Section 4-7 Solving Problems with Newton’s Laws; Free Body Diagrams

“It sounds like an implosion!”

• Forces are VECTORS!!

• Newton’s 2nd Law:

∑F = ma∑F = VECTOR SUM of all forces on mass m

VECTOR addition

is needed add forces in the 2nd Law!

– Forces add according to

the rules of VECTOR ADDITION! (Ch. 3)

Problem Solving Procedures

1. Draw a sketch. For each object separately, sketch a free-body diagram, showing all the forces acting on that object. Make the magnitudes & directions as accurate as you can. Label each force.

2. Resolve vectors into components.

3. Apply Newton’s 2nd Law separately to each object & for each vector component.

4. Solve for the unknowns.

Note that this often requires algebra, like solving 2 linear equations in 2

unknowns!

ExampleFind the resultant force FR

Example

FR = [(F1)2 + (F2)2](½) = 141 Ntanθ = (F2/F1) = 1, θ = 45º

Find the resultant force FR

Example 4-9

If the boat moves withacceleration a, ∑F = FR = ma

FRx = max, FRy = may

Find the resultant force FR

Example 4-9• Illustrates the procedures for Newton’s 2nd Law problems:

STEP 1: Sketch the situation!!

– Sketch a “Free Body” diagram for EACH body in problem & draw ALL forces acting on it.

• Part of your grade on exam & quiz problems!

STEP 2: Resolve the forces on each body into components

– Use a convenient choice of x,y axes• Use the rules for finding vector components from Ch. 3.

STEP 3: Apply N’s 2nd Law to EACH OBJECT SEPARATELY

∑F = ma, Note: This is the LAST step, NOT the first!

We NEED A SEPARATE equation like this for each object!

Resolved into components: ∑Fx = max, ∑Fy = may

Conceptual Example 4-10Moving at CONSTANT v, with NO friction,

which free body diagram is correct?

Example 4-11A box of mass m = 10 kg is pulled by an attached cord along a horizontal smooth (frictionless!) surface of a table. The force exerted is FP = 40.0 N at a 30.0° angle as shown. Calculate:

a. The acceleration of the box. b. The magnitude of the upward normal force FN exerted by the table on the box.

Free Body Diagram

The normal force, FN is NOT always equal & opposite to the

weight!!

Two boxes are connected by a lightweight (massless!) cord & are resting on a smooth (frictionless!) table. The masses are mA = 10 kg & mB = 12 kg. A horizontal force FP = 40 N is applied to mA. Calculate: a. The acceleration of the boxes. b. The tension in the cord connecting the boxes.

Example 4-12

Free Body Diagrams

Example 4-13 (“Atwood’s Machine”)Two masses suspended over a (massless frictionless) pulley by a flexible (massless) cable is an “Atwood’s machine”. Example: elevator & counterweight. Figure: Counterweight mC = 1000 kg. Elevator mE = 1150 kg. Calculate a. The elevator’s acceleration. b. The tension in the cable.

aE = - a

aC = a

a

a

Free Body Diagrams

Conceptual Example 4-14

mg = 2000 N

Advantage of a Pulley

A mover is trying to lift a piano (slowly) up to a second-story apartment. He uses a rope looped over 2 pulleys.

What force must he exert on the rope to slowly lift the piano’s mg = 2000 N weight?

Free Body Diagram

Example 4-15: Accelerometer

A small mass m hangs from a thin string & can swing like a pendulum. You attach it above the window of your car as shown. What angle does the string make

a. When the car accelerates at a constant a = 1.20 m/s2?

b. When the car moves at constant velocity, v = 90 km/h?

Free Body Diagram

Inclined Plane Problems

Understand ∑F = ma & how to resolve it into x,y components in the tilted coordinate system!!

Engineers & scientists MUST understand these!

a

The tilted coordinateSystem is convenient,

but not necessary.

A box of mass m is placed on a smooth (frictionless!) incline that makes an angle θ with the horizontal. Calculate: a. The normal force on the box. b. The box’s acceleration. c. Evaluate both for m = 10 kg & θ = 30º

Example 4-1b: Sliding Down An Incline

Free Body Diagram

Example= 300 N

FT2x = FTcosθFT2y = -FTsinθ

FT1x = -FTcosθFT1y = -FTsinθ

Free Body Diagram

Problem 32 Take up as positive!

m = 65 kg

mg = 637 N

FT + FT - mg = ma

2FT -mg = ma

FP = - FT

∑F = ma (y direction) on the woman + the bucket!

FTFT

FP

mg

aFree Body

Diagram

Newton’s 2nd Law

Newton’s 3rd Law!!

Problem 32 Solution Take up as positive!

m = 65 kg

mg = 637 N

FT + FT - mg = ma

2FT -mg = ma

FP = - FT

∑F = ma (y direction) on the woman + the bucket!

FTFT

FP

mg

aFree Body

Diagram

Newton’s 2nd Law

Newton’s 3rd Law!!

The window washer pulls down on the rope with a tension force sothe rope pulls up on her hands with a tension force The tension inThe rope is also applied at the other end of the rope, where itattaches to the bucket. So there is another force pulling up on thebucket. The bucket-washer combination has a net force upwards. The free body diagram shows only forces on the bucket-washer,not forces exerted by them (the pull down on the rope by theperson) or internal forces (normal force of bucket on person).(a) Write Newton’s second law in the vertical direction, with up aspositive. The net force must be zero if the bucket and washer haveconstant speed.

(b) Now the force is increased by 15%, so again write Newton’ssecond law, but with a non-zero acceleration.

Problem 32 Solution

T T T

21 1T 2 2

0 2

72 kg 9.80 m s 352.8 N 350 N

F F F mg F mg

F mg

mg

TF

TF

T T

2

2 2T

2 405.72 N 72 kg 9.80 m s21.47 m s 1.5m s

72 kg

F F F mg ma

F mga

m

Problem 33

FT1

a

m1g FT2

FT2

a

m2g

We draw free-body diagrams for each bucket. a. Since the buckets are at rest, their acceleration is 0. Write Newton’s 2nd Law for each bucket, calling UP the positive direction.

1 T1

2

T1

0

3.2 kg 9.80 m s 31N

F F mg

F mg

2 T2 T1

2

T2 T1

0

2 2 3.2 kg 9.80 m s 63N

F F F mg

F F mg mg

b. Now repeat the analysis, but with a non-zero acceleration. The free-body diagrams are unchanged.

1 T1

2 2

T1

2 T2 T1 T2 T1 T1

3.2 kg 9.80 m s 1.25m s 35.36 N 35 N

2 71N

F F mg ma

F mg ma

F F F mg ma F F mg ma F

General Approach to Problem Solving

1. Read the problem carefully; then read it again.

2. Draw a sketch, then a free-body diagram.

3. Choose a convenient coordinate system.

4. List the known & unknown quantities; find relationships between the knowns & the unknowns.

5. Estimate the answer.

6. Solve the problem without putting in any numbers (algebraically); once you are satisfied, put the numbers in.

7. Keep track of dimensions.

8. Make sure your answer is REASONABLE!

Chapter 4 Summary

• Newton’s 1st Law: If the net force on an object is zero, it will remain either at rest or moving in a straight line at constant speed.

• Newton’s 2nd Law:

• Newton’s 3rd Law:

• Weight is the gravitational force on an object.

• Free-body diagrams are essential for problem-solving. Do one object at a time, make sure you have all the forces, pick a coordinate system & find the force components, & apply Newton’s 2nd Law along each axis.