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Home , Equations of motion, Force, Particle acceleration, Scalar (physics)

’S LAWS OF , OF MOTION, & FOR A SYSTEM OF Today’s Objectives: Students will be able to: 1. Write the of motion for an accelerating body. In-Class Activities: 2. Draw the free-body and kinetic • Check Homework diagrams for an accelerating • Applications body. • Newton’s Laws of Motion • Newton’s Law of Gravitational Attraction • Equation of Motion For A or System of Particles • Group Problem Solving

APPLICATIONS

The motion of an object depends on the acting on it.

A parachutist relies on the atmospheric resistance to limit his .

Knowing the drag force, how can we determine the or velocity of the parachutist at any point in ?

1 APPLICATIONS (continued)

A freight elevator is lifted using a motor attached to a cable and system as shown.

How can we determine the force in the cable required to the elevator at a given acceleration?

Is the tension force in the cable greater than the of the elevator and its load?

NEWTON’S LAWS OF MOTION (Section 13.1) The motion of a particle is governed by Newton’s three laws of motion. First Law: A particle originally at rest, or moving in a straight line at constant velocity, will remain in this state if the acting on the particle is zero. Law: If the resultant force on the particle is not zero, the particle experiences an acceleration in the same direction as the resultant force. This acceleration has a proportional to the resultant force. Third Law: Mutual forces of and between two particles are equal, opposite, and collinear.

2 NEWTON’S LAWS OF MOTION (continued) The first and third laws were used in developing the concepts of . Newton’s second law forms the of the study of .

Mathematically, Newton’s second law of motion can be written F = ma where F is the resultant unbalanced force acting on the particle, and a is the acceleration of the particle. The positive m is called the of the particle. Newton’s second law cannot be used when the particle’s approaches the , or if the of the particle is extremely small (~ size of an ).

NEWTON’S LAW OF GRAVITATIONAL ATTRACTION

Any two particles or bodies have a mutually attractive gravitational force acting between them. Newton postulated the law governing this gravitational force as

2 F = G(m1m2/r )

where F = force of attraction between the two bodies, G = universal constant of gravitation ,

m1, m2 = mass of each body, and r = between centers of the two bodies. When near the surface of the , the only gravitational force having any sizable magnitude is that between the earth and the body. This force is called the weight of the body.

3 MASS AND WEIGHT

It is important to understand the difference between the mass and weight of a body!

Mass is an absolute property of a body. It is independent of the gravitational in which it is measured. The mass provides a measure of the resistance of a body to a change in velocity, as defined by Newton’s second law of motion (m = F/a). The weight of a body is not absolute, since it depends on the in which it is measured. Weight is defined as W = mg where g is the acceleration due to .

UNITS: SI SYSTEM VS. FPS SYSTEM SI system: In the SI system of units, mass is a base unit and weight is a derived unit. Typically, mass is specified in (kg), and weight is calculated from W = mg. If the gravitational acceleration (g) is specified in units of m/s2, then the weight is expressed in newtons (N). On the earth’s surface, g can be taken as g = 9.81 m/s2. W (N) = m (kg) g (m/s2) => N = kg·m/s2

FPS System: In the FPS system of units, weight is a base unit and mass is a derived unit. Weight is typically specified in pounds (lb), and mass is calculated from m = W/g. If g is specified in units of ft/s2, then the mass is expressed in slugs. On the earth’s surface, g is approximately 32.2 ft/s2. m (slugs) = W (lb)/g (ft/s2) => = lb·s2/ft

4 EQUATION OF MOTION (Section 13.2) The motion of a particle is governed by Newton’s second law, relating the unbalanced forces on a particle to its acceleration. If more than one force acts on the particle, the equation of motion can be written

∑F = FR = ma where FR is the resultant force, which is a vector summation of all the forces. To illustrate the equation, consider a particle acted on by two forces. First, draw the particle’s free- body diagram, showing all forces acting on the particle. Next, draw the kinetic diagram, showing the inertial force ma acting in the same direction as

the resultant force FR.

