Dynamics: Newton’s Laws of M iMotion

Force is a vector having both magnitude andForce is a vector, having both magnitude anddirection. The magnitude of a force can bemeasured using a spring scalemeasured using a spring scale.

This is Newton’s first law, which is often called the law of inertia:

Every object continues in its state of rest, or of uniform velocity in a straight line, as long as no net force acts on it.

F represents the force applied by the person and Ffr represents the force of friction.

Newton’s First LawIf the no force acts on a body the body’s velocityIf the no force acts on a body, the body s velocitycannot change; that is the body cannot accelerate

Note: If several forces act on a body

(say , , and ) the net force

is defined as: A B C net

net A B C

F F F F

F F F F= + +

i.e. is the vector sum of

, , and ne

C

t

A B

F

F F F, , CA B

Example : Newton’s first law.Example : Newton s first law.

A school bus comes to a sudden stop, and all of thebackpacks on the floor start to slide forward Whatbackpacks on the floor start to slide forward. Whatforce causes them to do that?

N f th b k k ti i tilNo force; the backpacks continue moving untilstopped by friction or collision.

Mass

Mass is the measure of inertia of an object, sometimesunderstood as the quantity of matter in the object. Inth SI t i d i kilthe SI system, mass is measured in kilograms.

Mass is not weight.

Mass is a property of an object. Weight is the forceexerted on that object by gravity.j y g y

If you go to the Moon, whose gravitational accelerationis about 1/6 g, you will weigh much less. Your mass,is about 1/6 g, you will weigh much less. Your mass,however, will be the same.

Newton’s Second Law of Motion

Newton’s second law is the relation between accelerationand force. Acceleration is proportional to force andi l ti l tinversely proportional to mass.

It takes a force to change eitherthe direction or the speed of anobject. More force means more

l ti th facceleration; the same forceexerted on a more massiveobject will yield lessobject will yield lessacceleration.The bobsled accelerates because the team exerts a force.

Newton’s Second Law

The results of the discussions on the relations between the net force Fnet applied on an object of mass m and the resulting acceleration acan be summarized in the following statement known as: “Newton’scan be summarized in the following statement known as:  Newton s second law”

The net force on a body is equal to the product of the body’s mass and its acceleration

In equation form Newton’s second law can be written as:

FnetF ma=

The above equation is a compact way of summarizing three separate equations, f h di t ione for each coordinate axis:

,net x xF ma= ,net y yF ma= ,net z zF ma=

The time rate of change of momentum of agparticle is equal to the resultant external forceacting on the particle:

F = dp/dt = d(mv)/dt = mdv/dt = ma

We place an object of mass m = 1 kg on a frictionlesssurface and measure the acceleration a that results fromthe application of a force F. The force is adjusted so thata = 1 m/s2. We then say that F = 1 newton (symbol: N)

The unit of force in the SI system is the newton (N).The unit of force in the SI system is the newton (N).

Note that the pound is a unit of force, not of mass, andcan therefore be equated to newtons but not tocan therefore be equated to newtons but not tokilograms.

Example : Force to accelerate a fast car.

h f l ( ) kEstimate the net force needed to accelerate (a) a 1000‐kg car at ½ g;(b) a 200‐g apple at the same rate.

Use Newton’s second law acceleration is about 5 m/s2 so F is aboutUse Newton’s second law: acceleration is about 5 m/s2, so F is about5000 N for the car and 1 N for the apple.

Example : Force to stop a carExample : Force to stop a car.

What average net force is required to bring a 1500‐kg car to restfrom a speed of 100 km/h within a distance of 55 m?from a speed of 100 km/h within a distance of 55 m?

First, find the acceleration (assumed constant) from the initialand final speeds and the stopping distance; a = ‐7.1 m/s2. Thenuse Newton’s second law: F = ‐1.1 x 104 N.

Newton’s Third Law of Motion

Whenever one object exerts a force on a second object,the second exerts an equal force in the opposite directionon the first.

If your hand pushes against theedge of a desk (the forceedge of a desk (the forcevector is shown in red), thedesk pushes back against yourhand.

Newton’s third law:when an ice skaterwhen an ice skaterpushes against the wall,the wall pushes backthe wall pushes backand this force causesher to accelerate awayher to accelerate away.

Newton’s Third Law:

Wh b di i b i f hWhen two bodies interact by exerting forces on eachother, the forces are equal in magnitude and opposite indirection

For example consider a book leaning against a bookcase. We label the force BCF

exerted the book the case. Using the same convention we label the force exerted the cas

on by on be thy e book.

CBF Newton's third law can be written as:

The book together with the bookcase are known asthird-

law force p

a" ir"a

BC CBF F=−

Rocket propulsion can also be explained using Newton’s third law:hot gases from combustion spew out of the tail of the rocket at high

d h f h l h kspeeds. The reaction force is what propels the rocket.

