Ch.05 Equilibrium of a Rigid Body

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05. Equilibrium of a Rigid Body

HCM City Univ. of Technology, Faculty of Mechanical Engineering Nguyen Tan Tien

Engineering Mechanics – Statics 5.01 Equilibrium of a Rigid Body

Chapter Objectives

• To develop the equations of equilibrium for a rigid body

• To introduce the concept of the free-body diagram for a rigid

body

• To show how to solve rigid-body equilibrium problems using

the equations of equilibrium



§1. Conditions for Rigid-Body Equilibrium

- The force and couple system acting on a body can be reduced

to an equivalent resultant force and resultant couple moment

at an arbitrary point 𝑂

- A rigid body is in equilibrium ⟺ 𝐹 𝑅 = ∑𝐹 𝑖 = 0, 𝑀𝑅𝑂= ∑𝑀𝑂 = 0

• The sum of the forces acting on the body is equal to zero

• The sum of the moments of all forces in the system about

point 𝑂, added to all the couple moments, is equal to zero



⟺

§1. Conditions for Rigid-Body Equilibrium

- The two equations of equilibrium for a rigid body

𝐹 𝑅 = ∑𝐹 = 0

𝑀𝑅𝑂= ∑𝑀𝑂 = 0

where 𝑂 is an arbitrary point

- Equilibrium in two dimensions

𝐹𝑥 = 0

𝐹𝑦 = 0

𝑀𝑧 = 0

- Note

𝐹 𝑑𝑖𝑟𝑒𝑐𝑡𝑖𝑜𝑛, 𝑠𝑒𝑛𝑠𝑒𝑚𝑎𝑔𝑛𝑖𝑡𝑢𝑑𝑒

known value −

unknown value ?



§2. Free-Body Diagram

- A Free-Body Diagram: a sketch of the isolated or free body

which shows all the pertinent weight forces, the externally

applied loads, and the reaction from its supports and

connections acting upon it by the removed elements

- General rules for support reactions

• If a support prevents the translation of a body in a given

direction, then a force is developed on the body in that

direction

• If rotation is prevented, a couple moment is exerted on the

body




- Support Reactions

• Roller: prevents the beam from translating in the vertical

direction, the roller will only exert a force on the beam in this

direction



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- Support Reactions

• Roller: prevents the beam from translating in the vertical

direction, the roller will only exert a force on the beam in this

direction

𝐹 ⊥ 𝑏𝑒𝑎𝑚

?




• Pin: prevents translation of the beam in any direction




• Pin: prevents translation of the beam in any direction

𝐹 ??= 𝐹 𝑥

∥ 𝑂𝑥?

+ 𝐹 𝑦 ∥ 𝑂𝑦?




• Fixed Support: prevents both translation and rotation of the

beam

𝐹 ??= 𝐹 𝑥

∥ 𝑂𝑥?

+ 𝐹 𝑦 ∥ 𝑂𝑦?

𝑀 −?




- Supports for Rigid Bodies Subjected to 2D Force Systems

• Cable: the reaction is a tension force which acts away from

the member in the direction of the cable

𝐹 −?

• Weightless link: the reaction is a force which acts along the

axis of the link

𝐹 −?




• Roller: the reaction is a force which acts perpendicular to the

surface at the point of contact

𝐹 −?

• Roller or pin in confined smooth slot: the reaction is a force

which acts perpendicular to the slot

𝐹 −?



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• Rocker: the reaction is a force which acts perpendicular to

the surface at the point of contact

𝐹 −?

• Smooth contacting surface: the reaction is a force which acts

perpendicular to the surface at the point of contact

𝐹 −?




• Member pin connected to collar on smooth rod: the reaction

is a force which acts perpendicular to the rod

𝐹 −?

• Smooth pin or hinge: the reactions are two components of

force, or the magnitude and direction of the resultant force

𝐹 ??




