Physics 170 Week 3, Lecture 2

http://www.phas.ubc.ca/∼gordonws/170

Physics 170 203 Week 3, Lecture 2 1

Text Chapter 4: Section 4.1-2

Physics 170 203 Week 3, Lecture 2 2

Learning Goals:

• Brief review of the condition for static equilibrium.

• Understand the concept of moment

• Learn how to calculate moment by taking a cross product ofvectors.

Physics 170 203 Week 3, Lecture 2 3

Review:

• The condition for equilibrium of a particle is∑

i

~Fi = 0

• If some components of ~Fi are not known, the equilibriumcondition can be used to find them. It contains three equations– therefore it can be used to find three unknown quantities.

• To solve a problem, one should begin by drawing a free bodydiagram.

• Analysis includes solving linear equations.

Physics 170 203 Week 3, Lecture 2 4

Rigid bodies: We will progress from considering statics of“particles”, where the extension of objects is not an importantfactor to considering the statics of extended rigid objects wheresome features of the shape of the objects is important.

We will study the effect of forces acting on rigid objects in staticsituations.

Physics 170 203 Week 3, Lecture 2 5

Moment of a force about a point is the tendency of a rigid bodyto rotate as a result of the force.

The tendency to rotate is always about a particular directionThe tendency to rotate depends on the direction and point ofaction of the force.

Physics 170 203 Week 3, Lecture 2 6

Moment of a force

The moment ~MO of a force ~F about a point O is a vector.

• The magnitude | ~MO| is the tendency of ~F to cause a system torotate about a base point O.

– The moment is larger if ~F is larger and it is smaller if |~F | issmaller.

– The moment is larger if ~F acts farther away from O and itis smaller if ~F acts closer to O.

• The direction of ~MO is the axis of the rotation that it wouldcause.

Physics 170 203 Week 3, Lecture 2 7

Moment of a force

To find the moment:

1. Find the magnitude |~F | of the force

2. Find the smallest distance d between the line of action of theforce and the center O

3. The magnitude of the moment is given by

| ~MO| = d|~F |

4. The direction of the moment is

(a) Perpendicular to the force vector

(b) Perpendicular to the displacement vector from O to thepoint of action of the force

(c) Determined by the right-hand-rule (displacement - force -moment)

Physics 170 203 Week 3, Lecture 2 8

Moment of a force

The moment ~M of a force ~F about a point O is equal to the crossproduct of the force ~F and the displacement, ~r, from O to theposition where the force acts,

~M = ~r × ~F

Physics 170 203 Week 3, Lecture 2 9

Cross Product

The cross product of two vectors ~u and ~v is denoted by ~u×~v. It is avector whose magnitude is given by the product of the magnitudesof the two vectors and the sine of the angle between them,

|~u× ~v| = |~u||~v| sin θ

Physics 170 203 Week 3, Lecture 2 10

Cross Product

The cross product of two vectors ~u and ~v is denoted by ~u×~v. It is avector whose magnitude is given by the product of the magnitudesof the two vectors and the sine of the angle between them,

|~u× ~v| = |~u||~v| sin θ

⊥ to ~u and ~v

Physics 170 203 Week 3, Lecture 2 11

Properties of cross product

• The direction of the cross product is determined by theright-hand-rule.

• The cross product is anti-symmetric

~r1 × ~r2 = −~r2 × ~r1

Direction is that of your thumb when you wrap your right-handfingers from ~r1 to ~r2.

• The cross product of parallel vectors is zero. θ = 0, sin θ = 0.

i× i = 0 , j × j = 0 , k × k = 0

• The cross product of perpendicular vectors is equal to theproduct of the magnitudes of the vectors. θ = 90, sin 90 = 1

i× j = k , j × k = i , k × i = j

Physics 170 203 Week 3, Lecture 2 12

Cross product in components

~r1 = x1i + y1j + z1k , ~r2 = x1i + y2j + z2k

~r1 × ~r2 = (y1z2 − y2z1)i + (z1x2 − z2x1)j + (x1y2 − x2y1)k

= det

i j k

x1 y2 z1

x2 y2 z2

When the determinant is evaluated using the method of minors andcofactors, it yields the above expression.