INERTIAL

This equation of motion is only valid if the acceleration is measured in a Newtonian or inertial frame of reference. What does this mean? For problems concerned with at or near the earth’s surface, we typically assume our “inertial frame” to be fixed to the earth. We neglect any acceleration effects from the earth’s .

For problems involving satellites or , the inertial frame of reference is often fixed to the stars.

5 EQUATION OF MOTION FOR A SYSTEM OF PARTICLES (Section 13.3) The equation of motion can be extended to include systems of particles. This includes the motion of solids, liquids, or gas systems. As in statics, there are internal forces and external forces acting on the system. What is the difference between them?

Using the definitions of m = ∑mi as the total mass of all particles and aG as the acceleration of the G of

the particles, then maG = ∑miai .

The text shows the details, but for a system of particles: ∑F = maG where ∑F is the sum of the external forces acting on the entire system.

KEY POINTS 1) Newton’s second law is a “Law of Nature”--experimentally proven and not the result of an analytical proof.

2) Mass (property of an object) is a measure of the resistance to a change in velocity of the object.

3) Weight (a force) depends on the local gravitational field. Calculating the weight of an object is an application of F = ma, i.e., W = m g.

4) Unbalanced forces cause the acceleration of objects. This condition is fundamental to all dynamics problems!

6 PROCEDURE FOR THE APPLICATION OF THE EQUATION OF MOTION 1) Select a convenient inertial . Rectangular, /tangential, or cylindrical coordinates may be used.

2) Draw a free-body diagram showing all external forces applied to the particle. Resolve forces into their appropriate components. 3) Draw the kinetic diagram, showing the particle’s inertial force, ma. Resolve this vector into its appropriate components. 4) Apply the equations of motion in their scalar component form and solve these equations for the unknowns. 5) It may be necessary to apply the proper kinematic relations to generate additional equations.

EXAMPLE

Given: A crate of mass m is pulled by a cable attached to a truck. The coefficient of kinetic between the crate and

road is μk. Find: Draw the free-body and kinetic diagrams of the crate. Plan: 1) Define an inertial coordinate system. 2) Draw the crate’s free-body diagram, showing all external forces applied to the crate in the proper directions. 3) Draw the crate’s kinetic diagram, showing the inertial force vector ma in the proper direction.

7 EXAMPLE : (continued) 1) An inertial x-y frame can be defined as fixed to the ground. 2) Draw the free-body diagram of the crate: y W = mg T The weight force (W) acts through the crate’s center of mass. T is the tension x 30° force in the cable. The (N) is perpendicular to the surface. The friction force (F = u N) acts in a direction F = uKN K N opposite to the motion of the crate. 3) Draw the kinetic diagram of the crate: The crate will be pulled to the right. The ma acceleration vector can be directed to the right if the truck is speeding up or to the left if it is slowing down.

GROUP PROBLEM SOLVING

Given: Each block has a mass m. The coefficient of kinetic friction at all surfaces of contact is μ. A horizontal force P is applied to the bottom block.

Find: Draw the free-body and kinetic diagrams of each block.

Plan: 1) Define an inertial coordinate system. 2) Draw the free-body diagrams for each block, showing all external forces. 3) Draw the kinetic diagrams for each block, showing the inertial forces.

8 GROUP PROBLEM SOLVING Solution: (continued) 1) An inertial x-y frame can be defined as fixed to the ground. 2) Draw the free-body diagram of each block: Block A: y Block B: y W = mg N A WB = mg B x x FfB = μNB T P

FfA = μNA FfB = μNB N B NA The friction forces oppose the motion of each block relative to the surfaces on which they slide. 3) Draw the kinetic diagram of each block: Block B: Block A: ma maB = 0 A

EQUATIONS OF MOTION: RECTANGULAR COORDINATES

Today’s Objectives: Students will be able to: 1. Apply Newton’s second law In-Class Activities: to determine forces and for particles in • Applications rectilinear motion. • Equations of Motion Using Rectangular (Cartesian) Coordinates • Group Problem Solving

9 APPLICATIONS

If a man is pushing a 100 lb crate, how large a force F must he exert to start moving the crate? What would you have to know before you could calculate the answer?