• Member fixed connected to collar on smooth rod: the

reactions are the couple moment and the force which acts

perpendicular to the rod

𝐹 −? , 𝑀

−?

• Fixed support: the reactions are the couple moment and the

two force components, or the couple moment and the

magnitude and direction of the resultant force

𝐹 ??

, 𝑀 −?




- Some typical examples of actual supports




- Internal Forces




- Weight and the Center of Gravity

𝑊 = 𝑚𝑔

𝑚: mass, 𝑘𝑔

𝑔 : gravity acceleration, 𝑚/𝑠2



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- Idealized Models




- Example 5.1 Draw the free-body diagram of the uniform

beam. The beam has a mass of 100𝑘𝑔

Solution




- Example 5.2 Draw the free-body diagram of the foot lever.

The operator applies a vertical force

to the pedal so that the spring is

stretched 36𝑚𝑚. and the force in the

short link at 𝐵 is 90𝑁

Solution

Spring force

𝐹𝑠 = 36 × 3.5 = 126𝑁




- Example 5.3 Two smooth pipes, each having a mass of

300𝑘𝑔, are supported by the forked tines of the tractor. Draw

the free-body diagrams for each pipe and both pipes together

Solution




- Example 5.4 Draw the free-body diagram of the unloaded

platform that is suspended off the edge of the oil rig. The

platform has a mass of 200𝑘𝑔

Solution



Problems

- Prob.5.1 Draw the free-body diagram of the 50𝑘𝑔 paper roll

which has a center of mass at 𝐺 and rests on the smooth

blade of the paper hauler. Explain the significance of each

force acting on the diagram



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Problem

- Prob.5.2 Draw the free-body diagram of member 𝐴𝐵, which

is supported by a roller at 𝐴 and a pin at 𝐵. Explain the

significance of each force on the diagram



Problem

- Prob.5.3 Draw the free-body diagram of the dumpster 𝐷 of

the truck, which has a weight of 5000𝑁 and a center of gravity

at 𝐺 . It is supported by a pin at 𝐴 and a pin-connected

hydraulic cylinder 𝐵𝐶 (short link). Explain the significance of

each force on the diagram



Problem

- Prob.5.4 Draw the free-body diagram of the beam which

supports the 80𝑘𝑔 load and is supported by the pin at 𝐴 and a

cable which wraps around the pulley at 𝐷 . Explain the

significance of each force on the diagram



Problem

- Prob.5.5 Draw the free-body diagram of the truss that is

supported by the cable 𝐴𝐵 and pin 𝐶. Explain the significance

of each force acting on the diagram



Problem

- Prob.5.6 Draw the free-body diagram of the bar, which has a

negligible thickness and smooth points of contact at 𝐴, 𝐵, and

𝐶. Explain the significance of each force on the diagram



§3. Equations of Equilibrium

- The conditions for equilibrium in two dimensions

∑𝐹𝑥 = 0

⟹ ∑𝐹𝑦 = 0

∑𝑀𝑂 = 0

- Alternative sets of equilibrium equations

• The first alternative set

∑𝐹𝑥 = 0

∑𝑀𝐴 = 0

∑𝑀𝐵 = 0

• The second alternative set

∑𝑀𝐴 = 0

∑𝑀𝐵 = 0

∑𝑀𝐶 = 0



∑𝐹 = 0

∑𝑀 = 0

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- Example 5.5 Determine the horizontal and vertical

