1. Applications & Examples of Newton’s 2nd Law

2. The normal force, FN is NOT always equal & opposite to the weight!!
Example 4-11
A 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
Figure Caption: (a) Pulling the box, Example 4–11; (b) is the free-body diagram for the box, and (c) is the free-body diagram considering all the forces to act at a point (translational motion only, which is what we have here).
Answer: (a) The free-body diagram is shown in part (b); the forces are gravity, the normal force, and the force exerted by the person. The forces in the vertical direction cancel – that is, the weight equals the normal force plus the vertical component of the external force. The horizontal component of the force, 34.6 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.
The normal force, FN is NOT
always equal & opposite to the weight!!

3. Example 4-12
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.
Figure Caption: Example 4–12. (a) Two boxes, A and B, are connected by a cord. A person pulls horizontally on box A with force FP = 40.0 N. (b) Free-body diagram for box A. (c) Free-body diagram for box B.
Answer: 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 boxes have the same acceleration.
The acceleration is the external force divided by the total mass, or 1.82 m/s2.
The tension in the cord is the mass of box B multiplied by the acceleration, or 21.8 N.
Free Body
Diagrams

4. 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.
Figure Caption: Example 4–13. (a) Atwood’s machine in the form of an elevator–counterweight system. (b) and (c) Free-body diagrams for the two objects.
Answer: Each mass has two forces on it, gravity pulling downward and the tension in the cable pulling upward. The tension in the cable is the same for both, and both masses have the same acceleration. Writing Newton’s second law for each mass gives us two equations; there are two unknowns, the acceleration and the tension. Solving the equations for the unknowns gives a = 0.68 m/s2 and FT = 10,500 N. a   a
aE = - a
Free Body
Diagrams
aC = a

5. Conceptual Example 4-14 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?
mg = 2000 N
Free Body Diagram

6. Example 4-15 = 300 N Newton’s 3rd Law  FBR = -FRB, FCR = -FRC
FRBx = -FRBcosθ
FRBy = -FRBsinθ
FRCx = FRCcosθ
FRCy = -FRCsinθ

7. 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?
Figure 4-25.
Answer: (a) The acceleration of the mass is given by the horizontal component of the tension; the vertical component of the tension is equal to the weight. Writing these two equations and combining them gives tan θ = a/g, or θ = 7.0°.
(b) When the velocity is constant, the string is vertical.
Free Body Diagram

8. General Approach to Problem Solving
Read the problem carefully; then read it again.
Draw a sketch, then a free-body diagram.
Choose a convenient coordinate system.
List the known & unknown quantities; find relationships between the knowns & the unknowns.
Estimate the answer.
Solve the problem without putting in any numbers (algebraically); once you are satisfied, put the numbers in.
Keep track of dimensions.
Make sure your answer is REASONABLE!

9. for EACH bucket separately!!!
Problem 25
 FT1
Newton’s 2nd Law
∑F = ma (y direction)
for EACH bucket separately!!!
 a
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
m1g 
 FT2
FT2 
 a
 m2g

10. Problem 29 FT + FT - mg = ma 2FT - mg = ma  FP = - FT
Take 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 
 FT
 FP
 a
 a
 mg