APPLICATIONS (continued)

Objects that move in any have a drag force acting on them. This drag force is a of velocity.

If the ship has an initial velocity vo and the magnitude of the opposing drag force at any instant is half the velocity, how long it would take for the ship to come to a stop if its stop?

10 RECTANGULAR COORDINATES (Section 13.4) The equation of motion, F = m a, is best used when the problem requires finding forces (especially forces perpendicular to the path), accelerations, or mass. Remember, unbalanced forces cause acceleration! Three scalar equations can be written from this vector equation. The equation of motion, being a vector equation, may be expressed in terms of its three components in the Cartesian (rectangular) coordinate system as

∑F = ma or ∑Fx i + ∑Fy j + ∑Fz k = m(ax i + ay j + az k)

or, as scalar equations, ∑Fx = max , ∑Fy = may , and ∑Fz = maz .

PROCEDURE FOR ANALYSIS

• Establish your coordinate system and draw the particle’s free body diagram showing only external forces. These external forces usually include the weight, normal forces, friction forces, and applied forces. Show the ‘ma’vector (sometimes called the inertial force) on a separate diagram.

Make sure any friction forces act opposite to the direction of motion! If the particle is connected to an elastic , a spring force equal to ks should be included on the FBD.

11 PROCEDURE FOR ANALYSIS (continued) • Equations of Motion

If the forces can be resolved directly from the free-body diagram (often the case in 2-D problems), use the scalar form of the equation of motion. In more complex cases (usually 3-D), a Cartesian vector is written for every force and a vector analysis is often best.

A Cartesian vector formulation of the second law is ∑F = ma or

∑Fx i + ∑Fy j + ∑Fz k = m(ax i +ay j +az k) Three scalar equations can be written from this vector equation. You may only need two equations if the motion is in 2-D.

PROCEDURE FOR ANALYSIS (continued) •

The second law only provides for forces and accelerations. If velocity or have to be found, kinematics equations are used once the acceleration is found from the equation of motion.

Any of the learned in Chapter 12 may be needed to solve a problem. Make sure you use consistent positive coordinate directions as used in the equation of motion part of the problem!

12 EXAMPLE

Given: WA = 10 lb WB = 20 lb voA = 2 ft/s μk = 0.2

Find: vA when A has moved 4 feet.

Plan: Since both forces and velocity are involved, this problem requires both the equation of motion and kinematics. First, draw free body diagrams of A and B. Apply the equation of motion . Using dependent motion equations,

derive a relationship between aA and aB and use with the equation of motion formulas.

EXAMPLE (continued)

Solution: 2T Free-body and kinetic diagrams of B: =

WB mBaB

+↓∑ Fy = ma y Apply the equation of W − 2 T = m a motion to B: B B B 20 − 2T = 20 a 32.2 B (1)

13 EXAMPLE (continued) y Free-body and kinetic diagrams of A:

x WA mAaA T =

F = m N N k Apply the equations of motion to A: ←+ +↑ ∑ Fy = may = 0 ∑ Fx = ma x 10 lb F = m a N = WA = − T A A F μ = k N = 2 lb 10 2 − T = a (2) 32. 2 A

EXAMPLE (continued) Now consider the kinematics. Constraint equation:

sA + 2 sB = constant sA Datums or v + 2 v = 0 A A B Therefore s B aA + 2 aB = 0

aA = -2 aB (3)

(Notice aA is considered B positive to the left and aB is positive downward.)

14 EXAMPLE (continued) Now combine equations (1), (2), and (3).

T = 22 = 7. 33 lb 3 ft ft 2 2 aA = −17.16 s = 17.16 s →

Now use the kinematic equation: 2 v 2 + 2 a (s − ) vA = oA A A soA 2 2 vA = 2 + 2(17.16)(4) ft vA = 11. 9 s →

GROUP PROBLEM SOLVING Given: The 400 kg mine car is hoisted up the incline. The force in the cable is F = (3200t2) N. The car has an initial velocity of

vi = 2 m/s at t = 0.