components of reaction on the beam caused by the pin at 𝐵

and the rocker. Neglect the weight of the beam

Solution

Free-body Diagram

Equations of Equilibrium

+→ ∑𝐹𝑥: 600𝑐𝑜𝑠450 − 𝐵𝑥 = 0

+ ↑ ∑𝐹𝑦: 𝐴𝑦 − 600𝑠𝑖𝑛450

−100 + 𝐵𝑦 = 0

+↺ ∑𝑀𝐵: −7𝐴𝑦 +600𝑠𝑖𝑛450 ×5

−600𝑐𝑜𝑠450 ×0.2

+100×2 = 0

⟹𝐴𝑦 = 319𝑁, 𝐵𝑥 = 424𝑁, 𝐵𝑦 = 405𝑁




- Example 5.6 The cord supports a force of 100𝑁 and wraps

over the frictionless pulley. Determine the tension in the cord at

𝐶 and the horizontal and vertical components of reaction at pin 𝐴

Solution

Free-body Diagram


+→ ∑𝐹𝑥: −𝐴𝑥 + 𝑇𝑠𝑖𝑛300 = 0

+ ↑ ∑𝐹𝑦: 𝐴𝑦−𝑇𝑐𝑜𝑠450−100=0

+↺ ∑𝑀𝐴: 100×0.5−𝑇×0.5= 0

⟹𝑇 = 100𝑁, 𝐴𝑥 = 50𝑁, 𝐴𝑦 = 187𝑁

Note: The tension remains

constant as the cord passes

over the pulley




- Example 5.7 The member is pin-connected at 𝐴 and rests

against a smooth support at 𝐵. Determine the horizontal and

vertical components of reaction at the pin 𝐴

Solution

Free-body Diagram


+→ ∑𝐹𝑥: 𝐴𝑥 −𝑁𝐵𝑠𝑖𝑛300 = 0

+ ↑ ∑𝐹𝑦: 𝐴𝑦 −𝑁𝐵𝑐𝑜𝑠300 − 60 = 0

+↺ ∑𝑀𝐴: 𝑁𝐵×0.75−60×1−90=0

⟹𝑁𝐵 = 200𝑁

𝐴𝑥 = 100𝑁

𝐴𝑦 = 233𝑁




- Example 5.8 The box wrench is used to tighten the bolt at 𝐴.

If the wrench does not turn when the load is applied to the

handle, determine the torque or moment applied to the bolt

and the force of the wrench on the bolt

Solution

Free-body Diagram


+→ ∑𝐹𝑥: 𝐴𝑥 − 525

13+ 30𝑐𝑜𝑠300 = 0

+ ↑ ∑𝐹𝑦: 𝐴𝑦 − 5212

13− 30𝑠𝑖𝑛600 = 0

+↺ ∑𝑀𝐴: 𝑀𝐴 − 5212

13× 0.3 −

30𝑠𝑖𝑛600 × 0.7 = 0

⟹ 𝑀𝐴 = 32.6𝑁𝑚, 𝐴𝑥 = 5𝑁, 𝐴𝑦 = 74𝑁




- Example 5.9 Determine the horizontal and vertical components

of reaction on the member at the pin 𝐴 , and the normal

reaction at the roller 𝐵

Solution

Free-body Diagram


+→ ∑𝐹𝑥: 𝐴𝑥 −𝑁𝐵𝑠𝑖𝑛300 = 0

+ ↑ ∑𝐹𝑦: 𝐴𝑦 − 500 + 𝑁𝐵𝑐𝑜𝑠300 = 0

+↺ ∑𝑀𝐴: −500 × 3 + 𝑁𝐵𝑐𝑜𝑠300 × 6

−𝑁𝐵𝑠𝑖𝑛300 × 2 = 0

⟹ 𝑁𝐵 = 536𝑁

𝐴𝑥 = 268𝑁

𝐴𝑦 = 286𝑁




- Example 5.10 The uniform smooth rod is subjected to a force

and couple moment. If the rod is supported at 𝐴 by a smooth wall

and at 𝐵 and 𝐶 either at the top or bottom by rollers, determine

the reactions at these supports. Neglect the weight of the rod

Solution

Free-body Diagram


+→ ∑𝐹𝑥: 𝐶𝑦′𝑠𝑖𝑛300+𝐵𝑦′𝑠𝑖𝑛30

0−𝐴𝑥 = 0

+ ↑ ∑𝐹𝑦: −300+𝐶𝑦′𝑐𝑜𝑠300+𝐵𝑦′𝑐𝑜𝑠30

0 =0

+↺ ∑𝑀𝐴: −𝐵𝑦′ × 2 + 4000 − 𝐶𝑦′ × 6

+300𝑐𝑜𝑠300 × 8 = 0

⟹ 𝐵𝑦′ = −1000.0𝑁, 𝐶𝑦′ = 1346.4𝑁

𝐴𝑥 = 173𝑁



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- Example 5.11 The uniform truck ramp has a weight of 400𝑁

