Today’s Objectives:

Students will be able to:

1. Apply the three equations of

motion for a rigid body in

planar motion.

2. Analyze problems involving

translational motion.

In-Class Activities:

• Check Homework

• Reading Quiz

• Applications

• FBD of Rigid Bodies

• EOM for Rigid Bodies

• Translational Motion

• Concept Quiz

• Group Problem Solving

• Attention Quiz

PLANAR KINETIC EQUATIONS OF MOTION:

TRANSLATION

READING QUIZ

1. When a rigid body undergoes translational motion due to

external forces, the translational equations of motion (EOM)

can be expressed for _____________.

A) the center of rotation B) the center of mass

C) any arbitrary point D) All of the above

2. The rotational EOM about the mass center of the rigid body

indicates that the sum of moments due to the external loads

equals _____________.

A) IG B) m aG

C) IG + m aG D) None of the above.

APPLICATIONS

The boat and trailer undergo

rectilinear motion. In order to

find the reactions at the trailer

wheels and the acceleration of

the boat, we need to draw the

FBD and kinetic diagram for

the boat and trailer.

How many equations of motion do we need to solve this

problem? What are they?

APPLICATIONS (continued)

As the tractor raises the

load, the crate will undergo

curvilinear translation if the

forks do not rotate.

If the load is raised too quickly, will the crate slide to the left or

right?

How fast can we raise the load before the crate will slide?

PLANAR KINETIC EQUATIONS OF MOTION

(Section 17.2)

• We will limit our study of planar kinetics to rigid bodies that are

symmetric with respect to a fixed reference plane.

• As discussed in Chapter 16, when a body is subjected to general

plane motion, it undergoes a combination of translation and

rotation.

• First, a coordinate system with

its origin at an arbitrary point P

is established.

The x-y axes should not rotate

and can either be fixed or

translate with constant velocity.

• If a body undergoes translational motion, the equation of motion

is F = m aG . This can also be written in scalar form as

Fx = m(aG)x and Fy = m(aG)y

EQUATIONS OF TRANSLATIONAL MOTION

(continued)

• In words: the sum of all the

external forces acting on the

body is equal to the body’s mass

times the acceleration of it’s mass

center.

EQUATIONS OF ROTATIONAL MOTION

We need to determine the effects caused by the moments of

an external force system.

The moment about point P can be written as:

(ri Fi) + Mi = r maG + IG

Mp = (Mk)p

where r = x i + y j and Mp is the resultant moment about P

due to all the external forces.

The term (Mk)p is called the kinetic moment about point P.

If point P coincides with the mass center G, this equation reduces

to the scalar equation of MG = IG .

In words: the resultant (summation) moment about the mass

center due to all the external forces is equal to the moment of

inertia about G times the angular acceleration of the body.

Thus, three independent scalar equations of motion may be used

to describe the general planar motion of a rigid body. These

equations are:

EQUATIONS OF ROTATIONAL MOTION (continued)

Fx = m(aG)x

Fy = m(aG)y

and MG = IG or Mp = (Mk)p

When a rigid body undergoes only translation, all the particles of

the body have the same acceleration so aG = a and = 0.

The equations of motion become:

Note that, if it makes the problem easier, the moment equation

can be applied about another point instead of the mass center.

For example, if point A is chosen,

MA = (m aG ) d .

EQUATIONS OF MOTION: TRANSLATION

(Section 17.3)

Fx = m(aG)x

Fy = m(aG)y

MG = 0

When a rigid body is subjected to

curvilinear translation, it is best to use

an n-t coordinate system.

Then apply the equations of motion, as

written below, for n-t coordinates.

EQUATIONS OF MOTION: TRANSLATION

(continued)

Fn = m(aG)n

Ft = m(aG)t

MG = 0 or

MB = e[m(aG)t] – h[m(aG)n]

PROCEDURE FOR ANALYSIS

Problems involving kinetics of a rigid body in only translation

should be solved using the following procedure:

1. Establish an (x-y) or (n-t) inertial coordinate system and specify

the sense and direction of acceleration of the mass center, aG.