Find: The velocity when t = 2 s. Plan: Draw the free-body diagram of the car and apply the equation of motion to determine the acceleration. Apply kinematics relations to determine the velocity.

15 GROUP PROBLEM SOLVING (continued) Solution: 1) Draw the free-body and kinetic diagrams of the mine car:

W = mg F ma y = x θ N

Since the motion is up the incline, rotate the x-y axes.

θ = tan-1(815) = 28.07°

Motion occurs only in the x-direction.

GROUP PROBLEM SOLVING (continued) 2) Apply the equation of motion in the x-direction:

+ ∑ Fx = max => F – mg(sinθ) = max

=> 3200t2 – (400)(9.81)(sin 28.07°) = 400a

=> a = (8t2 – 4.616) m/s2 3) Use kinematics to determine the velocity: a = dv/dt => dv = a dt v t

2 ∫ dv = ∫ (8t – 4.616) dt, v1 = 2 m/s, t = 2 s v1 0 2 v – 2 = (8/3t3 – 4.616t)⎜ = 12.10 => v = 14.1 m/s 0

16 READING QUIZ

1. In dynamics, the friction force acting on a moving object is always A) in the direction of its motion. B) a kinetic friction. C) a static friction. D) zero.

2. If a particle is connected to a spring, the elastic spring force is expressed by F = ks . The “s” in this equation is the A) spring constant. B) un-deformed length of the spring. C) difference between deformed length and un-deformed length. D) deformed length of the spring.

17 ATTENTION QUIZ 1. Determine the tension in the cable when the 400 kg box is moving upward with a 4 m/s2 T acceleration. 60 A) 2265 N B) 3365 N a = 4 m/s2 C) 5524 N D) 6543 N

2. A 10 lb particle has forces of F1= (3i + 5j) lb and

F2= (-7i + 9j) lb acting on it. Determine the acceleration of the particle. A) (-0.4 i + 1.4 j) ft/s2 B) (-4 i + 14 j) ft/s2 C) (-12.9 i + 45 j) ft/s2 D) (13 i + 4 j) ft/s2

18 READING QUIZ 1. Newton’s second law can be written in mathematical form as ΣF = ma. Within the summation of forces ΣF, ______are(is) not included. A) external forces B) weight C) internal forces D) All of the above.

2. The equation of motion for a system of n-particles can be

written as ΣFi = Σ miai = maG, where aG indicates ______. A) summation of each particle’s acceleration B) acceleration of the center of mass of the system C) acceleration of the largest particle D) None of the above.

CONCEPT QUIZ

1. The block (mass = m) is moving upward with a speed v.

Draw the FBD if the kinetic friction coefficient is μk. mg mg v

A)μkN B) μkN

N N mg

C)μkmg D) None of the above.

N

19 CONCEPT QUIZ (continued) 2. Packaging for oranges is tested using a that exerts 2 2 ay = 20 m/s and ax = 3 m/s , simultaneously. Select the correct FBD and kinetic diagram for this condition. y

W may x A) B) W

= • max = • max

Rx Rx Ry Ry

C) D) may W may

= • = • max

Ry Ry

ATTENTION QUIZ 1. Internal forces are not included in an equation of motion analysis because the internal forces are_____. A) equal to zero B) equal and opposite and do not affect the calculations C) negligibly small D) not important

2. A 10 lb block is initially moving down a ramp with a velocity of v. The force F is applied to F bring the block to rest. Select the correct FBD.

10 F 10 F 10 F v

A) μ k10 B) μ k10 C) μkN

N N N

20 CONCEPT QUIZ

1. If the cable has a tension of 3 N, determine the acceleration of block B.

2 2 A) 4.26 m/s B) 4.26 m/s 10 kg

μk=0.4 C) 8.31 m/s2 D) 8.31 m/s2 4 kg

2. Determine the acceleration of the block. 30•° A) 2.20 m/s2 B) 3.17 m/s2

2 2 C) 11.0 m/s D) 4.26 m/s 60 N 5 kg

21