Net Force...... The Tutorial
By Devin Burke, Winter 1999
Table of Contents

X Displacement

t Time

V_{o} Initial velocity

V_{a} Average Velocity

V Final Velocity

a Acceleration

F Force

m Mass

g Gravity
These quantities are defined and explained on other pages except for Average
Velocity
which is explained below
New Quantities
And here are the formulas that we have so far:

X = V_{a}t = (V^{2}  V_{o}^{2})/_{2}

X = V_{o}t + (0.5)at^{2}

V_{a} = at = (V + V_{o})/_{2}

F = ma
Defining Net Force
Depending on how many times you've watched Star Wars or its ilk, you would
intuitively
think of force simply as a push or a pull exerted on an object. Or,
if you are just super, you
might think of force the way Sir Isaac Newton did in his famous three
laws of motion, the first
two of which will be stated here not only for your convenience in this
tutorial but because they are
fundamental laws which should be recited every night before bedtime
for the rest of your life.
Newton's first law of motion is as follows:
Every body (or object) continues
in its state of rest or of uniform speed in a straight line
unless it is compelled to
change that state by a NET FORCE acting on it.
So already, although this first law deals primarily with inertia (not
the subject of this tutorial),
you can see where NET FORCE is important, because Newton mentions it
in not only
his first famous law of motion, but his second as well:
The acceleration of an object
is directly proportional to the NET FORCE acting on it and is
inversely proportional to
its mass. The direction of the acceleration is in the direction
of the applied NET FORCE.
To know what Newton is actually saying in these two laws, you obviously
have to know what
NET FORCE is first. Once you have already learned that force is mass
times acceleration (F = ma),
then you are merely a hop, skip, and a jump from the concept of NET
FORCE, which is simply the
vector sum of all forces acting on an object.
Now, when you begin to use formulas like F = ma, you are entering
a world of physics called
dynamics. Newton, and by extension you, will be defining motion in
terms of mass. But mass is
a tricky part of this equation, and it is always worth remembering
that mass is not the same as weight.
If you think of weight as a force (which, according to the laws of
physics, you should) and mass as
the amount of matter contained in an object, measured in kilograms
(kg), then you will be much better
off when doing your problems. Thusly, weight is a force, which is measured
in Newtons with an (N) for units.
So, F = ma is as simple an equation as they come. One way in
which it will almost always be applied
is to find the weight of an object, and this is simply F_{weight}
= mg, where g is gravity and is a downward
acceleration of 9.8 m/s/s.
Net force can be found when you take all the forces exerted on an object
at one time and add them
together. If this is difficult to picture, think about three guys trying
to push their pickup truck out of a
ditch. While all three are pushing on the back of the truck Bobby Joe
is exerting one force, while
Billyum is exerting a separate force, and Uncle Ricky is pushing with
yet another force. But the truck
only goes in one direction, as if it were being pushed with only one
force. That is because the three men's
pushings all add up to a single NET FORCE. You have to convert the
forces to vector form in order
to add them, and don't stress, it's not that hard.
Formulas
So now we have all the formulas we need for solving net
force problems with
uniform mass and acceleration:
# F = ma
# F_{weight} = mg
Go back to: Table of Contents
General Problem Solving Strategy:

Read the problem.

Identify all the forces acting on the object in the question.

Draw a diagram of the object and calculate the magnitude of any force that
you can, (e.g. the weight, or the force of friction) indicating this with
a value and an arrow on the diagram..

Solve for any other quantities thatn you can (e.g. acceleration, mass)

Set the vector sum of all the forces equal to mass times acceleration.
(F = ma) F must be the vector sum of
all the forces.
Go back to: Table of Contents
Example problem 1
If you exert a force of 100 N on the world's largest wad of aluminum
foil, the wad moves at a constant velocity. What is it's mass if you exert
150 N on it, and it accelerates from rest to 3 m/s in 75 seconds?

NET FORCE = 150 N

m = ?

a = ?

V_{o} = 0 m/s

V = 3 m/s

t = 75 seconds
To start out, you can easily find acceleration by use of your a = (V V_{o})/_{2}
formula. This gives you
a = ( VV_{o})/_{t} = (3 m/s0 m/s)/_{75} =
0.4 m/s/s
Now, from the problem, you know that the first 100 Newtons of force
have nothing to do with acceleration; they simply keep the wad of aluminum
going at a constant velocity. Since acceleration is not involved, you can't
use it in your F = ma formula. This leaves you with
150 N  100 N = 50 N
to play with. Thusly therefore,
F = ma = 50 N = m(0.4 m/s/s)
m = (50 N)/ (0.4 m/s/s) = 1250 kg
The wad of aluminum has a mass of 1250 kg.
Go back to: Table of Contents
Example problem 2
A rocket named Apollo 33 1/3 shoots upward with an acceleration of
29 m/s. The rocket has a weight of 98 Newtons.
What is the rocket's thrust?
Well, you know that....

