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

2. 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 the 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 Dimensional Motion

5. Strings exert a force on the objects they are connected to
Transmitting Forces
Strings exert a force on the objects they are connected to
This also applies to cables, ropes, etc.
The mass of the cable may have to be taken into account
Pulleys can redirect forces
Forces can be amplified

6. T is the tension in the string.
Strings exert a force on the objects they are connected to
Cables & ropes act the same way
The strings exert force
due to their tension
The ends of the string
both exert a force of magnitude T on the supports where they
are connected.
T is the tension in the string.

7. Tension Example – Elevator Cable
Two forces are acting on the compartment
Gravity acting downward
Tension in cable acting upward, T
Assume an acceleration upward
Applying Newton’s Second Law gives
T- mg = ma

8. Applying Newton’s 2nd Law gives: TC = T
Now consider the cable
Assume the cable is massless
Applying Newton’s 2nd Law gives: TC = T
Tension is the same for all points along the cable
True for all massless cables
Tension has force units

9. Cables with Mass T1 -T2 - mcable g = 0
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, Newton’s 2nd Law is
T1 -T2 - mcable g = 0
We can assume a massless cable if the mass of the cable is small compared to the other masses in the problem.

10. Single Pulleys We often need to change the direction of the force.
A simple pulley changes the direction of the force, but not the magnitude
See diagram
Assume the rope and pulley are both massless
Assume the cable does not slip on the pulley

11. Pulleys To Amplify Forces
The person exerts a force T on the rope.
The rope exerts a force 2T on the pulley.
This force 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”)
Two masses suspended over a (massless frictionless) pulley by a flexible (massless) cable  “Atwood’s machine”. Example: Elevator & counterweight. Figure: Counterweight mC = 1000 kg. Elevator mE = kg. Calculate a) Elevator’s acceleration. b) Tension in the cable.
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
Free Body
Diagrams

13. 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!

14. 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

15. 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

16. 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