EquilibriumForces and Torques

9/11/07

Topics to Cover

• Components of forces and trigonometry

• Force examples

• Center of Mass

• Torques/Moments

• Torque Examples

• Equilibrium

• Terminology

http://archone.tamu.edu/rburt/cosc321/cosc321whole.htm

Force Components

x

y

yx

y

x

F

F

FFF

FF

FF

tan

sin

cos

22

soh sine (angle) = opposite/hypotenusecah cosine (angle) = adjacent/hypotenusetoa tangent (angle) = opposite/adjacent

Trigonometry

• Fx is negative

– 90° to 270°

• Fy is negative

– 180° to 360°• tan is positive

– quads I & III• tan is negative

– quads II & IV

Components of Force

• F = 10 lbs

• Fx = F * cos (30°)

• Fx = 10 * 0.87 = 8.7 lbs

• Fy = F * sin (30°)

• Fy = 10 * 0.5 = 5 lbs

10 lbs

30°

8.7 lbs

5 lbs

x

y

Think – Pair - Share

• F = 20 lbs

• Find Fx

• Fx = 20 * cos (60°) = 10 lbs

• Find Fy

• Fy = 20 * sin (60°) = 17 lbs

20 lbs

60°

10 lbs

17 lbs

x

y

Components of ForceSimilar Triangles

• Find Fx Find Fy

• Hypotenuse – Use pythagorean theorem– √(32 +42) = 5

lbsF

F

F

F

y

y

y

y

165

5

5

204

5204

20

54

**

**

20 lbs

3

4

x

y

lbsF

F

F

F

x

x

x

x

125

5

5

203

5203

20

53

**

**

Think - Pair - Share

• Find Fx Find Fy

• Hypotenuse– √(32 +72) = 7.6

lbsF

F

F

F

y

y

y

y

41867

67

67

207

67207

20

677

..

*.

.

*

*.*

.

20 lbs

3

7

x

y

lbsF

F

F

F

x

x

x

x

867

67

67

203

67203

20

673

.

*.

.

*

*.*

.

How can we tell if forces are balanced?

• The x and y components cancel

Do these forces balance?

• No No

20 lbs

3

7

x

y

30°

10 lbs

15 lbs

x

y

10 lbs

Do these forces balance?

• X components cancel• y components: • 10 * sin (30°) = 5 • (* 2 for each force)

• YES!10 lbs

x

y

10 lbs 10 lbs

30°30°

What is the balancing force?• 10 lb force

– Fx = 10 * cos (30°) = -8.7 lbs– Fy = 10 * sin (30°) = 5 lbs

• 20 lb force– Fx = 3*20/7.6 = 8 lbs– Fy = 7*20/7.6= 18.4 lbs

• Combine components– X: -8.7 + 8 = -0.7 lbs– Y: 5 + 18.4 = 23.4 lbs

• Negative of these components represent balancing force– X = 0.7 lbs, y = -23.4 lbs– Resultant = √(.72 + 23.42) = 23.4 lbs– Angle: tan-1(-23.4/0.7) = 88° in fourth quadrant

20 lbs

3

7

x

y

30°

10 lbs

88°

23 lbs

Force Example

• Find the MAGNITUDE, DIRECTION & SENSE of the 3 Coplanar Forces

– Find the hypotenuse of the 2 triangles using pythagorean theory.

Force Example – con’t

10 # Force

Fy = Sin 30º * 10Fy = 0.5 * 10Fy = + 5 #

Fx = Cos 60º * 10Fx = 0.866 * 10Fx = - 8.66 #

20 # ForceFy = (4/5) * 20Fy = 0.8 * 20Fy = 16 #

Fx = (3/5) * 20Fx = 0.6 * 20Fx = 12 #

30 # Force

Fy = (7/7.616) * 30Fy = 0.919 * 30Fy = - 27.57 #

Fx = (3/7.616) * 30Fx = 0.394 * 30Fx = 11.82 #

Force Example – con’t

Sum up all the components Ry = 5 + 16 - 27.57 Rx = -8.66 + 12 + 11.82Ry = -6.57 # Rx = 15.16 #

#52.1699.27216.1557.6 22 R

The Resultant Force (R) is found using Pythagorean theory.