F = ?

Weight = 98 Newtons

a = 29 m/s/s

g = 9.8 m/s/s
Although it's not a mass, don't back away from that weight of 98 Newtons.
Since it's in Newtons, you already have one component of your net force.
However, in order to find the force required to accelerate the rocket,
you do have to find it's mass. Easier done than said:
F = ma = 98 Newtons = m(9.8 m/s/s)
(98 N) / (9.8 m/s/s) = 10 kg
To find the acceleration force,
F = ma = (10 kg)(29 m/s/s) = 290 Newtons
At this point, you might be wondering whether you should add or subtract
the two forces. Ah ha! If you are, that's good, especially if you have
drawn a picture with vector arrows pointing in the direction of the acceleration.
This usually help with confusing problems, and is necessary as you add
more and more forces to your equation. But here, we only have two forces,
the downward pointing gravity, and the rocket acceleration. You might ask
whether the rocket's force vector points upward because it's in the direction
of the rocket's acceleration. But this would be WRONG! This is because
the direction of the thrust is pointing downward, and interestingly enough,
when you get in to conservation of momentum, the rocket actually propels
itself off the particles being blown out its backside. But that's another
story. So, since you have two downward pointing forces, you add them to
get
NET FORCE = 98 N + 290 N = 388 Newtons
Go back to: Table of Contents
Sample Problems
The answers to each problem follow it in parentheses. They also link to a solution to
the problem. Try the problem, check your answer, and go to the solution if you do not
understand.
1.
A new Weight Watchers ad company tries to fix its definition of weight
in its ad campaign so it will cut down on the phone calls from angry physics
teachers. So, instead of showing a slimmeddown woman with the sign "I
lost x amount of kilograms," they show the woman with a sign saying "I
lost over 588 Newtons!!!" How much mass, in kilograms, are the ad companies
claiming the woman lost with Weight Watchers? And for an extra bonus, you
can find out how much mass, in pounds, they claim she lost. (1 Kilogram
= 2.2 lbs.) (60 kg) and (132 lbs.)
2.
At a carnival, a man named Jesse "The Dentures" Ventura is able to lift
a 1500 kg car using only the strength of his teeth. If he suspends the
car in the air with no movement up or down, how much force is he exerting
on the car? (14700 Newtons)
3.
It's crucifixion time again in the Holy Roman Empire. Julius Iglesias is
a Christian who is sentenced to carry a cross which has a mass of 50 kg.
If, from the crack of the whip he speeds up from 0 m/s to 2 m/s in 4 seconds
while bearing the crucifix, disregarding friction, how much force is Julius
Iglesias exerting on the cross?
(515 Newtons)
4.
The anchor of the Titanic has a mass of 20000 kg (neglect water). What
is the acceleration of the anchor if the chain is under 200000 Newtons
of tension? 182000 Newtons? (0.2 m/s/s, 0.7 m/s/s)
5.
In a Looney Tunes cartoon, Elmer Fudd is just about to pounce on Bugs Bunny
but Bugs steps out of the way at the last moment to reveal an ejector seat.
Poor Elmer fall into the seat, rests just long enough to say, "I'll get
you for dis, you siwwy wabbit," then is ejected into the air with an acceleration
of 40 m/s/s. Elmer, being a healthy little fellow, has a mass of 25 kg.
What is the thrust of Bugs Bunny's ejector seat? Say that the ejector seat
has a mass of
25 kg also. (2490 Newtons) Go back to: Table
of Contents
6.
According to a Greek myth, Sisyphus was a Greek king who was sentenced
by the gods in the underworld to roll a rock up a hill forever. As soon
as Sisyphus got close to pushing the stone over the hill, the stone always
fell back to the bottom of the hill, and Sisyphus had to start over. If
Sisyphus exerts 250 N to push the 700 kg stone up the hill at a constant
velocity, then decides to push 350 N in order to get the stone going fast
enough to go over the hill, what is the acceleration of the stone? (0.143
m/s/s)
7.
What force is needed to drag a 5 kg light sabre with the power of your
mind across the ground with an acceleration of 2.5 m/s/s when the friction
force is 25 Newtons? Remember, you are fighting Darth Vader, so you can't
waste time when calulating the force.
(37.5 Newtons)
8.
Arnold Schwarzenegger is holding on to a cable attached to the helicopter
of a mean, evil, bad guy. Due to Mr. Schwarzenegger's considerable bulk
mass, plus the innocent hostage clinging to his leg, the chopper, while
trying to fly away, exerts 2500 Newtons with its motor, yet does not move
from its hover position. What is the tension on the cable while the chopper
is hovering, and what is the acceleration of the chopper if the motor exerts
3500 N on Mr. Schwarzenegger and his cable? Remember, for the sake of science,
the weight of the helicopter is neglible. (2500 N, 3.92 m/s/s)
9.
If you push a truck out of a ditch, the truck weighing 1000 kg, and the
truck accelerates from 0 m/s to 12 m/s in 6 seconds, what is the force
required in such a situation? Imagine your name is Billyum. (2000
Newtons)
10.
A rocket thrusts up into space with a thrust of 150 N. What
is the mass as it accelerates upward at an acceleration of 15 m/s/s? For
an extra bonus chocolate chip cookie, tell me what famous physicist was
the unit Newton named for? (6.05 kg)
Go back to: Table of Contents
Solutions to Sample Problems
1.
How much did a woman who loses 588 Newtons lose in terms of mass, both
in kilograms and pounds? Well, you start off with your formula of
F = ma. What you know is:

