MOMENTS

Created by The North Carolina School of Science and Math. Copyright 2012. North Carolina Department of Public Instruction.

If our parrot isn’t sitting in the center, how do I calculate T1 and T2?

W

T2T1

We need another equation!

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Static Equilibrium

• If a system is in static equilibrium– Σ Fx = 0

– Σ Fy = 0

and …– Σ M = 0

• Just like with forces,moments havedirection.

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Moments• A moment of a force is a measure of

its tendency to cause a body to rotate about a point or axis.

• A moment (M) is calculated using the formula:

Moment = Force * Distance

M = F * D

Always use the perpendicular distance between the force and the point!

Moment = Force * Perpendicular Distance M = F * D

What distance would you use here?

L

F = 20 lb

L = 2 ft

Find the moment about point O:

M = F * D

M = (20 lb) * (2 ft)

M = 40 ft-lb clockwise

Moments

Moment = Force * Perpendicular Distance

M = F * D

Can you visualize the result of the force acting on the beam?

The beam has a tendency to rotate clockwise about point O.

Moments

Moment = Force * Distance

M = F * D

w

L

F = 20 lbf

L = 0.5 ft

Find the moment about point O:

M = F * D

M = (20 lb) * (0.5 ft)

M = 10 ft-lb clockwise

Moments

Moment = Force * Distance

M = F * D

Can you visualize the result of the force acting on the beam?

The beam has a tendency to rotate clockwise about point O.

Moments

Moment = Force * Perpendicular Distance M = F * D What distance would you use here?

L

Moments – Direct Method

d'

M = F * d’Use trigonometry to find d’

θ

1 2

θ d‘ = L cos(θ)

M = F * L cos(θ)

Moment = Force * Perpendicular Distance M = F * D Alternative Method …

L

Moments – Component Method

M = Fx * Dx + Fy * Dyθ

Let’s find the components

Fx = F sin(θ) Fy = F cos(θ)

Dx = 0 Dy = L

M = Fx * Dx + Fy * Dy

M = F sin(θ) * 0 + F cos(θ) * L

M = F * L cos(θ) (same as before)

Fx

Fy

Advantages of Component Method

• Use static equilibrium equations directly to find reaction forces.

• Find reaction forces for the beam in the prior slide.– Σ Fx = 0: Fpx – Fx = 0 -> Fpx = Fx = F sin(θ)

– Σ Fy = 0: Fpy + Fy = 0 -> Fpy = - F cos(θ)

– Σ M = 0: Mp + Fy L + Fx 0 = 0 -> Mp = - F L cos(θ)

L θ

Fx

Fy

𝐹𝑃𝑋

𝐹𝑃𝑌

𝑀𝑃

HOW TO SOLVE A MOMENT PROBLEM

• Draw a free body diagram that shows all forces.

• List the knowns and unknowns• Use the following three formulas to determine

unknowns (system is in static equilibrium)– Σ Fx = 0

– Σ Fy = 0

– Σ M = 0

Moment Problem, example

• Use static equilibrium equations directly to find reaction forces.

• Find reaction forces for the beam in the prior slide.– Σ Fx = 0: Fpx – Fx = 0 -> Fpx = Fx = F sin(θ)

– Σ Fy = 0: Fpy – Fy = 0 -> Fpy = - F cos(θ)

– Σ M = 0: Mp – Fy L + Fx 0 = 0 -> Mp = - F L cos(θ) (CCW)

L θ

Fx

Fy

𝐹𝑃𝑋

𝐹𝑃𝑌

𝑀𝑃

If our parrot is not sitting in the center, how do I calculate T1 and T2?

W

T2T1

Parrot on the Perch: Revisited

41

𝑾 −𝟓∗𝑻 𝟐=𝟎𝟓∗𝑻 𝟐=𝑾𝑻 𝟐=

𝑾𝟓

Σ M = 0 (about where?):

Σ Fy = 0

𝑻 𝟏+𝑻𝟐=𝑾 𝑻 𝟏=𝟒𝑾𝟓

• The sum of moments is zero around any point!• Choose any location, but the same for the whole problem.• Some places are easier than others.

W

T2T1

Where to take the moment

41

𝑻 𝟏∗𝟏−𝑻𝟐∗𝟒−𝟎∗𝑾=𝟎Σ M = 0 (this time about W)

Σ Fy = 0𝑻 𝟏+𝑻𝟐=𝑾

𝑻 𝟏−𝟒𝑻𝟐=𝟎𝑻 𝟏=𝟒𝑻𝟐

Solve:𝟒𝑻 𝟐+𝑻𝟐=𝑾 5

𝑻 𝟐=𝑾𝟓

Supports are translated into forces and moments in a free body diagrams. The following are three common supports and the forces and moments used to replace them.

Roller:

Pin Connection:

Fixed Support:

Fy

Fy

Fx

Fx

Fy

Mo

Review: Types of Supports

8kN 9kN

5kN compressed spring

ΣFX = 0, therefore RAx = 0

ΣFy = 0 = RAy + RB - 8kN + 5kN - 9 kN

RAy + RB = 12kN -> RAy = 12kN - RB

ΣMA = 0 = -8kN·2m + 5kN·4m + RB·5m - 9kN·7m

(16 - 20 + 63) kN-m = RB·5m

RB = 11.8 kN

Put into ΣFy result: RAy = 22kN-11.8 kN = 10.2 kN

RAx

RAy RB

8kN 5kN9kN

2m 2m 1m 2m

2m 2m 1m 2m

C

A

B30°

30°

3m

2m

3kN

The rod AC is hinged at point A and is suspended by a cable (BD) at Point B. A weight of 3kN hangs off the rod at Point C. Calculate the reaction forces at Point A and the tension in the cable BD.

D

C

A

B30°

30°

3m

2m

3kN

RAy

RAx

FBy

FBx

5 sin(30)=2.5m

3 sin(30)= 1.5m

3 cos(30)=2.6m

5 cos(30) = 4.3m

D

FreeBodyDiagram

3kN

RAy

RAx

FBy

FBx

5 sin(30)=2.5m

3 sin(30)= 1.5m

3 cos(30)=2.6m

5 cos(30) = 4.3m

FBy = Fb sin(30°) = 0.5 FB

FBx = Fbcos(30°) = 0.87 FB

ΣMA = 0 = -3kN·4.3m + FBy·2.6m + FBx·1.5m

12.9kN-m = 0.5 FB·2.6m + 0.87 FB·1.5m

FB = 4.95 kN

3kN

RAy

RAx

FBy

FBx

5 sin(30)=2.5m

3 sin(30)= 1.5m

3 cos(30)=2.6m

5 cos(30) = 4.3m

FB = 4.95 kN

FBy = Fb sin(30°) = 0.5 FB = 0.50*4.95 kN = 2.5 kN FBx = Fbcos(30°) = 0.87FB= 0.87*4.95 kN = 4.3 kN

ΣFx = 0 = RAx – FBx Rax = FBx = 4.3 kN ΣFy = 0 = RAy + FBy - 3kN

Ray = FBx = 3kN – Fby= 3kN – 2.5kN = 0.5 kN