and is pinned to the body of the truck at each side and held in

the position shown by the two side cables. Determine the

tension in the cables

Solution

Free-body Diagram


∑𝑀𝐴 =−𝑇𝑠𝑖𝑛100×0.165+400×0.125𝑐𝑜𝑠300 =0⟹𝑇′ =𝑇/2=755.6𝑁




- Example 5.12 Determine the support reactions on the

member in the figure. The collar at 𝐴 is fixed to the member

and can slide vertically along the vertical shaft

Solution

Free-body Diagram


+→ ∑𝐹𝑥: 𝐴𝑥 = 0

+ ↑ ∑𝐹𝑦: 𝑁𝐵 − 900 = 0

+↺ ∑𝑀𝐴: 𝑀𝐴 − 500 − 900 × 1.5

+𝑁𝐵 × (1𝑐𝑜𝑠450 + 3) = 0

⟹ 𝐴𝑋 = 0

𝑁𝐵 = 900𝑁

𝑀𝐴 = −1.49𝑘𝑁𝑚 = 1.49𝑘𝑁 ↻



§4. Two- and Three-Force Members

- The solutions to some equilibrium problems can be simplified by

recognizing members that are subjected to only two or three forces

- Two-Force Members

• Forces applied at only two points on the member

• Force equilibrium: 𝐹 𝐴 = −𝐹 𝐵

• Moment equilibrium: ∑𝑀𝐴 = 0 or ∑𝑀𝐵 = 0

⟹ 𝐹 𝐴 ↑↓ 𝐹 𝐵

|𝐹 𝐴| = |𝐹 𝐵|




- Three-Force Members

• A member is subjected to only three forces

• Moment equilibrium can be satisfied only if the three forces

form a concurrent or parallel force systems

• If the lines of action of 𝐹 1 and 𝐹 2 intersect at point 𝑂, then the

line of action of 𝐹 3 must also pass through point 𝑂 so that the

forces satisfy: ∑𝑀𝑂 = 0

• If the three forces are all parallel, the location of the point of

intersection 𝑂 will approach infinity




- Example 5.13 The lever 𝐴𝐵𝐶 is pin supported at 𝐴 and

connected to a short link 𝐵𝐷. If the weight of the members is

negligible, determine the force of the pin on the lever at 𝐴

Solution

Free-body Diagram


𝜃 = 𝑡𝑎𝑛−1(0.7/0.4) = 60.30

+→ ∑𝐹𝑥: 𝐹𝐴𝑐𝑜𝑠𝜃 −𝐹𝑐𝑜𝑠450 +400 = 0

+ ↑ ∑𝐹𝑦: 𝐹𝐴𝑠𝑖𝑛𝜃 −𝐹𝑠𝑖𝑛450 = 0

⟹ 𝐹𝐴 = 1.07𝑘𝑁

𝐹 = 1.32𝑘𝑁



Fundamental Problems

- F5.1 Determine the horizontal and vertical components of

reaction at the supports. Neglect the thickness of the beam



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- F5.2 Determine the horizontal and vertical components of

reaction at the pin 𝐴 and the reaction on the beam at 𝐶




- F5.3 The truss is supported by a pin at 𝐴 and a roller at 𝐵.

