1. Solving Problems with Newton’s Laws
“It sounds like an implosion!”

2. Newton’s 2nd Law: ∑F = ma ∑F = VECTOR SUM of all forces on mass m
Note that forces are VECTORS!!
Newton’s 2nd Law: ∑F = ma
∑F = VECTOR SUM of all forces on mass m
 We need VECTOR addition to add forces in the 2nd Law!
Forces add according to the rules of
VECTOR ADDITION!
(next chapter)
In this chapter, we consider only 1 dimensional motion & therefore only 1 dimensional Force vectors

3. Problem Solving Procedures
Make a sketch. For
Each Object Separately,
sketch a free-body diagram, showing all forces acting on that object. Make magnitudes & directions as accurate as you can. Label each force.
Apply Newton’s 2nd Law
Separately to Each Object.
Solve for the unknowns.
Note that this often requires algebra, such as solving 2 linear equations in 2 unknowns!

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

5. we will assume massless cables, strings & wires in this course.
Transmitting Forces
Strings exert a Force on objects they are connected to
This also applies to cables, ropes, etc.
To be realistic, the mass of the string may have to be taken into account, but
we will assume massless cables,
strings & wires in this course.
Pulleys can redirect forces & can also amplify them.

6. T is the Tension Force in the string.
Strings exert a Force on the objects they are connected to
Cables & ropes act similarly
Strings exert a force due to their Tension
For massless strings, the ends both exert the same force of magnitude T on the supports where they
are connected.
T is the Tension Force in the string.

7. Briefly: Cables with Mass
Apply Newton’s 2nd Law to the cable. To support the cable, the upper tension, T1 must be larger than the tension from the box, T2
If there is no acceleration, for the cable, Newton’s 2nd Law is
T1 -T2 - mcable g = 0
So, if the cable is massless,
mcable = 0, then
T1 = T2 = T
So, we can assume mcable = 0
if it is small compared to the other masses in the problem.

8. Tension Example – Elevator Cable
Two Forces act on
the elevator
Gravity, Fgrav = mg
acts downward
The Tension Force T in the cable acts upward
Assume an upward acceleration
Apply Newton’s 2nd Law & get:
T- mg = ma

9. Briefly consider the cable by itself
Assume it is massless
Apply Newton’s 2nd Law & get: TC = T
So, the Tension T is the same for all points in the cable
True for all massless cables
Tension T is a force with units of Newtons.

10. Single Pulleys Assume:
We often need to change the direction of a force.
A Simple Pulley changes the direction of a force, but not the magnitude (diagram).
Assume:
The rope & the pulley are both massless
The cable does not slip on the pulley
There is no friction on axis of the pulley.

11. Pulleys Amplify Forces
A person exerts a force T on the rope. So, the rope exerts a force 2T on the pulley. This can be used to lift an object.
More complex sets of pulleys can amplify an applied force by greater factors.
The distance decreases to
compensate for the increase in force

12. Example (“Atwood’s Machine”) Elevator + Counterweight
Two masses suspended over a (massless frictionless)
pulley by a flexible (massless) cable  an
“Atwood’s Machine”
Example:
Elevator + Counterweight
It can be shown that a
massless cable means also
that, in the figure:
aE = - a & aC = a
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

13. Example: Elevator & Counterweight.
Counterweight mC = 1000 kg.
Elevator mE = 1150 kg.
Calculate
a) The elevator’s acceleration a.
b) The tension in the cable FT.
aE = - a
aC = a
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.
Work out in detail on
the white board!

14. Example: Elevator & Counterweight.
Counterweight mC = 1000 kg.
Elevator mE = 1150 kg. Calculate
a) The elevator’s acceleration a.
b) The tension in the cable FT.
aE = - a
aC = a a 
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
Work out in detail on the white board!
Free Body
Diagrams

15. 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!

16. Work out in detail on the white board!
Example
Two boxes connected by a lightweight (massless!)
cord are resting on a smooth (frictionless!) table.
mA = 10 kg, mB = 12 kg.
A horizontal force FP = 40 N is applied to mA.
Calculate:
a. The acceleration a of the boxes.
b. The tension T 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.
Work out in detail on the white board!

17. Example
Two boxes connected by a lightweight (massless!) cord are resting on a smooth (frictionless!) table. 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 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

18. Problem Calculate FT1 & FT2  FT1  a m1g   FT2 FT2   a  m2g
m1 = m2 = 3.2 kg, m1g = m2g = 31.4 N
Acceleration a = 2.0 m/s2
Calculate FT1 & FT2
 FT1
 a
m1g 
 FT2
FT2 
 a
 m2g

19. Problem Calculate FT1 & FT2 Use Newton’s 2nd Law: ∑Fy = ma
m1 = m2 = 3.2 kg, m1g = m2g = 31.4 N
Acceleration a = 2.0 m/s2
Calculate FT1 & FT2
Use Newton’s 2nd Law: ∑Fy = ma
for EACH bucket separately!!!
Take up as positive.
Bucket 1: FT1 - FT2 - m1g = m1a (1)
Bucket 2: FT2 - m2g = m2a (2)
 FT1
 a
m1g 
 FT2
FT2 
 a
 m2g

20. Problem Calculate FT1 & FT2 Use Newton’s 2nd Law: ∑Fy = ma
m1 = m2 = 3.2 kg, m1g = m2g = 31.4 N
Acceleration a = 2.0 m/s2
Calculate FT1 & FT2
Use Newton’s 2nd Law: ∑Fy = ma
for EACH bucket separately!!!
Take up as positive.
Bucket 1: FT1 - FT2 - m1g = m1a (1)
Bucket 2: FT2 - m2g = m2a (2)
From (2), FT2 = m2(g + a) = (3.2)( )
or FT2 = N
Put this into (1)
FT1 - m2(g + a) - m1g = m1a
Gives: FT1 = m2(g + a) + m1(g+a)
or FT1 = (m2+m1) (g + a) = 75.5 N
 FT1
 a
m1g 
 FT2
FT2 
 a
 m2g

21. Problem Acceleration a = 2.0 m/s2 m = 65 kg, mg = 637 N
Calculate FT & FP
FT 
 FT
 FP
 a
 a
 mg

22. (y direction) on woman + bucket!
Problem
Acceleration a = 2.0 m/s2
m = 65 kg, mg = 637 N
Calculate FT & FP
Take up as positive.
Newton’s 2nd Law: ∑F = ma
(y direction) on woman + bucket!
FT + FT - mg = ma
2FT - mg = ma solve & get
FT 
 FT
 FP
 a
 a
 mg

23. Problem  FP = - FT = -383.5 N FT = (½)m(g + a) = 383.5 N
Acceleration a = 2.0 m/s2
m = 65 kg, mg = 637 N
Calculate FT & FP
Take up as positive.
Newton’s 2nd Law: ∑F = ma
(y direction) on woman + bucket!
FT + FT - mg = ma
2FT - mg = ma
FT = (½)m(g + a) = N
Also, Newton’s 3rd Law says that
 FP = - FT = N
FT 
 FT
 FP
 a
 a
 mg