Applications & Examples of Newton’s 2nd Law

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. Acceleration of the boxes. b. 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= 300 N

FRBx = -FRBcosθFRBy = -FRBsinθ

Newton’s 3rd Law FBR = -FRB, FCR = -FRC

FRCx = FRCcosθFRCy = -FRCsinθ

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

General Approach to Problem Solving1. 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!

Problem 25 FT1

m1g FT2

FT2

a

m2g

Take up as positive.

m1 = m2 = 3.2 kg

m1g = m2g = 31.4 N

Bucket 1: FT1 - FT2 - m1g = m1a

Bucket 2: FT2 - m2g = m2a

Newton’s 2nd Law

∑F = ma (y direction)

for EACH bucket separately!!! a

Problem 29Take up as positive.

Newton’s 2nd Law: ∑F = ma (y direction) on woman + bucket!

m = 65 kg, mg = 637 N

FT + FT - mg = ma

2FT - mg = ma

Also, Newton’s 3rd Law:

FP = - FT

FT

a

mg

FP

FT

a