Determine the support reactions




- F5.4 Determine the components of reaction at the fixed

support 𝐴. Neglect the thickness of the beam




- F5.5 The 25𝑘𝑔 bar has a center of mass at 𝐺 . If it is

supported by a smooth peg at 𝐶, a roller at 𝐴, and cord 𝐴𝐵,

determine the reactions at these supports




- F5.6 Determine the reactions at the smooth contact points 𝐴,

𝐵, and 𝐶 on the bar



§5. Free-Body Diagrams (3D)

- Supports for Rigid Bodies Subjected to 3D Force Systems

• Cable: the reaction is a force which acts away from the

member in the known direction of the cable

𝐹 −?

• Smooth surface support: the reaction is a force which acts

perpendicular to the surface at the point of contact

𝐹 −?



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• Roller: the reaction is a force which acts perpendicular to the

surface at the point of contact

𝐹 −?

• Ball and socket: the reactions are three rectangular force

components

𝐹 = 𝐹 𝑥 −? + 𝐹 𝑦

−? + 𝐹 𝑧

−?




• Single journal bearing: the reactions are two force and two

couple-moment components which act perpendicular to the

shaft

𝐹 = 𝐹 𝑥 −? + 𝐹 𝑧

−?

𝑀 = 𝑀𝑥 −? +𝑀𝑧

−?

• Single journal bearing with square shaft: the reactions are

two force and three couple-moment components

𝐹 = 𝐹 𝑥 −? + 𝐹 𝑧

−?

𝑀 = 𝑀𝑥 −? +𝑀𝑦

−? +𝑀𝑧

−?




• Single thrust bearing: the reactions are three force and two

couple-moment components

𝐹 = 𝐹 𝑥 −? + 𝐹 𝑦

−? + 𝐹 𝑧

−?

𝑀 = 𝑀𝑥 −? +𝑀𝑧

−?

• Single smooth pin: the reactions are three force and two


𝐹 = 𝐹 𝑥 −? + 𝐹 𝑦

−? + 𝐹 𝑧

−?

𝑀 = 𝑀𝑦 −? +𝑀𝑧

−?




• Single hinge: The reactions are three force and two couple-

moment components

𝐹 = 𝐹 𝑥 −? + 𝐹 𝑦

−? + 𝐹 𝑧

−?

𝑀 = 𝑀𝑥 −? +𝑀𝑧

−?

• Fixed support: the reactions are three force and three


𝐹 = 𝐹 𝑥 −? + 𝐹 𝑦

−? + 𝐹 𝑧

−?

𝑀 = 𝑀𝑦 −? +𝑀𝑧

−?




- Some typical examples of actual supports

• Free-body Diagrams



ball-and-socket joint journal bearing thrust bearing pin


- Example 5.14 Consider the two rods and plate, along with

their associated free-body diagrams. The 𝑥 ,𝑦 ,𝑧 axes are

established on the diagram and the unknown reaction

components are indicated in the positive sense. The weight is

neglected

Solution



Properly aligned journal bearings at 𝐴, 𝐵, 𝐶

The force reactions developed by the bearings are sufficient for equilibrium since they prevent the shaft from rotating about each of the coordinate axes

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Pin at 𝐴 and cable 𝐵𝐶 Moment components are developed by the pin on the rod to prevent rotation about the 𝑥 and 𝑧 axes

Only force reactions are developed by the bearing and hinge on the plate to prevent rotation about each coordinate axis. No moments at the hinge are developed

Properly aligned journal bearing at 𝐴 and hinge at 𝐶. Roller at 𝐵

§6. Equations of Equilibrium (3D)