F = 588 N

a = g = 9.8 m/s/s

m = ?
Since this is a problem dealing with weight, you know that the acceleration
is going to be gravity, which you know is 9.8 m/s/s. Normally, since gravity
is a downward force, it would have an acceleration of (9.8 m/s/s), but
since a person can't have a negative mass and since direction of motion
is not important in this problem, the acceleration of gravity can be positive.
So you start with 588 N = m(9.8 m/s/s). Using division, 588 N/ 9.8 m/s/s
= m = 60 kg. Then, you use (60 kg)(2.2) to get 132 lbs.
Go to: ProblemFormulasTable
of Contents
2.
How much force is required to suspend a 1500 kg car using only one's teeth?
Using F = ma, you know this:

F = ?

m = 1500 kg

a = 9.8 m/s/s
This problem is basically the same problem as sample problem 1. Even though
the car is suspended in the air, it does not move up or down, so there
is no acceleration that the car itself is doing. That leaves you with an
acceleration of 9.8 m/s/s due to gravity, which you can then plug and chug
with the formula to get F = ma = (1500 kg)(9.8 m/s/s) = 14700 Newtons.
Go to: ProblemFormulasTable
of Contents
3.
How much force does Julius Iglesias exert on a cross of mass 50 kg when
he speeds up from 0 m/s to 2 m/s in 4 seconds?
First of all, we know:

F = ?

m = 50 kg

a = ?

V_{o} = 0 m/s

V = 2 m/s

t = 4 seconds

g = 9.8 m/s/s
So, with these tools, we can start by finding acceleration. We know from
earlier investigation that (V V_{o})/_{2 }= a, so (2 m/s
 0 m/s)/_{4 seconds} =
0.5 m/s/s. So one element of the net force will be the cross's acceleration,
which is F = ma = (50 kg)(0.5 m/s/s) = 25 Newtons. However, you can't forget
weight, because not only does it take Newtons of force to accelerate an
object, but it also takes Newtons to simply get the thing moving. So you
have to add the force of the weight to the force with which Julius is accelerating
the cross. So F = (50 kg)(9.8 m/s/s) = 490 Newtons. Then, since NET FORCE
requires the sum of all forces, (490 Newtons) + (25 Newtons) = 515 Newtons.
Go to: Problem FormulasTable
of Contents
4.
Titanic problem:
OK, so spill your guts, give me everything you've got. You know:

F_{1} = 200000 N

F_{2} = 182000 N

m = 20000 kg

a_{1} = ?

a_{2} = ?

g = 9.8 m/s/s
You know that the force required to lift the anchor against gravity is
F = ma = (20000 kg)(9.8 m/s/s) = 196000 Newtons. This means that whatever
is left out of the total net force is what does the acceleration. So, (200000
N)  (196000 N) = 4000 N. Plug this in to find 4000 N = ma =
(20000 kg)(a_{1}) and you can use algebra to get (4000 N) /
(20000 kg) = 0.2 m/s/s. In the same manner, (182000 N)  (196000
N) = 14000 N. So,
(14000 N ) = (20000 kg)(a_{2}), and (14000 N) / (20000 kg)
= (0.7 m/s/s). Go to: ProblemFormulasTable
of Contents
5.
What is the thrust of Bug's Bunny's 25 kg ejector seat, in which Elmer
Fudd, who has a mass of 25 kg, is now accelerating at a rate of 40 m/s/s?