Physics 170 203 Week 3, Lecture 2 13

Minors and cofactors

The minors and cofactors technique begins with

det

i j k

x1 y2 z1

x2 y2 z2

= i det

[y1 z1

y2 z2

]+j det

[z1 x1

z2 x2

]+k det

[x1 y1

x2 y2

]

Then, we use the formula for the determinant of a 2× 2 matrix,

det[

y1 z1

y2 z2

]= y1z2 − z1y2 , det

[z1 x1

z2 x2

]= z1x2 − x1z2

det[

x1 y1

x2 y2

]= x1y2 − y1x2

This gives the formula for the cross product

~r1 × ~r2 = i(y1z2 − z1y2) + j(z1x2 − x1z2) + k(x1y2 − y1x2)

Physics 170 203 Week 3, Lecture 2 14

To calculate the moment of a force ~F acting on an object atpoint ~r about a point O with coordinates vector ~rO we must findthe cross product

~M = (~r − ~rO)× ~F

Physics 170 203 Week 3, Lecture 2 15

Vector addition of moments:

If a number of forced ~F1, ~F2, ..., ~Fk act at different points ~r1, ~r2, ...,~rk on an object, the total moment of these forces about the pointwith position vector ~rO is given by the sum of the individualmoment vectors

~MO = ~MO1 + ~MO2... + ~MOk ≡k∑

i=1

~MOi

where~MOi = (~ri − ~rO)× ~Fi

Physics 170 203 Week 3, Lecture 2 16

Example

Physics 170 203 Week 3, Lecture 2 17

Example:

Find the projection of the moment due to the force ~F whosemagnitude is given and whose direction is that of the unit vectorn = 1√

3

(i− j + k

)about the point O in the x-z plane.

Physics 170 203 Week 3, Lecture 2 18

Strategy:

• Find a mathematical expression for the force vector as well asthe position vector of the point ~r where the force acts.

• Find the displacement vector ~d from O to ~r.

• Find the moment using ~MO = ~d× ~F .

• Find the y-component of ~MO.

Physics 170 203 Week 3, Lecture 2 19

Free Body Diagram

Physics 170 203 Week 3, Lecture 2 20

Force vector

~F = (20)n lb. =20√

3

(i− j + k

)lb.

Force acts at the point

~r =(6i + 14j + 10k

)in.

Displacement from O to point of action of force is

~d = ~r − (0, 0, 0) =(6i + 14j + 10k

)in.

Moment is the cross product

~MO = ~d× ~F =(6i + 14j + 10k

)× 20√

3

(i− j + k

)in.lb.

Physics 170 203 Week 3, Lecture 2 21

~MO = ~d× ~F =(6i + 14j + 10k

)× 20√

3

(i− j + k

)in.lbs.

We need to calculate the cross-product.

(6i + 14j + 10k

)× 20√

3

(i− j + k

)=

i j k

6 14 1020√3

− 20√3

20√3

=(

(14)(

20√3

)+ (10)

(20√

3

))i−

((6)

(20√

3

)− (10)

(20√

3

))j

+(

(6)(− 20√

3

)− (14)

(20√

3

))k

=20√

3

(24i + 4j − 20k

)

Physics 170 203 Week 3, Lecture 2 22

~MO = ~d× ~F =(6i + 14j + 10k

)× 20√

3

(i− j + k

)in.lb.

=20√

3

(24i + 4j − 20k

)in.lb.

Projection in x− z plane is the y-component:

~MOy = ~MO · j =(20)(4)√

3in.lb.

~MOy =(20)(4)12√

3ft.lb. = 3.85 ft.lb.

~MOy = 3.85 ft.lb.

Physics 170 203 Week 3, Lecture 2 23

For the next Lecture:

Read Chapter 4, Section 4.3-4

Electronic Assignment due 3pm tomorrow!

New Electronic Assignment available at 3pmtomorrow

Physics 170 203 Week 3, Lecture 2 24