- Vector Equations of Equilibrium

∑𝐹 = 0

∑𝑀𝑂 = 0

- Scalar Equations of Equilibrium

∑𝐹 = ∑𝐹𝑥𝑖 + ∑𝐹𝑦𝑗 + ∑𝐹𝑧𝑘 = 0

∑𝑀𝑂 = ∑𝑀𝑥𝑖 + ∑𝑀𝑦𝑗 + ∑𝑀𝑧𝑘 = 0

or

∑𝐹𝑥 = 0, ∑𝐹𝑦 = 0, ∑𝐹𝑧 = 0

∑𝑀𝑥 = 0, ∑𝑀𝑦 = 0, ∑𝑀𝑧 = 0



§7. Constrains and Statical Determinacy

- To ensure the equilibrium of a rigid body, it is not only

necessary to satisfy the equations of equilibrium, but the body

must also be properly held or constrained by its supports

- Redundant constraints: when a body has redundant supports,

that is, more supports than are necessary to hold it in

equilibrium, it becomes statically indeterminate

- Statically indeterminate: there will be more unknown loadings

on the body than equations of equilibrium available for their

solutions




- Example: the beam is shown together with its free-body

diagram

The beam is statically indeterminate because of additional (or

redundant) supports reactions

There are five unknown 𝑀𝐴, 𝐴𝑥, 𝐴𝑦, 𝐵𝑦, 𝐶𝑦 for which only three

equilibrium equations can be written

∑𝐹𝑥 = 0, ∑𝐹𝑦 = 0, ∑𝑀𝑂 = 0




- Example: the pipe is also statically indeterminate because of

additional (or redundant) supports reactions

The pipe assembly has eight unknowns, for which only six

equilibrium equations can be written




Statically indeterminate

the number of unknown reactive forces > the number of the

derived static equilibrium equations

How to solve ?

- The additional equations needed to solve statically

indeterminate problems are generally obtained from the

deformation conditions at the points of supports

- This is done in courses dealing with “Mechanics of Materials”



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- Improper constraints: having the same number of unknown