The Direction of the Force (q ) is found using trigonometry

Force – Final Result• The Magnitude of the Resultant is 16.52 #

• The Direction is 23.41o to the horizontal (X) axis

• The Sense of the force is way from the origin

Lamp Post Forces Example

• Cable Tension = mg / sin • Pull of cable = push of strut• Cable is in Tension, the strut in Compression!

cable

strut

Lamp post

mg

tensioncable

mg

sin

Extended Objects – Center of Mass

• Draw the forces (the free-body diagram) on a stack of books. – How much does a book weigh? – What is diagram for the top book? For middle? For

bottom? • Note that the bottom of the structure must support the most

weight!

• A book applies a force over some area (distributed load). Why am I justified in drawing only one arrow?

• Center of Mass! top

middle

bottom

N

mg

Center of mass of triangular wedge

• Toward the thicker end - more mass on that side. – exact value is a third of the way from thicker end

• For an ideal, rigid object that does not change shape or

break, we can say that gravity just acts at one point, the center of mass (or center of gravity, or centroid).

Center of mass con’t

• Where can you hold a meter stick without causing a rotation? What if there is a weight on one end?

• How about a see-saw? How to balance?

Mechanical Equilibrium

• Mechanical equilibrium– Total translational and rotational forces equal

zero– No acceleration– Static equilibrium means not moving– Dynamic equilibrium means constant velocity,

i.e. not accelerating

Translational Equilibrium

• In order to be in translational equilibrium, the total forces of the object must be zero.

• Set the horizontal and vertical components of the forces to zero.

0 0 yx FF

Rotational EquilibriumTorque or Moment

• In order to be in rotational equilibrium, the total torques acting on an object must be zero.

dFMordF

Mor

0 0

Moments – rotation

• Defined by magnitude and direction• Units N*m or ft*lb• Direction – represented by arrows

– counterclockwise positive– clockwise negative

• Value found from F and perpendicular distance– M = F*d– d is also called “lever” or “moment” arm

Another Torque Example

• Same F

• Which torque is larger? Why?

Tools that use torque

• Torques allows us to move objects heavier than we can actually lift

• What if we needed 1000 pounds of force to raise a block? – None of us can just bend down and lift with that force– What tool would we use?

• Problem:– Say we dig a little hole, and slide our 2x4 under the

block, and then place a small wedge 6” away from the end (board is 6’, remember). How much can we exert now?

Pole Vaulter

The pole vaulter is holding pole so that it does not rotate and does not move. It is in equilibrium.

c.g.

2 ft.

FG= 5 lbsFR

FL

pivot

6 ft.

•Find the forces FL and FR.

Pole vaulter – con’t

0

net

GGLLnet

GLnet

FrFr

0

net

GRLnet

F

FFFFRotation = 0 Translation = 0

Pole vaulter – con’t

lbsF

lbftFft

L

L

GLnet

15

5620

lbsF

lbsFlbs

FFFF

R

R

GRLnet

10

5150

c.g.

2 ft.

FG= 5 lbs

FL=15 lbs

pivot

6 ft.FR=10 lbs

Torques or Moments - Rotation

• Torques are also arrows, just like forces– clockwise or counterclockwise

cable

strut

Lamp post

Activity 2 - Equilibrium

a) Class 1 b) Class 2

Terminology

• A metal column deforms when it supports weight, but it is so small you can’t see it.

• This deformation is called COMPRESSION if it pushes inwards and TENSION if it pulls out, and SHEAR if it makes different parts of the object go different ways.

Terminology - con’t“Loads”

• “Static loads” are approximately constant

• Some of these are “dead loads” and don’t vary at all – Weight of the structure

• “Live loads” vary, but only change very slowly– snow– number of people – Furnishings

Terminology - con’tDynamical loads

• Dynamic loads change rapidly– Wind and Earthquakes– Can be very dangerous to buildings

• Demo: Impact load. Weigh an object on a scale, show that dropping from equilibrium gives double the force of the settled value.

• Oscillations – Everything has a natural frequency it will prefer to

oscillate at. If driven near this frequency, the response is large.