F = ?

m of Mr. Fudd = 25 kg

m of ejector seat = 25 kg

a = 40 m/s/s
First, add the masses. 25 kg + 25 kg = 50 kg. Then, find the two
forces of gravity and of the acceleration. This F = (50 kg)(9.8 m/s/s)
= 490 N
and F = (50 kg)(40 m/s/s) = 2000 N. Then, you add the two forces together
to find net force and your answer comes out to be 2000 N + 490 N = 2490
N. Go to: ProblemFormulasTable
of Contents
6.
The problem of Sisyphus trying to push a 700 kg stone up a hill for eternity.
Sisyphus exerts 250 N to keep the stone moving at a constant velocity,
When he exerts 350 N, what's the stones acceleration?
This problem (or, if you want to think of it in less scary terms, this
question) is very much like Example Problem 1. You know:

F = 350 N

m = 700 kg

a = ?
To begin with, the 350 N is the net force, because that is the total force
Sisyphus is using to not only get the stone moving but also to accelerate
it. Since you are only interested in the acceleration, get rid of the other
force used to keep the stone at constant velocity, which is 250 N. Doing
this would be 350 N  250 N = 100 N. This part of the force is what goes
into your formula to find F = ma. 100 N = (700 kg)(a), and (100 N) / (700
kg) = 0.143 m/s/s. Go to: Problem FormulasTable
of Contents
7.
Vow never to turn to the Dark Side! Long live the Force! Ok, you're dragging
the 5 kg light sabre across the ground that has a friction force of 25
N and causing the sabre to accelerate at 2.5 m/s/s. What's the net force?
So you know:

F_{net} = ?

F_{friction} = 25 N

m = 5 kg

a = 2.5 m/s/s
First, find the force used to accelerate, which is F = ma = (5 kg)(2.5
m/s/s) = 12.5 N. This is added to the friction force to find the NET FORCE,
and you get 12.5 N + 25 N = 37.5 N. Go to: ProblemFormulasTable
of Contents
8.
Arnold Schwarzenegger is cling to the chopper while the hovering chopper
is exerting a force of 2500 N. What is the cable's tension and what is
the chopper's acceleration when it exerts a total force of 3500 N.

F_{net} = 3500 N

F_{weight} = 2500 N

m = ?

a = ?
Since, the helicopter does not move while exerting 2500 N of force,
you know that gravity is only being cancelled out. thus, you know that
the upward force of the chopper motor is exactly the same as the force
of weight pulling down on it, which is only the weight of Arnie and his
hostage if you disregard the helicopter's weight. So, the tension in the
cable is also 2500 N. Now, this leaves you with 1000 N out of the net force
(3500 N  2500 N) that deals with acceleration. From the 2500 N weight
of Srnie and his hostage, you can find that (2500 N) = (m)(9.8 m/s/s).
Then (2500 N) / (9.8 m/s/s) = 255.1 kg. Now you can find (1000 N) = (255.1
kg)(a). Finally, (1000 N) / (255.1 kg) = 3.92 m/s/s.
Go to: ProblemFormulasTable
of Contents
9.
You are pushing the 1000 kg truck out of the ditch while speeding up the
truck from 0 m/s to 12 m/s in 6 seconds. You know:

F = ?

m = 1000 kg

a = ?

V_{o} = 0 m/s

V = 12 m/s

t = 6 seconds
For a problem #9, this is not that hard, I admit. Acceleration is (V 
V_{o})/_{t} = (12 m/s  0 m/s)_{6 s }= 2 m/s/s.
So,
F = ma = (1000 kg)(2 m/s/s) = 2000 N.Go to: ProblemFormulas
Table of Contents
10.
What is the mass of a rocket that accelerates upward with an acceleration
of 15 m/s/s with the use of 150 N of thrust?

F = 150 N

m = ?

a = 15 m/s/s

g = 9.8 m/s/s
Since you that the force of gravity added to the froce of acceleration
gives you the net force, you can show this with
(F_{gravity}) + (F) = NET FORCE. You can substitute in this
euation to get (mg) + (ma) = NET FORCE. Then, you can distribute to get
the equation (g + a)(m) = NET FORCE. Quatitatively, you can find the mass,
by
(9.8 m/s/s + 15 m/s/s)(m) = 150 N. This is (24.8 m/s/s)(m) = 150 N,
and (150 N) / (24.8 m/s/s) = m = 6.05 kg.
Go to: Problem FormulasTable
of Contents