Another example of Newton’s third law: the launch of a rocket.The rocket engine pushes the gases downward and the gasesThe rocket engine pushes the gases downward, and the gasesexert an equal and opposite force upward on the rocket,accelerating it upward.

Conceptual Example : What exerts the force to move acar?

Response: A common answer is that the engine makesthe car move forward But it is not so simple Thethe car move forward. But it is not so simple. Theengine makes the wheels go around. But if the tires areon slick ice or deep mud, they just spin. Friction isp , y j pneeded. On firm ground, the tires push backwardagainst the ground because of friction. By Newton’sthird law, the ground pushes on the tires in theopposite direction, accelerating the car forward.

Helpful notation: the first subscript isHelpful notation: the first subscript isthe object that the force is beingexerted on; the second is the source.

We can walk forward because, when one foot pushes backwardagainst the ground, the ground pushes forward on that footagainst the ground, the ground pushes forward on that foot(Newton’s third law). The two forces shown act on different objects.

Applying Newton’s Laws / Free body Diagrams

Part of the procedure of solving a mechanics problem usingNewton’s laws is drawing a free body diagram. This meansg y gthat among the many parts of a given problem we chooseone which we call the “system”. Then we choose axes andenter all the forces that are acting on the system andomitting those acting on objects that were not included inthe systemthe system.

An example is given in the figure below. This is a problem that involves two

bl k l b l d "A" d "B" hi h l f i dFblocks labeled "A" and "B" on which an external force is exerted. We have the following "system" choices:

a System

appF

The only horizontal force is= block A + block B Fa. System . The only horizontal force is

b. . There are now two horizontal forces: and

c The only horizontal force

= block A + block B

System = block A

System = block B is

app

app AB

F

F F

Fc. . The only horizontal forceSystem = b lock B is BAF

Weight—the Force of Gravity; and the Normal Force

Weight is the force exerted on an object by gravity. Closeg j y g yto the surface of the Earth, where the gravitational forceis nearly constant, the weight of an object of mass m is:

hwhere

An object at rest must have no net force on it. If it is sitting on atable, the force of gravity is still there; what other force is there?, g y ;

The force exerted perpendicular to a surface is

called the normal force. It is exactly asylarge as needed to balance the forcefrom the object. (If the required forcegets too big something breaks!)gets too big, something breaks!)

Example : Weight, normal force, and a box.

A f i d h i i l ift b f 10 0 k ithA friend has given you a special gift, a box of mass 10.0 kg with a mystery surprise inside. The box is resting on the smooth (frictionless) horizontal surface of a table. 

(a) Determine the weight of the box and the normal force exerted on it by the table. 

(b) Now your friend pushes down on the box with a force of 40.0(b) Now your friend pushes down on the box     with a force of 40.0 N. Again determine the normal force exerted on the box by the table. 

(c) If your friend pulls upward on the box with a force of 40 0 N(c) If your friend pulls upward on the box with a    force of 40.0 N, what now is the normal force exerted on the box by the table?

The box is always at rest, so the forces must always add to zero. In (a), the normal force equals the weight; in (b), the normal force equals the weight plus the 40 0 N downwardnormal force equals the weight plus the 40.0 N downward force; in (c), the normal force equals the weight minus the 40.0 N upward force.

Example : Accelerating the box.

What happens when a person pullsWhat happens when a person pullsupward on the box in the previousexample with a force greater than thebox’s weight, say 100.0 N?

Th b l d b FThe box accelerates upward because FP >mg.Answer: Now there is a net upward force onthe box of 2.0 N, and the box acceleratesupward at a rate of 0.20 m/s2.

Example : Apparent weight loss.

A 65‐kg woman descends in an elevator thatbriefly accelerates at 0.20g downward. Shestands on a scale that reads in kgstands on a scale that reads in kg.

(a) During this acceleration, what is her weightand what does the scale read?

(b) What does the scale read when the elevatordescends at a constant speed of 2.0 m/s?

Th l ti t i h i ld t di ti i h it f thThe acceleration vector is shown in gold to distinguish it from thered force vectors.

(a) Her weight is always mg, but the normal force is 0.8 mg, whichthe scale reads as 52 kg.

(b) The scale reads her true weight, or a mass of 65 kg.

Solving Problems with Newton’s Laws: Free‐Body Diagramsg

1. Draw a sketch.

2. For one object, draw a free‐body diagram,showing all the forces acting on the object. Makethe magnitudes and directions as accurate asyou can. Label each force. If there are multipleobjects, draw a separate diagram for each one.

3. Resolve vectors into components.

4. Apply Newton’s second law to each component.

5. Solve.

Two force vectors act on a boat.In this example, the net force on the boat has amagnitude of 53.3 N and acts at an 11° angle to thex axis.

