PLANAR KINETIC EQUATIONS of MOTION: TRANSLATION Today’S Objectives: Students Will Be Able To: 1
PLANAR KINETIC EQUATIONS OF MOTION: TRANSLATION Today’s Objectives: Students will be able to: 1. Apply the three equations of In-Class Activities: motion for a rigid body in planar motion. • Applications 2. Analyze problems involving • FBD of Rigid Bodies translational motion. • EOM for Rigid Bodies • Translational Motion • Group Problem Solving
APPLICATIONS The boat and trailer undergo rectilinear motion. In order to find the reactions at the trailer wheels and the acceleration of the boat its center of mass, we need to draw the FBD for the boat and trailer.
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How many equations of motion do we need to solve this problem? What are they?
1 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?
EQUATIONS OF TRANSLATIONAL 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.
2 EQUATIONS OF TRANSLATIONAL MOTION (continued) • If a body undergoes translational motion, the equation of motion
is ΣF =maG . This can also be written in scalar form as
Σ Fx = m(aG)x and Σ Fy = m(aG)y • 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.
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EQUATIONS OF ROTATIONAL MOTION
We need to determine the effects caused by the moments of the external force system. The moment about point P can be written as
Σ (ri × Fi) + Σ Mi = rG × maG + IGα
Σ Mp = Σ( Mk )p
where 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.
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3 EQUATIONS OF ROTATIONAL MOTION (continued) If point P coincides with the mass center G, this equation reduces to the scalar equation of Σ MG = IGα . 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: Σ Fx = m(aG)x
Σ Fy = m(aG)y
and Σ MG = IGα or Σ Mp = Σ (Mk)p
EQUATIONS OF MOTION: TRANSLATION ONLY (Section 17.3) 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:
Σ Fx = m(aG)x
Σ Fy = m(aG)y
Σ MG = 0 Note that, if it makes the problem easier, the moment equation can be applied about other points instead of the mass center. In this case,
ΣMA = (m aG ) d .
4 EQUATIONS OF MOTION: TRANSLATION ONLY (continued) 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.
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 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. 4. Apply the three equations of motion:
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 5. Remember, friction forces always act on the body opposing the motion of the body.
5 EXAMPLE
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 Solution: (continued) 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 .
Applying the equations of motion:
Σ Fx = m(aG)x: 600 – 0.2 Nc = 50 aG Nc = 490 N
Σ Fy = m(aG)y: Nc – 490.5 = 0 Î x = 0.467 m 2 Σ MG = 0: -600(0.3) + Nc(x)-0.2 Nc (0.5) = 0 aG = 10.0 m/s
6 EXAMPLE (continued)
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.
CONCEPT QUIZ 1. A 2 lb disk is attached to a uniform 6 lb B rod AB with a frictionless collar at B. If the disk rolls without slipping, select the correct FBD. A
N N A) Nb B)b C) b 2 lb 2 lb 6 lb 8 lb 6 lb F Fs s
Na Na Na
7 CONCEPT QUIZ
2. A 2 lb disk is attached to a uniform 6 lb B rod AB with a frictionless collar at B. If the disk rolls with slipping, select the correct FBD. A
Nb N A) Nb B) C) b 2 lb 2 lb 6 lb 8 lb 6 lb Fk μ N μs Na k a N Na a Na
GROUP PROBLEM SOLVING
Given: A uniform connecting rod BC has a mass of 3 kg. The crank is rotating at a constant angular velocity of
ωAB = 5 rad/s. Find: The vertical forces on rod BC at points B and C when θ = 0 and 90 degrees. Plan: Follow the procedure for analysis.
8 GROUP PROBLEM SOLVING (continued) Solution: Rod BC will always remain horizontal while moving along a curvilinear path. The acceleration of its mass center G is the same as that of points B and C. 2 2 When θ = 0º, the FBD is: mrω = (3)(0.2)(5 )
2 Note that mrω is Bx the kinetic force Cx due to the body’s G acceleration Cy 350mm 350mm B (3)(9.81) N y Applying the equations of motion:
Σ MC = Σ (Mk)C Σ Fy = m(aG)y (0.7)By – (0.35)(3)(9.81) = -0.35(15) Cy + 7.215 – (3)(9.81) = -15 By = 7.215 N Cy = 7.215 N
GROUP PROBLEM SOLVING (continued) When θ = 90º, the FBD is:
mr ω 2 = (3)(0.2)(52) Bx Cx G
Cy 350 mm 350 mm B (3)(9.81) N y Applying the equations of motion:
Σ MC = Σ (Mk)C Σ Fy = m(aG)y (0.7)By – (0.35)(3)(9.81) = 0 Cy + 14.7 – (3)(9.81) = 0 By = 14.7 N Cy = 14.7 N
9 ATTENTION QUIZ
1. As the linkage rotates, box A A undergoes
A) general plane motion. 1.5 m
B) pure rotation. ω = 2 rad/s C) linear translation. D) curvilinear translation. 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
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