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 QUIZ1. 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 α B) m aG

C) IG α + 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 ΣF = m aG . This can also be written in scalar form as

Σ Fx = m(aG)x and Σ 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:

Σ (ri × Fi) + Σ Mi = r × maG + IGαΣ Mp = Σ(Mk)p

where r = x i + y j and Σ Mp is the resultant moment about P due to all the external forces. The term Σ(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 ΣMG = IG α .

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)

Σ Fx = m(aG)x

Σ Fy = m(aG)y

and Σ MG = IGα or Σ Mp = Σ (Mk)p

When a rigid body undergoes only translation, all the particles of the body have the same acceleration so aG = a and α = 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,

ΣMA = (m aG ) d .

EQUATIONS OF MOTION: TRANSLATION(Section 17.3)

Σ Fx = m(aG)x

Σ Fy = m(aG)y

Σ 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)

Σ Fn = m(aG)n

Σ Ft = m(aG)t

Σ MG = 0 or

Σ MB = e[m(aG)t] – h[m(aG)n]

PROCEDURE FOR ANALYSISProblems 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):

Σ Fx = m(aG)x Σ Fy = m(aG)y Σ Fn = m(aG)n Σ Ft = m(aG)t

Σ MG = 0 or Σ MP = Σ (Mk)P Σ MG = 0 or Σ MP = Σ (Mk)P

Given:A 50 kg crate rests on a horizontal surface for which the kinetic friction coefficient μk = 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 FBDare as shown. The weight of (50)(9.81) N is applied at the center of mass and the normal force Ncacts 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:

Σ Fx = m(aG)x: 600 – 0.2 Nc = 50 aG

Σ Fy = m(aG)y: Nc – 490.5 = 0Σ MG = 0: -600(0.3) + Nc(x) – 0.2 Nc(0.5) = 0

Nc = 490 Nx = 0.467 maG = 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

B1. 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 lb2 lb

Na

Nb

Fs

8 lb

Na

Nb

6 lb2 lb

Na

NbA) B) C)

Fs

CONCEPT QUIZ

A

B2. 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 lb2 lb

Na

Nb

μk Na

8 lb

Na

NbA) B) C)

6 lb2 lb

Na

Nb

μs NaFk

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:

←Σ Fx = m(aG)x50 cos 60° = 200 aGaG = 0.125 m/s2

+

ΣMG = 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)

+↑Σ Fy = 0 ⇒ NA + NB –1962 –50 sin 60° = 0NA + NB = 2005 N (1)

GROUP PROBLEM SOLVING (continued)

=x

y

Using Eqs. (1) and (2), solve for the reactions, NA and NBNA = 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 motionB) pure rotationC) linear translationD) curvilinear translation

A

1.5 m

ω = 2 rad/s