CH 3 Statics Problems 2-D and 3-D

Engr221 Chapter 3 1

Announcements

Equilibrium of a Particle in 2-D

Today’s Objectives• Draw a free body diagram (FBD)

• Apply equations of equilibrium to solve a 2-D problem

Class Activities• Applications

• What, why, and how of a Free Body Diagram

• Equations of equilibrium in 2-D

• Springs and pulleys

• Examples

Engr221 Chapter 3 2

Applications

For a spool of given weight, what are the forces in cables ABand AC?

Applications - continued

For a given cable strength, what is the maximum weight that can be lifted?

Engr221 Chapter 3 3

For a given engine weight, what are the forces in the cables? What size of cable should you use?

Applications - continued

Equilibrium of a Particle in 2-D

This is an example of a 2-D, or coplanar force system. If the whole assembly is in equilibrium, then particle A is also in equilibrium.

To determine the tensions in the cables for a given weight of the engine, we need to learn how to draw a Free Body Diagram and apply equations of equilibrium.

Engr221 Chapter 3 4

The What, Why, and How of a Free Body Diagram

Free Body Diagramsare one of the most important things for you to know how to draw, and how to use.

What is it? - A drawing that shows all external forces acting on a particle.

Why draw it? - It helps you write the equations of equilibrium used to solve for the unknowns (usually forces or angles).

How? • Imagine the particle to be isolated or cut free from its

surroundings

• Show all the forces that act on the particle- Active forces: They tend to move the particle

- Reactive forces: They tend to resist the motion

• Identify each force and show all known magnitudes and directions as variables

FBD at A Engine mass = 250 Kg

A

Engr221 Chapter 3 5

Equations of 2-D Equilibrium

Or written in scalar form,

ΣFx = 0 and ΣFy = 0

These are two scalar equations of equilibrium (E of E). They can be used to solve for up to two unknowns.

Since particle A is in equilibrium, the net force at A is zero.

ΣΣΣΣF = 0 (vector equation)

or TB + TD + W = 0FBD at A

A

In general, for a particle in equilibrium, ΣΣΣΣF = 0 orΣFx i + ΣFy j = 0 = 0 i + 0 j (A vector equation)

A

Example A

Write the scalar Equations of Equilibrium (E of E):

+ → ΣFx = TB cos 30º – TD = 0

+ ↑ ΣFy = TB sin 30º – 2.452 kN = 0

Solving the second equation gives: TB = 4.90 kNFrom the first equation, we get: TD = 4.25 kN

Engine mass = 250 Kg FBD at A

Engr221 Chapter 3 6

Springs

Spring Force = spring constant × deformation

F = k × s

Cables and Pulleys

With a frictionless pulley, tensions are equal

Engr221 Chapter 3 7

Given: Sack A weighs 20 lb with geometry as shown

Find: Forces in the cables and weight of sack B

Plan:

1. Draw a FBD at point E

2. Apply E of E at point E to solve for the unknowns (TEG & TEC)

3. Repeat this process at point C

Example B

The scalar E of E are:+ → ΣFx = TEGsin 30º – TEC cos 45º = 0+ ↑ ΣFy = TEGcos 30º – TEC sin 45º – 20 lbs = 0

Solving these two simultaneous equations for the two unknowns yields: TEC = 38.6 lbTEG = 54.6 lb

A FBD at E should look like the one to the left. Note the assumed directions for the two cable tensions.

Example B - continued

Engr221 Chapter 3 8

+ → Σ Fx = 38.64 cos 45° – (4/5) TCD = 0+ ↑ Σ Fy = (3/5) TCD + 38.64 sin 45° – WB = 0

Solving the first equation and then the second yields:TCD = 34.2 lband WB = 47.8 lb

The scalar E of E are:

Now, move on to ring C. A FBD for C should look like the one to the left.

Example B - continued

Example C

Given: The car is towed at constant speed by the 600 lb force The angle θ is 25°

Find: The forces in the ropes ABand AC

Plan:

1. Draw a FBD at point A

2. Apply the E of E to solve for the forces in ropes AB and AC

Engr221 Chapter 3 9

30°25°

600 lb

FABFAC

AFBD at point A

Applying the E of E at point A, we get:

+ → ∑ Fx = FAC cos 30° – FAB cos 25° = 0

+ ↑ ∑ Fy = -FAC sin 30° – FAB sin 25° + 600 = 0

Solving the above equations, we get:

FAB = 634 lbFAC = 664 lb

Example C - continued

Questions

1) When a particle is in equilibrium, the sum of forces acting on it equals ___

A) a constant B) a positive number C) zero

D) a negative number E) an integer

2) For a frictionless pulley and cable, tensions in the cable (T1 and T2) are related as _____

A) T1 > T2

B) T1 = T2

C) T1 < T2

D) T1 = T2 sin θ

Engr221 Chapter 3 10

Question

A

30°

40°

100 lb

Select the correct FBD of particle A.