reactive forces as available equations of equilibrium does not

always guarantee that a body will be stable when subjected to

a particular loading

- For example, the pin support at 𝐴 and the roller support at 𝐵

for the beam are placed in such away that the lines of action

the reactive forces are concurrent at point 𝐴

- Consequently, the applied loading 𝑃 will cause the beam to

rotates lightly about 𝐴 , and so the beam is improperly

constrained




- In three dimensions, a body will be improperly constrained if

the lines of action of all the reactive forces intersect a common

axis

- For example, the reactive forces at the ball-and-socket

supports at 𝐴 and 𝐵 all intersect the axis passing through 𝐴

and 𝐵

- Note: Since the moments of these forces about 𝐴 and 𝐵 are all

zero, then the loading 𝑃 will rotate the member about the 𝐴𝐵

axis




- Another way in which improper constraining leads to instability

occurs when the reactive forces are all parallel

- Note: the summation of forces along the 𝑥 axis will not be

equal zero




- Example 5.15 The homogeneous plate has a mass of 100𝑘𝑔

and is subjected to a force and

couple moment along its edges. If it

is supported in the horizontal plane

by a roller at 𝐴 , a ball-and-socket

joint at 𝐵, and a cord at 𝐶, determine

the components of reaction at these

supports

Solution

Free-body Diagram





∑𝐹𝑥 = 0: 𝐵𝑥 = 0

∑𝐹𝑦 = 0: 𝐵𝑦 =0

∑𝐹𝑧 = 0: 𝐴𝑧 +𝐵𝑧 +𝑇𝐶 −300−981= 0

∑𝑀𝑥 = 0: 𝑇𝐶 ×2+981×1+𝐵𝑧 ×2= 0

∑𝑀𝑦 = 0: 300×1.5+981×1.5

−𝐵𝑧 ×3−𝐴𝑧 ×3−200= 0

⟹ 𝐴𝑧 = 790𝑁

𝐵𝑥 = 0

𝐵𝑦 = 0

𝐵𝑧 = −217𝑁

𝑇𝐶 = 707𝑁




- Example 5.16 Determine the components of reaction that the

ball-and-socket joint at 𝐴, the smooth journal bearing at 𝐵, and

the roller support at 𝐶 exert on the rod assembly

Solution

Free-body Diagram


∑𝐹𝑥 = 0: 𝐴𝑥 + 𝐵𝑥 = 0

∑𝐹𝑦 = 0: 𝐴𝑦 = 0

∑𝐹𝑧 = 0: 𝐴𝑧 − 900 + 𝐵𝑧 + 𝐹𝐶 = 0

∑𝑀𝑥 = 0: −900×0.4+𝐵𝑧×0.8+𝐹𝐶×1.2=0

∑𝑀𝑦 = 0: −900 × 0.4 + 𝐹𝐶 × 0.6 = 0

∑𝑀𝑧 = 0: 𝐵𝑥 × 0.8 = 0

⟹ 𝐴𝑦 = 0, 𝐴𝑥 = 0, 𝐴𝑧 = 750𝑁, 𝐵𝑥 = 0, 𝐵𝑧 = −450𝑁, 𝐹𝐶 = 600𝑁



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- Example 5.17 The boom is used to support the 75𝑁 flowerpot.

Determine the tension developed in

wires 𝐴𝐵 and 𝐴𝐶

Solution

Free-body Diagram


𝐹 𝐴𝐵 = 𝐹𝐴𝐵𝑟 𝐴𝐵𝑟𝐴𝐵

= 𝐹𝐴𝐵2𝑖 − 6𝑗 + 3𝑘

22 + (−6)2+32

=2

7𝐹𝐴𝐵𝑖 −

6

7𝐹𝐴𝐵𝑗 +

3

7𝐹𝐴𝐵𝑘

𝐹 𝐴𝐶 = 𝐹𝐴𝐶𝑟 𝐴𝐶𝑟𝐴𝐶

= 𝐹𝐴𝐶−2𝑖 − 6𝑗 + 3𝑘

(−2)2+(−6)2+32

= −2

7𝐹𝐴𝐶𝑖 −

6

7𝐹𝐴𝐶𝑗 +

3

7𝐹𝐴𝐶𝑘




𝐹 𝐴𝐵 =2


6

7𝐹𝐴𝐵𝑗 +

3

7𝐹𝐴𝐵𝑘

𝐹 𝐴𝐶 = −2


6

7𝐹𝐴𝐶𝑗 +

3

7𝐹𝐴𝐶𝑘

𝑊 = −75𝑘

∑𝑀𝑂 = 0: 𝑟 𝐴 × 𝐹 𝐴𝐵 + 𝐹 𝐴𝐶 +𝑊 = 0

⟹6𝑗 × 2


6

7𝐹𝐴𝐵𝑗 +

3

7𝐹𝐴𝐵𝑘 +

−2


6

7𝐹𝐴𝐶𝑗 +

3

7𝐹𝐴𝐶𝑘 −75𝑘 = 0

⟹18

7𝐹𝐴𝐵 +

18

7𝐹𝐴𝐶 −450 𝑖

+ −12

7𝐹𝐴𝐵+

12

7𝐹𝐴𝐶 𝑘 = 0




18

7𝐹𝐴𝐵 +

18

7𝐹𝐴𝐶 − 450 𝑖

+ −12

7𝐹𝐴𝐵 +

12

7𝐹𝐴𝐶 𝑘 = 0

⟹ ∑𝑀𝑥 = 0: 18

7𝐹𝐴𝐵 +

18

7𝐹𝐴𝐶 − 450 = 0

∑𝑀𝑦 = 0: 0 = 0

∑𝑀𝑧 = 0: −12

7𝐹𝐴𝐵 +

12

7𝐹𝐴𝐶 = 0

⟹ 𝐹𝐴𝐵 = 87.5𝑁

𝐹𝐴𝐶 = 87.5𝑁




- Example 5.18 Rod 𝐴𝐵 is subjected to the 200𝑁 force.

