Equilibrium

FORCES ARE VECTORS

THEREFORE WE NEED

TO USE THE

TECHNIQUES OF

VECTOR ALGEBRA

Collinear Force Systems

EQUILIBRIUM EQUATIONS

CONDITIONS OF EQUILIBRIUM

Only ONE unknown (Force component) can be found

Collinear Force Systems

Example:

P

Determine the value of the force P so as to

satisfy the equilibrium?

F 0 -350+250-80+P=0 P=180 kN x

Collinear Force Systems

EQUILIBRIUM EQUATIONS

CONDITIONS OF EQUILIBRIUM

Two unknowns (Force components) can be found

2D Concurrent at a point Force System

Equilibrium condition equation of a particle subjected to concurrent forces in the x-y plane can be written as:

Fx i + Fy j = 0

Apparently, above vector equation implies that both algebraic sums ( in X and Y directions) should be equal to zero.

+→ Fx = 0 F1x + F2x + ….. = 0

+↑ Fy = 0 F1y + F2y + ….. = 0

0F

Only TWO unknowns can be found

2D Concurrent at a point Force System

0F

0 xF 0 yF

0 jFiF yx

and

Two Force component unknowns can

be found

2D Concurrent at a point Force System

• Resolve the given forces into i and j components and apply the equilibrium

+→ ∑Fx = 0 +↑ ∑Fy = 0 • Scalar equations of equilibrium require that the algebraic sum of the x and y components to equal to zero.

2D Concurrent at a point Force System

Determine the magnitudes of F1 and F2 for

equilibrium. Set θ=60°.

Example:

2D Concurrent at a point Force System

F1=1.827 kN F2=9.596 kN

Only TWO unknowns can be found

2D Concurrent at a point Force System

Coplanar Concurrent at a point Forces

2D Concurrent at a point Force System

Coplanar Concurrent at a point Forces

Exercise for

lecture 2:

Example

Determine the tension in

cables AB and AD for

equilibrium of the 250kg

engine.

2D Concurrent at a point Force System

Coplanar Concurrent at a point Forces

SINCE the mass of the engine is given i.e. unit is „kg‟ (scalar) and not the weight (FORCE)

the calculations should be corrected to a vector having a unit of Newton.

(mass * gravity )

2D Concurrent at a point Force System

Coplanar Concurrent at a point Forces

Procedure for Analysis 1. Free-Body Diagram

- Establish the x, y axes in any suitable orientation

- Label all the unknown and known forces magnitudes and directions

- Sense of the unknown force can be assummed

2D Concurrent at a point Force System

Coplanar Concurrent at a point Forces

Procedure for Analysis

2. Equations of Equilibrium

- Apply the equations of equilibrium

+→ ∑Fx = 0 +↑ ∑Fy = 0

- Components are positive if they are directed along the positive axis and negative, if directed along the negative axis

2D Concurrent at a point Force System

Coplanar Concurrent at a point Forces

2D Concurrent at a point Force System

Coplanar Concurrent at a point Forces

Solution

FBD at Point A - Initially, two forces acting, forces of cables AB and AD - Engine Weight [W=m.g] = (250kg)(9.81m/s2) = 2.452 kN supported by cable CA

- Finally, three forces acting, forces TB and TD and engine weight on cable CA

FBD of the ring A

2D Concurrent at a point Force System

Coplanar Concurrent at a point Forces

Solution

+→ ∑Fx = 0; TB cos30° - TD = 0

+↑ ∑Fy = 0; TB sin30° - 2.452 = 0

Solving,

TB = 4.904 kN

TD = 4.247 kN

*Note: Neglect the weights of the cables since they are small compared to the weight of the engine

FBD of the ring A

2D Concurrent at a point Force System

Coplanar Concurrent at a point Forces

Example

If the sack at A has a weight

of 20 N , determine

the weight of the sack at B

and the force in each cord

needed to hold the system in

the equilibrium position shown.

2D Concurrent at a point Force System

Coplanar Concurrent at a point Forces

Solution

TEC

2D Concurrent at a point Force System

Coplanar Concurrent at a point Forces

TEC

TEC

TEC

FBD of the ring E

FBD of the ring C

TEC

2D Concurrent at a point Force System

Coplanar Concurrent at a point Forces

Solution

FBD at Point E.

Three forces acting,

forces of cables EG

and EC and the weight

of the sack on cable EA

FBD of the ring E

2D Concurrent at a point Force System

Coplanar Concurrent at a point Forces

Solution

Use equilibrium at the ring to determine tension in

CD and weight of B with TEC known

+→ ∑Fx = 0; TEG sin30° - TECcos45° = 0

+↑ ∑Fy = 0; TEG cos30° - TECsin45° - 20 = 0

Solving,

TEC = 38.637 N

TEG = 54.641 N

2D Concurrent at a point Force System

Coplanar Concurrent at a point Forces

FBD of the ring E FBD of the ring C

2D Concurrent at a point Force System

Coplanar Concurrent at a point Forces

Solution

FBD at Point C

- Three forces acting, forces by cable CD

and EC (known) and

weight of sack B on

cable CB.

