Applications & Examples of Newton’s Laws

• Forces are VECTORS!!

• Newton’s 2nd Law: ∑F = ma

∑F = VECTOR SUM of all forces on mass m

Need VECTOR addition to add forces in the 2nd Law!– Forces add according to rules of VECTOR

ADDITION! (Ch. 3)

• Newton’s 2nd Law problems:

• STEP 1: Sketch the situation!!– Draw 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 Newton’s 2nd Law to

EACH BODY SEPARATELY:

∑F = ma

– A SEPARATE equation like this for each body!– Resolved into components:

∑Fx = max ∑Fy = may

Notice that this is the LAST step, NOT the first!

Conceptual ExampleMoving at constant v, with NO friction,

which free body diagram is correct?

Example Particle in Equilibrium

“Equilibrium” ≡ The total force is zero. ∑F = 0 or ∑Fx = 0 & ∑Fy = 0

Example (a) Hanging lamp (massless chain).

(b) Free body diagram for lamp.∑Fy = 0 T – Fg = 0; T = Fg = mg

(c) Free body diagram for chain.∑Fy = 0 T – T´ = 0; T´ = T = mg

Example Particle Under a Net Force

Example (a) Crate being pulled to right across a floor.

(b) Free body diagram for crate.∑Fx = T = max ax = (T/m)

ay = 0, because of no vertical motion.

∑Fy = 0 n – Fg = 0; n = Fg = mg

Example Normal Force Again

“Normal Force” ≡ When a mass is in contact with a surface, the Normal Force n = force perpendicular to (normal to)

the surface acting on the mass. Example Book on a table. Hand pushing down.

Book free body diagram. ay = 0, because of no vertical motion (equilibrium).

∑Fy = 0 n – Fg - F = 0

n = Fg + F = mg + F Showing again that the normal force is not always = & opposite to the weight!!

ExampleA 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

Free Body Diagrams

Example 5.4: Traffic Light at Equilibrium

(a) Traffic Light, Fg = mg = 122 Nhangs from a cable, fastened to a support. Upper cables are weaker than vertical one. Will break if tension exceeds100 N. Does light fall or stay hanging? (b) Free body diagram for light. ay = 0, no vertical motion.

∑Fy = 0 T3 – Fg = 0 T3 = Fg = mg = 122 N(c) Free body diagram for cable junction (zero mass). T1x = -T1cos(37°), T1y = T1sin(37°) T2x = T2cos(53°), T2y = T2sin(53°), ax = ay = 0. Unknowns are T1 & T2.

∑Fx = 0 T1x + T2x = 0 or -T1cos(37°) + T2cos(53°) = 0 (1)∑Fy = 0 T1y + T2y – T3 = 0 or T1sin(37°) + T2sin(53°) – 122 N = 0 (2)

(1) & (2) are 2 equations, 2 unknowns. Algebra is required to solve for

T1 & T2! Solution: T1 = 73.4 N, T2 = 97.4 N

Example 5.6: Runaway Car

Example 5.7: One Block Pushes Another

Example 5.8: Weighing a Fish in an Elevator

Example 5.9: Atwood Machine

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

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: 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

Example= 300 N

FT2x = FTcosθFT2y = -FTsinθ

FT1x = -FTcosθFT1y = -FTsinθ

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: Sliding Down An Incline

Free Body Diagram

Example 5.10Inclined Plane, 2 Connected Objects