 40 m, and d is 10 m.

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Will the Ladder Slip?A 3.0-m-long ladder leans against a frictionlessWall at angle of 60o. What is the minimum valueof mS, the coefficient of static friction with theground, that will prevent the ladder from slipping?

Using the dimensions given in the figure, find the torque of force f3 with the bottom of the ladder as the axis.

f3a +f3a f3a sin(q) +f3a cos(q) +f3a sin(q)qf3a=33

Using the dimensions given in the figure, find the torque of persons weight mg with the bottom of the ladder as the axis.

mgb +mgb mgb sin(q) +mgb cos(q) +mgb sin(q)qf3ab

What is the torque of tension T at point A as the pivot?

hT sin(q) +hT sin(q) +hT hT cos(q) +hT cos(q)Th=q35

LWpWhat is the torque of the poles weight WP at point A as the pivot?

(L/2) WP sin(q) +(L/2) WP sin(q) +(L/2) WP (L/2) WP cos(q) +(L/2) WP cos(q)qq

Traffic Light Pole A traffic light hangs from a pole as shown in the figure below. The uniform aluminum pole AB is 7.50 m long and has a mass of 12.0 kg. The mass of the traffic light is 21.5 kg. Determine the tension in the horizontal massless cable CD.Solution: Net torque at pivot A= 3.8*T (L/2 cos(37)) (12g) (L cos(37))(21.5*g)=0 3.8*T = (L cos(37))*g*(6+21.5) T= 425 N. (D)Center of gravity = Center of mass

The center of mass is the mass-weighted center of the object.

The Center of Mass of a Rod Find the center of mass of a thin, uniform rod of length L and mass M. Use this result to find the tangential acceleration of one tip of a 1.60-m-long rod that rotates about its center of mass with an angular acceleration of 6.0 rad/s2.

Divide the rod into many small cellsof mass dm.One such cell, at position x.The cells width is dx.

2010 Pearson Education, Inc.39Answer: BBalanceFor an object to balance, its center of gravity must reside over its base of support. That way gravity does not exert a torque.

Base of supportGravity acts at the center of gravity.Line of actionThis force exerts no torque about her toes.40To maintain stability, we unconsciously moveour arms and legs to keep our center of gravityover our base of support.In Fig. a, the center of gravity is well abovethe base of support (the foot), ensuring stability.In Fig. b, when we stand on tiptoes, we needto lean forward to maintain stability.

Stability and Balance of the Human Bodya.b.

Stable: if a vertical line through objects center of gravity passes through its base of support.Unstable: if such a line lies outside its base of support.

Sedan

The maximum possible angle of tilt (qc) is determined by the ratio oftrack width (t) to the height of center-of-gravity (h).

The larger t/h the more stable is the car.

SUVPutting c.o.g. above the base point:

An object will be stable if

its center of gravity is below its highest point.its center of gravity lies over its base of support. its center of gravity lies outside its base of support.the height of its center of gravity is less than 1/2 its total height.

44Answer: BElasticity

Copyright 2008 Pearson Education, Inc., publishing as Pearson Addison-Wesley.45Elasticity and Fracture

The SI units of stress are N/m2 and 1 N/m2 = PaPa stands for pascals.Elasticity and Fracture

A stress on a rod causes a strain.ElasticityF/A is proportional to L/L. We can write the proportionality as The proportionality constant Y is called Youngs modulus. The quantity F/A is called the tensile stress. The quantity L/L, the fractional increase in length, is called strain.

With these definitions, we can writestress = Y strainCopyright 2008 Pearson Education, Inc., publishing as Pearson Addison-Wesley.

Copyright 2008 Pearson Education, Inc., publishing as Pearson Addison-Wesley.

Stretching a Wire A 2.0-m-long, 1.0-mm-diameter wire is suspended from the ceiling. Hanging a 4.5 kg mass from the wire stretches the wires length by 1.0 mm. What is Youngs modulus for this wire? Can you identify the material?

Referring to Table 15.3, we see that the wire is made of copper.Copyright 2008 Pearson Education, Inc., publishing as Pearson Addison-Wesley.The rod fractures at some maximum tensile stress known as thetensile strength sten. Experiments show that tensile strength depends on the material of the rod.

A microscopic model illustrates the strain on a rod as it reacts to a stress exerted on it by a hanging load. Each molecular bond stretches vertically, obeying Hookes law.Shear Deformation

Shear stress t results when a force is appliedtangentially to one surface of a stationary object:

G is called the shear modulus.The dedication plate mounted to the base of a building was deformed during an earthquake. The plate, made from a metal alloy of shear modulus G=4.0010 Pa, was originally 80.0 cm high, 50.0 cm long, and 5.00 mm thick. The earthquake shifted the top surface of the plate 0.100 mm relative to the bottom surface. What shear force did the plate experience during the earthquake?

Shear strength is the strength of a material or component against the type of yield or structural failure where the material or component fails in shear. Shear strength of a material is the maximum shear stress that the material can support without yielding in shear. Shear strength = max

A shear load is a force that tends to produce a sliding failure on a material along a plane that is parallel to the direction of the force. When a paper is cut with scissors, the paper fails in shear.Given total force at failure (F) and the force-resisting area (e.g. the cross-section of a bolt loaded in shear), Ultimate Shear Strength () is:

What force is required to punch a 20-mm-diameter hole in a plate that is 25 mm thick? The shear strength of the plate is 350 MN/m2.

FThe resisting area is the shaded area along thePerimeter:

Given total force at failure (F) and the force-resisting area (e.g. the cross-section of a bolt loaded in shear), Ultimate Shear Strength () is:

The shear force F is equal to the punching force.Seeing Isnt Always BelievingIn the figure, a car seems to be hanging from some gooey substance.Estimate the Youngs modulus for this unidentified material.

Applications

Copyright 2008 Pearson Education, Inc., publishing as Pearson Addison-Wesley.