Solving Problems with Newton’s Laws
“It sounds like an implosion!”
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
Problem Solving Procedures
1. 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.
2. Apply Newton’s 2nd Law separately to
each object.
3. Solve for the unknowns.
Note that this often requires algebra, such as
solving 2 linear equations in 2 unknowns!
Applications & Examples of Newton’s 2nd
Law in 1 Dimensional Motion
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
Tension
• 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.
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
• 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
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,
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.
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
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
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 =
1150 kg. Calculate a) Elevator’s acceleration. b) Tension in the cable.
aE = - a
aC = a
a


a
Free Body
Diagrams
General Approach to Problem Solving
1.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!
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.
Free Body
Diagrams
Problem
 FT1
m1 = m2 = 3.2 kg, m1g = m2g = 31.4 N
Acceleration a = 2.0 m/s2
a
m1g   FT2
for EACH bucket separately!!!
Take up as positive.
Bucket 1: FT1 - FT2 - m1g = m1a (1)
Bucket 2: FT2 - m2g = m2a
(2)
FT2 
a
 m2g
Calculate FT1 & FT2
Use Newton’s 2nd Law: ∑Fy = ma
From (2), FT2 = m2(g + a) = (3.2)(9.8 + 2.0)
or
FT2 = 37.76 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
Problem
FT 
 FT
 FP
a
a
 mg
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) = 383.5 N
Also, Newton’s 3rd Law says that
 FP = - FT = -383.5 N