FBD of the ring C

2D Concurrent at a point Force System

Coplanar Concurrent at a point Forces

Solution

+→ ∑Fx = 0; 38.637cos45° - (4/5)TCD = 0

+↑ ∑Fy = 0; (3/5)TCD + 38.637sin45° – WB = 0

Solving,

TCD = 34.151 N

WB = 45.534 N

*Note: components of TCD are proportional to the slope of the cord by the 3-

4-5 triangle

2D Concurrent at a point Force System

Coplanar Concurrent at a point Forces

Example: The 50-kg homogenous smooth sphere rests on the 30°

incline A and bears against the smooth vertical wall B. Calculate the contact forces at A and B?

30° A

B

2D Concurrent at a point Force System

Coplanar Concurrent at a point Forces

FBD of the sphere

30° A

B

30°

A

B

30°

RA

RB

C

W

2D Concurrent at a point Force System

Coplanar Concurrent at a point Forces

Example:

y A A

+

x B B

W 50x9.81 490.5

F 0 F cos30 -490.5= 0 F 566.381 N

(assumed direction correct)

F 0 566.381sin30 -F 0 F 283.191 N

(assumed direction correct)

2D Concurrent at a point Force System

Coplanar Concurrent at a point Forces

Example:

N

EQUILIBRIUM EQUATIONS

CONDITIONS OF EQUILIBRIUM

3 unknowns

Three-D Force Systems Concurrent at a point

when a particle is subjected to concurrent forces in the x-y-z axes, its “equilibrium condition equation” can be written as:

Fx i + Fy j + Fz k = 0

the above vector equation implies that each of the “x”, the “y” and the “z” components should be equal to zero separately. Hence,

+→ Fx = 0 F1x + F2x + ….. = 0

+↑ Fy = 0 F1y + F2y + ….. = 0

+ Fz = 0 F1z + F2z + ….. = 0

∑ 0=F

Obviously THREE unknowns can be found

Three-D Force Systems Concurrent at a point

When a system of external 3D forces acts on a particle in equilibrium, we should have:

F = (Fx) i + (Fy) j + (Fz) k = 0

so each component of this equation must be determined separately:

Fx =0, Fy =0, Fz =0.

Three-D Force Systems Concurrent at a point

• Resolve the given forces into i, j and k components and apply the equilibrium condition

+→ ∑Fx = 0 +↑ ∑Fy = 0

+ ∑Fz = 0 • Equations of equilibrium require that the algebraic

sum of x, y and z components be equal to zero.

Three-D Force Systems Concurrent at a point

The 100-kg cylinder is suspended from the ceiling by cables attached at points B, C and D. What are the tensions in cables AB, AC & AD ?

Note that:

the gravity effect is in –ve y direction.

Example:

Three-D Force Systems Concurrent at a point

Solution Strategy:

• Isolate the part of the “cable system” near point A,

• Obtain a free-body diagram subjected to forces due to

the tensions in the cables.

• Because the sums of the external forces in the x, y, and

z directions must be IN BALANCE, obtain 3

INDEPENDENT equations for the three unknown cables

that are in tension.

• To do so, express the forces exerted by the tensions in

terms of their components.

Three-D Force Systems Concurrent at a point

Drawing the Free-Body Diagram and Applying the Equations

Three-D Force Systems Concurrent at a point

• Isolating the part of the cable system near point A and

show the forces exerted by the tensions in the cables.

The sum of the forces must equal zero:

F = TAB + TAC + TAD (981 N)j = 0 • Writing the Forces in Terms of their Components

• Obtain a unit vector that has the same direction as the

force TAB by dividing the position vector rAB from point

A to point B by its magnitude.

rAB = (xB xA)i + (yB yA)j + (zB zA)k

= 4i + 4j +2k (m)

Three-D Force Systems Concurrent at a point

Solution

kjir

re 333.0667.0667.0

AB

ABABλ AB

Three-D Force Systems Concurrent at a point

Expressing the force TAB in terms of its components by

writing it as the product of the tension TAB in cable AB

and the unit vector eAB...

TAB = TABeAB == TAB (0.667 i + 0. 667 j + 0.333 k)

Express the forces TAC and TAD in terms of their

components using the same procedure.

TAC = TAC (0.408 i + 0.816 j 0.408 k)

TAD = TAD (0.514 i + 0.686 j + 0.514k )

λAB =

λAB

Three-D Force Systems Concurrent at a point

Substituting these expressions into the equilibrium equation

TAB + TAC + TAD (981 N)j = 0

Because the i, j, and k components must each equal to

zero, this results in three equations of:

i-component: 0.667TAB 0.408TAC 0.514TAD = 0

j-component: 0.667TAB + 0.816TAC + 0.686TAD = 981

k-component: 0.333TAB 0.408TAC + 0.514TAD = 0

Solving the system of 3 equations three unknowns will yield:

TAB = 519 N

TAC = 636 N

TAD = 168 N

Three-D Force Systems Concurrent at a point