Example : Pulling the mystery box.Suppose a friend asks to examine the 10.0‐kg

box you were given previously hoping tobox you were given previously, hoping toguess what is inside; and you respond, “Sure,pull the box over to you.” She then pulls theb b th tt h d d l th thbox by the attached cord along the smoothsurface of the table. The magnitude of theforce exerted by the person is FP = 40.0 N,and it is exerted at a 30.0° angle as shown.Calculate

(a) the acceleration of the box, and(a) the acceleration of the box, and(b) the magnitude of the upward force FN

exerted by the table on the box.

(a) The forces in the vertical direction cancel – that is, theweight equals the normal force plus the vertical componentof the external force. The horizontal component of the force,

234.6 N, accelerates the box at 3.46 m/s2.(b) The vertical component of the external force is 20.0 N; the

weight is 98.0 N, so the normal force is 78.0 N.

Example : Two boxes connected by a cordExample : Two boxes connected by a cord.

Two boxes, A and B, are connected by alightweight cord and are resting on a smoothg g gtable. The boxes have masses of 12.0 kg and10.0 kg. A horizontal force of 40.0 N is appliedto the 10 0 kg box Find (a) the acceleration ofto the 10.0‐kg box. Find (a) the acceleration ofeach box, and (b) the tension in the cordconnecting the boxes.

Free-body diagrams for both boxes are shown.The net force on box A is the external force minus the

tension; the net force on box B is the tension. Both boxeshave the same acceleration.

(a)The acceleration is the external force divided by the total mass, or 1.82 m/s2.(b) The tension in the cord is the mass of box B multiplied by the acceleration,

or 21.8 N.

Example : Elevator and counterweight (Atwood’smachine).

A system of two objects suspended over a pulleyby a flexible cable is sometimes referred to as anAtwood’s machine Here let the mass of theAtwood s machine. Here, let the mass of thecounterweight be 1000 kg. Assume the mass ofthe empty elevator is 850 kg, and its mass whencarrying four passengers is 1150 kg. For thelatter case calculate (a) the acceleration of theelevator and (b) the tension in the cable.( )

Each mass has two forces on it, gravity pullingdownward and the tension in the cable pullingupward The tension in the cable is the same forupward. The tension in the cable is the same forboth, and both masses have the sameacceleration. Writing Newton’s second law foreach mass gives us two equations; there are twog q ;unknowns, the acceleration and the tension.Solving the equations for the unknowns gives a =0.68 m/s2 and FT = 10,500 N.

Conceptual Example :                   The advantage of a pulleyThe advantage of a pulley.

A mover is trying to lift a piano( l l ) d(slowly) up to a second‐storyapartment. He is using a ropelooped over two pulleys as shownlooped over two pulleys as shown.What force must he exert on therope to slowly lift the piano’sp y p2000‐N weight?

“Slowly” means with a constant speed,so the tension in the rope must be halfthe piano’s weight, or 1000 N.

Example : Accelerometer.

A small mass m hangs from a thin stringA small mass m hangs from a thin stringand can swing like a pendulum. Youattach it above the window of your caras shown. What angle does the stringmake (a) when the car accelerates at aconstant a = 1.20 m/s2, and (b) when theconstant a 1.20 m/s , and (b) when thecar moves at constant velocity, v = 90km/h?

(a) The acceleration of the mass is given bythe horizontal component of the tension; thevertical component of the tension is equal tovertical component of the tension is equal tothe weight. Writing these two equations andcombining them gives tan θ = a/g, orθ = 7.0°.(b) When the velocity is constant, the stringis vertical.

Example : Box slides down an incline.

A box of mass m is placed on a smoothA box of mass m is placed on a smoothincline that makes an angle θ with thehorizontal. (a) Determine the normalforce on the box. (b) Determine the box’sacceleration. (c) Evaluate for a mass m =10 kg and an incline of θ = 30°.10 kg and an incline of θ 30 .

(a) The normal force is equal to the componentof the weight perpendicular to the incline, or mgcos θcos θ.(b) The force causing the acceleration is thecomponent of the weight parallel to the incline;therefore the acceleration is g sin θ.g(c) The normal force is 85 N and theacceleration is 4.9 m/s2.

Summary

• Newton’s first law If the net force on an object is zero• Newton’s first law: If the net force on an object is zero,it will remain either at rest or moving in a straight line atconstant speed.constant speed.

• Newton’s second law:

• Newton’s third law:

• Weight is the gravitational force on an object.e g t s t e g a tat o a o ce o a object

• Free‐body diagrams are essential for problem‐solving.Do one object at a time make sure you have all theDo one object at a time, make sure you have all theforces, pick a coordinate system and find the forcecomponents, and apply Newton’s second law along eachp , pp y gaxis.