Determine the reactions at the ball-and-socket joint 𝐴 and the

tension in the cables 𝐵𝐷 and 𝐵𝐸

Solution

Free-body Diagram


𝐹 𝐴 = 𝐴𝑥𝑖 + 𝐴𝑦𝑗 + 𝐴𝑧𝑘

𝑇𝐸 = 𝑇𝐸𝑖

𝑇𝐷 = 𝑇𝐷𝑗

𝐹 = −200𝑘




Applying the force equation of equilibrium

∑𝐹 = 0: 𝐹 𝐴 + 𝑇𝐸 + 𝑇𝐷 + 𝐹 = 0

∑𝑀𝐴 = 0: 𝑟 𝐶 × 𝐹 + 𝑟 𝐵 × 𝑇𝐸 + 𝑇𝐷 = 0

⟹ 𝐴𝑥 + 𝑇𝐸 𝑖 + 𝐴𝑦 + 𝑇𝐷 𝑗

+ 𝐴𝑧 − 200 𝑘 = 0

0.5𝑖 + 𝑗 − 𝑘 × −200𝑘

+ 𝑖 +2𝑗 −2𝑘 × 𝑇𝐸𝑖 +𝑇𝐷𝑗 = 0

⟹ 𝐴𝑥 + 𝑇𝐸 𝑖 + 𝐴𝑦 + 𝑇𝐷 𝑗

+ 𝐴𝑧 − 200 𝑘 = 0

2𝑇𝐷 − 200 𝑖 + −2𝑇𝐸 + 100 𝑗

+ 𝑇𝐷 −2𝑇𝐸 𝑘 = 0

⟹𝑇𝐷 = 100𝑁, 𝑇𝐸 = 50𝑁, 𝐴𝑥 =−50𝑁, 𝐴𝑦 =−100𝑁, 𝐴𝑧 = 200𝑁




- Example 5.19 The bent rod is supported at 𝐴 by a journal

bearing, at 𝐷 by a ball-and-socket joint, and

at 𝐵 by means of cable 𝐵𝐶. Using only one

equilibrium equation, obtain a direct solution

for the tension in cable 𝐵𝐶. The bearing at 𝐴

is capable of exerting force components

only in the 𝑧 and 𝑦 directions since it is

properly aligned on the shaft

Solution

Free-body Diagram



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The cable tension may be obtained directly

by summing moments about an axis that

passes through points 𝐷 and 𝐴

𝑢 =𝑟 𝐷𝐴𝑟𝐷𝐴

= −1

2𝑖 −

1

2𝑗 = −0.7071(𝑖 + 𝑗 )

The sum of the moments about this axis is zero

∑𝑀𝐷𝐴 = 𝑢∑ 𝑟 × 𝐹 = 0

⟹ 𝑢 𝑟 𝐵 × 𝑇𝐵 + 𝑟 𝐸 ×𝑊 = 0

−0.7071(𝑖 + 𝑗 ) −𝑗 ×𝑇𝐵𝑘 −0.5𝑗 × −981𝑘 = 0

−0.7071(𝑖 + 𝑗 ) −𝑇𝐵 + 490.5 𝑖 = 0

−0.7071 −𝑇𝐵 + 490.5 + 0 + 0 = 0

⟹ 𝑇𝐵 = 490.5𝑁




- F5.7 The uniform plate has a weight of 500𝑁. Determine the

tension in each of the supporting cables




- F5.8 Determine the reactions at the roller support 𝐴, the ball-

and-socket joint 𝐷, and the tension in cable 𝐵𝐶 for the plate




- F5.9 The rod is supported by smooth journal bearings at 𝐴, 𝐵

and 𝐶 and is subjected to the two forces. Determine the

reactions at these supports




- F5.10 Determine the support reactions at the smooth journal

bearings 𝐴, 𝐵, and 𝐶 of the pipe assembly




- F5.11 Determine the force developed in cords 𝐵𝐷, 𝐶𝐸, and 𝐶𝐹

and the reactions of the ball-and-socket joint 𝐴 on the block



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- F5.12 Determine the components of reaction that the thrust

bearing 𝐴 and cable 𝐵𝐶 exert on the bar



Ch.05 Equilibrium of a Rigid Body

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Transcript of Ch.05 Equilibrium of a Rigid Body