CHAPTER 4

EQUILIBRIUM OF COPLANER

FORCE SYSTEMS

STATIC EQUILIBRIUM

CONDTIONS FOR EQUILIBRIUMOF A RIGID OBJECT

• The (vector) sum of the external forces on the rigid object must equal zero:

ΣF = 0

10 lbs 10 lbs

5 lbs

5 lbs

we say the object is in Translational Equilibrium

• The sum of the external torques on the rigid object must equal zero.

ΣM = 0

we say the object is in Rotational Equilibrium

ΣF = 0 ΣM = 0+ = Static Equilibrium

ΣFx = 0 ΣFy = 0 ΣMz = 0

Since nearly all of the problems we will solve are two–dimensional problems, these equations reduce to:

Free body diagrams (1)

FBD is an essential step in the solution of all problems involving forces on bodies

it is a diagram of the external surface of the body - not interested in internal forces

all other bodies in contact with the one we are interested in are replaced by vectors

Free body diagrams (2)

Sketch of person standing

mg

R1 R2

F=ma

R1+R2-mg=ma, but no acceleration so,

R1+R2=mg

Free body diagrams (3)

sketch

mg

T

free body diagram

Free body diagrams (4)

Rules: clear sketches draw in the correct orientation show all forces acting on the body don’t show any internal forces between

different parts of the body show the forces not the components

TWO FORCE MEMBERS

CAN ONLY HAVE 2 FORCES ACTING ON THEM, AT THEIR ENDSFOR EQUILIBRIUM TO EXIST THE FORCES MUST BE:

EQUAL IN MAGNITUDEOPPOSITE IN SENSECOLINEARLINE OF ACTION PASSES THROUGH CENTER LINE

FORCE TRIANGLE METHOD

Move force A so that it's tail meets the head of force B Draw resultant from tail of B to Tip of A

F2

F3

100lbC

T

METHOD OF COMPONENTS

APPLY LAWS OF EQUILIBRIUM

ΣFH = 0 ΣFV=0

Write equation for Horizontal equilibrium

Write equation for Vertical equilibrium

Substitute for T from earlier equation

Note: the positive signs do not necessarilymean the forces act in a positive direction,only that the senses you assumed were correct

Strategy for solving problems in static equilibrium:

• Determine all the forces that are acting on the rigid body. They will come from the other objects with which the body is in contact (supports, walls, floors, weights resting on them) as well as gravity

• Draw a diagram and put in all the information you have about these forces: The points on the body at which they act, their magnitudes (if known), their directions (if known).

• Write down the equations for static equilibrium.

• Solve the equations!

APPLY EQUILIBRIUM EQUATIONS

SUSTITUTING

EQUILIBRIUM OF PARALLEL FORCE SYSTEMS

ΣF = 0 ΣM = 0+

PARALLEL SYSTEMS CANNOT BE DETERMINED BY ΣF ONLY

ΣM MUST BE INVOLVED

Assumptions of + and – are important

EQUILIBRIUM OF NONCONCURRENTFORCE SYSTEMS

ΣFH = 0 ΣFV=0 ΣM=0

Compute the Reactions at A and B on the truss below. There is a roller supportat A and a pin support at B

- force system is noncoplaner, nonconcurrent and nonparallel

ΣFx = 0 ΣFy = 0 ΣMz = 0

Write Equilibrium Equation forHorizontal Forces

RBH = 423 lbs

As RBH is positive, the assumed sense was correct!

Determine Reaction force at Aby summing moments about B

Determine Vertical Reaction force at Bby summing moments about A

Verify by Writing Equilibrium equationfor Vertical Forces.

Note: There a 4 unknown forces,acting on the bar, but we only have 3 equilibrium equations!!!! (Hint: Force N comes from the cylinder)

Knowing Force N allows us to use equilibrium equations

Example A 6 kN force is supported away from the wall.

Determine the reactions at the supports.

We can write the vertical force equilibrium equation to find RB.

sin 45 6kN 0 vertical BF R

8.485kNBR

cos45 0 horizontal B CF R R

6kNCR

Then, the horizontal force equilibrium equation gives us

Example 2

Determine the weight on front and rear wheels of an 8,000 lb van.

The wheelbase (the distance between the front and rear wheels) is 10 ft.

Assume that the center of gravity is close to the front tires and at a distance of one-third of the wheelbase from the front wheels.

Example 2

Step I: Draw the free-body diagram of the van

The free-body diagram shows two reactions (Rrear and Rfront) along with the weight of the van.

The weight of the van is placed at its center of gravity.

The distances between the forces are marked in terms of wheel base L.

This is a three-force system.