2. Draw a FBD and kinetic diagram showing all external forces,

couples and the inertia forces and couples.

3. Identify the unknowns.

5. Remember, friction forces always act on the body opposing the

motion of the body.

4. Apply the three equations of motion (one set or the other): Fx = m(aG)x Fy = m(aG)y Fn = m(aG)n Ft = m(aG)t MG = 0 or MP = (Mk)P MG = 0 or MP = (Mk)P

Given: A 50 kg crate rests on a

horizontal surface for

which the kinetic friction

coefficient k = 0.2.

Find: The acceleration of the

crate if P = 600 N.

Plan: Follow the procedure for analysis.

Note that the load P can cause the crate either to slide or to

tip over. Let’s assume that the crate slides.

We will check this assumption later.

EXAMPLE

The coordinate system and FBD

are as shown. The weight of

(50)(9.81) N is applied at the center

of mass and the normal force Nc

acts at O. Point O is some distance

x from the crate’s center line.

The unknowns are Nc, x, and aG .

EXAMPLE (continued)

Applying the equations of motion:

Solution:

Fx = m(aG)x: 600 – 0.2 Nc = 50 aG

Fy = m(aG)y: Nc – 490.5 = 0

MG = 0: -600(0.3) + Nc(x) – 0.2 Nc(0.5) = 0

Nc = 490 N

x = 0.467 m

aG = 10.0 m/s2

Since x = 0.467 m < 0.5 m, the crate slides as originally assumed.

If x was greater than 0.5 m, the problem would have to be

reworked with the assumption that tipping occurred.

EXAMPLE

(continued)

CONCEPT QUIZ

A

B 1. A 2 lb disk is attached to a uniform 6 lb

rod AB with a frictionless collar at B.

If the disk rolls without slipping, select

the correct FBD.

6 lb

2 lb

Na

Nb

Fs

8 lb

Na

Nb

6 lb

2 lb

Na

Nb A) B) C)

Fs

CONCEPT QUIZ

A

B 2. A 2 lb disk is attached to a uniform 6 lb

rod AB with a frictionless collar at B.

If the disk rolls with slipping, select the

correct FBD.

6 lb

2 lb

Na

Nb

k Na

8 lb

Na

Nb A) B) C)

6 lb

2 lb

Na

Nb

s Na Fk

GROUP PROBLEM SOLVING

Given: The handcart has a mass of

200 kg and center of mass at

G. A force of P=50 N is

applied to the handle.

Neglect the mass of the

wheels.

Find: The normal reactions at each of the two wheels at A and B.

Plan: Follow the procedure for analysis.

GROUP PROBLEM SOLVING (continued)

Solution: The cart will move along a rectilinear path.

Draw FBD and kinetic diagram.

=

x

y

Applying the equations of motion:

Fx = m(aG)x

50 cos 60 = 200 aG

aG = 0.125 m/s2

+

MG = 0

-(0.3)NA+(0.2)NB+0.3(50 cos 60) – 0.6(50 sin 60) = 0

− 0.3 NA+ 0.2 NB = 18.48 N m (2)

+↑ Fy = 0 NA + NB –1962 –50 sin 60 = 0

NA + NB = 2005 N (1)

GROUP PROBLEM SOLVING (continued)

=

x

y

Using Eqs. (1) and (2), solve for the reactions, NA and NB

NA = 765 N, NB = 1240 N

Applying the equations of motion:

ATTENTION QUIZ

2. The number of independent scalar equations of motion that

can be applied to box A is?

A) One B) Two

C) Three D) Four

1. As the linkage rotates, box A

undergoes ___________.

A) general plane motion

B) pure rotation

C) linear translation

D) curvilinear translation

A

1.5 m

= 2 rad/s