A) A

100 lb

B)

30°

40°

A

F1 F2

C) 30°A

F

100 lb

A

30° 40°F1 F2

100 lb

D)

F2

20 lb

F1

C

50°

Using the FBD for point C, the sum of forces in the x-direction (Σ FX) is ___ . Use a sign convention of + →A) F2 sin 50° – 20 = 0

B) F2 cos 50° – 20 = 0

C) F2 sin 50° – F1 = 0

D) F2 cos 50° + 20 = 0

Question


Textbook Problem 3-10

The 500-lb crate is hoisted using the ropes AB and AC. Each rope can withstand a maximum tension of 2500 lb before it breaks. If AB always remains horizontal, determine the smallest angle θ to which the crate can be hoisted.

For TAB = 2500 lb, θ = 11.31º

For TAC = 2500 lb, θ = 11.54º

• Draw a free body diagram (FBD)


Summary


Announcements

• F = mg

• F = ks

• Test Monday

Equations of 2-D Equilibrium - Review

Or written in scalar form,

ΣFx = 0 and ΣFy = 0

These are two scalar equations of equilibrium (E of E). They can be used to solve for up to two unknowns.

Since particle A is in equilibrium, the net force at A is zero.

ΣΣΣΣF = 0 (vector equation)

or TB + TD + W = 0FBD at A

A

In general, for a particle in equilibrium, ΣΣΣΣF = 0 orΣFx i + ΣFy j = 0 = 0 i + 0 j (A vector equation)

A


• Imagine the particle to be isolated or cut free from its surroundings

• Show all the forces that act on the particle- Active forces: They tend to move the particle

- Reactive forces: They tend to resist the motion

• Identify each force and show all known magnitudes and directions as variables

FBD at A Engine mass = 250 Kg

A

Free Body Diagrams - Review

The unstretched length of spring AB is 2 m. If the block is held in the equilibrium position shown, determine the mass of the block at D.


FAB = 90 N

FAC = 102 N

W = 126 N

Mass = 12.8 kg


Determine the un-stretched length of spring AC if a force P = 80 lb causes the angle θ = 60º for equilibrium. Cord AB is 2 ft long. Take k = 50 lb/ft.


Fs = 40 lb

Luns = 2.66 ft

Equations of 3-D Equilibrium

Or written in scalar form:

ΣFx = 0, ΣFy = 0, and ΣFz = 0

These are the three scalar equations of equilibrium. They can be used to solve for up to three unknowns.

Since the particle is in equilibrium, the net force at the origin is zero.

So F1 + F2 + F3 = 0

or ΣΣΣΣF = 0FBD at A

A

In general, for a particle in equilibrium in 3-D, ΣΣΣΣF = 0 orΣFx i + ΣFy j + ΣFz k = 0 = 0 i + 0 j + 0 k


Procedure for Analysis

Free-Body Diagram• Establish the x, y, and z axes in any suitable orientation

• Label all known and unknown force magnitudes and directions on the diagram

• Assume senses for unknown magnitudes

Equations of Equilibrium• Use scalar equations of equilibrium when it is easy to resolve

each force into its x, y, and z components (not often)

• If the 3-D geometry is difficult, express the forces in Cartesian form, use ΣΣΣΣF = 0, and equate thei, j, andk components to 0

• If the answer is negative, the assumed sense should be reversed

Example 3-D Problem

The 50-kg pot is supported from A by the three cables. Determine the force acting in each cable for equilibrium. Take d = 2.5 meters.

FAB = 580 N

FAC = 312 N

FAD = 312 N


Announcements

• Homework notes

• Test Monday– One crib sheet (given)

– Calculator

– Pencil

– Eraser

– FBD’s, units, process are grading considerations

Procedure for Analysis - Review

Free-Body Diagram• Establish the x, y, and z axes in any suitable orientation.

• Label all known and unknown force magnitudes and directions on the diagram.

• Assume senses for unknown magnitudes.

Equations of Equilibrium• Use scalar equations of equilibrium when it is easy to resolve

each force into its x, y, and z components.

• If the 3-D geometry is difficult, express the forces in Cartesian form, use ΣΣΣΣF = 0, and set thei, j, andk components to 0.

• If the answer is negative, the assumed sense should be reversed.


Textbook Example 3.8

Textbook Example 3.8 - continued


A small peg P rests on a spring that is contained inside the smooth pipe. When the spring is compressed so that s = 0.15 m, the spring exerts an upward force of 60 N on the peg. Determine the point of attachment A(x,y,0) of cord PA so that the tension in cords PB and PC equals 30 N and 50 N, respectively.

Example Problem

x = 0.1904 m

y = 0.0123 m

Determine the tension developed in cables OD and OB and the strut OC, required to support the 50-kg crate. The spring OA has an unstretched length of 0.8 m and a stiffness of kOA = 1.2 kN/m. The force in the strut acts along the axis of the strut.

Example Problem

Answers TBD


• Draw a free body diagram (FBD)



Chapter 3 Summary

CH 3 Statics Problems 2-D and 3-D

Documents

Transcript of CH 3 Statics Problems 2-D and 3-D