Step II: Apply the equilibrium equations

The force equilibrium equation can be written as

If we take the moment about the rear wheel, the reactions at the rear wheels (Rrear) do not create any moment. We get

Substituting the value of Rrear in the force equilibrium equation, we get

8000lb 0

8000lb

rear front

rear front

F R R

R R

2LL 8000lb 0

3 frontM R

5,333lbfrontR

2,667 lbrearR

Each front wheel carries 2667 lb (half of Rfront), and each rear wheel carries 1334 lb.

The results make sense because the front wheels, which are close to the load application, carry more weight than the rear wheels.

Auto designers arrange the subsystems in such a way that the center of gravity is closer to the drive wheels. As a result, the drive wheels carry more weight and, therefore, have increased traction (frictional force).

Determine the reactions at the supports. The length of the beam is L.

Step I: Draw the free-body diagram The pin support provides both horizontal and vertical

reaction.

The roller provides a vertical reaction.

Step II: Apply the equilibrium equations If we take the force equilibrium equation along the x-axis, we get

The moment equilibrium equation at point A yields

Similarly, the moment at point B gives us

0

0

x

Ax

F

R

L 0.7L 0.3L 0 ByRAM P P 0.4ByR P

L 0.7L 0.3L 0 AyRBM P P 0.4AyR P

Tail Tipping

Tail tipping is a phenomenon where an airplane tips over the rear wheels due to an increased load in the aft section.

When the airplane is approaching the tipping condition, the normal force acting on the nose wheel is greatly reduced.

An airplane close to tail tipping looses the traction force needed for steering during the take-off or landing.

Table Tipping

If the weight distribution is uniform, the weight the table (W) can be applied at the center of gravity The reactions at the supports will be

If only force F acting at the end of the table, the equilibrium equations are

0.5

0.5

A-W

B-W

R

R

W

W

2a b 0

b2a

b1 2a

about point B

A-F B-F

A-F

A-F

B-F

= R R

= R

R

R

F F

M F

F

F

Table Tipping

Using the principle of superposition, the reactions when both forces (W and F) are applied

b0.5 2a

b0.5 1 2a

A A-W A-F

B B-W B-F

R R R

R R R

FW

W F

Table Tipping

Increasing the magnitude of force F reduces the reaction at support A and increases the reaction at support B.

At a certain magnitude of force F, support A is about to loose contact with the ground.

This property is useful to experimentally determine the impending tipping condition by measuring the reaction force at the supports by sensors.

b0 a ARFW

Tail Tipping The second method for determining

the impending tipping condition uses the equilibrium equation.

The equilibrium equations for the free-body diagram

At tipping the magnitude of RA is equal to zero. Substituting the magnitude of RA, we get

Often used to calculate the magnitude of the tipping force.

2a a b 0about point B A= RM W F

baFW

Footprint

A third method for determining the tipping condition is the use of footprint.

Footprint is the minimum area by a closed curve with nonnegative (positive or zero) curvature at any point on the curve encompassing the base.

Imagine an elastic band stretched to encompass the object at its base.

The area encompassed by the band indicates the footprint.

The Leaning Tower of Pisa

The Tower of Pisa is the tallest self-supporting bell tower.

With a diameter of 15.5 m and height of 56 m, the tower weighs 15,000 metric tons.

Due to the foundation problems, the tower started leaning, and this design failure resulted in one of the Seven Wonders of the World.

The tower leans at an angle of 5.50.

The Leaning Tower of Pisa

To find the critical angle at which it will become unstable (from the rigid-body point of view), the Tower is approximated as a cylinder with uniform density.

The tower will tip-over when the weight passes outside the footprint. The critical angle can be calculated as

0

7.75 mtan θ

28 m

θ 15.47

The crane shown in the figure below weighs 10,000 lb and its boom weighs 1000 lb.

The centers of gravity of the crane and the boom are known.

Determine the maximum load the crane can carry at a boom angle of 450.

For tipping, the moment caused by the load W and the boom weight overcomes the moment caused by the weight of the crane.

At the point of tipping, the reaction at the support B is zero.

Taking the moments about A, we get

(12cos 45 3) 1000 (9cos 45 3) 10000 6 W

10,325lbW

For the step ladder shown, sides AC and EC are 8 ft long and hinged at C. TieRod BD is 2.5 ft long and halfway up. A man weighing 192 lb climbs 6 feet up the ladder. Assume floor is frictionless, neglect weight of ladder. 1. Find tensionin Rod BD. 2. Reaction forces at points A and E.

Hint: consider the sides of ladderas separate objects

FX

FY

ΣFH = 0

ΣFY = 0

ΣM about C = 0

C

1

2

3

4

5From equations 2 & 5

From equation 3

1 2

3 4

5