Force Table
4. Force Table*
st
In this unit you learn about forces as vectors and apply Newton’s 1 law.
Learning Objectives:
1.
To be able to recognize and operate on
forces as vectors.
2.
To understand that an object has constant
velocity if and only if the vector sum of
forces on it is zero.
3.
Be able to solve simple problems involving
a sum of forces.
Force table showing ring around central peg and
three strings attached from ring to mass hangers.
Reading Assignment:
Read the following sections. (Section numbers may be slightly different depending on the edition of your
textbook: Check the section titles.)Study the following sections of your textbook before coming to lab.
There will be a check to see that you have done this.
Knight Jones & Field (161): 4.1 What Causes Motion?, 4.2 Force, 4.3 A Short Catalog of Forces, 4.4
Identifying Forces, 4.5 What Do Forces Do?, 4.6 Newton's Second Law, 4.7 Free-Body Diagrams, 5.:
Equilibrium
Serway and Vuille (211): 4.1 Forces, 4.2 Newton's First Law, 4.3 Newton's Second Law
Serway and Jewett (251): 5.1 The concept of Force, 5.3 Mass, 5.4 Newton’s Second Law
Newton’s second law says that the acceleration of an object times the mass of the object equals the sum
of the forces acting on the object. The essential point is that acceleration and force are vector quantities.
Thus Newton’s second law says
 F  m a,
where  indicates a sum of vectors. So if four forces act on the object, then

F1

F3

F4

F2

 


F1 + F2 + F3 + F4 = m a .
Recall that an arrow can represent a vector pictorially.
Hence the forces acting on the small object at the left
are shown as arrows. The length of the arrow is
proportional to the magnitude of the force (in Newtons,
for instance) while the direction of the arrow indicates
the direction of the force. To “add” the forces, recall
that we mean vector addition, so we translate each
vector (slide it parallel to itself) and combine them
head to tail, as shown below.
____________________________________________________________________________
*© William A Schwalm 2012
4-1
Force Table
The sum of the forces in this particular case (NOT
in general) turns out to be zero, since starting from

F1

F1 at the origin and following around to

the tip of F4 one returns to the origin. Thus for

F2
the tail of
the particular case shown, the object has zero
acceleration, since the net force acting is

0.

F4

F3
When the vector sum of all forces acting on a certain object is zero, then the object does not accelerate.
Thus its velocity vector is constant (not necessarily zero, so the object need not be at rest). This special
case of Newton’s second law is also covered by Newton’s first law (the “law of inertia”).
Prediction problem: (You are going to be doing this in class, so think about it ahead of time.) Discuss
st
amongst your group members and figure out a formula for Newton’s 1 law, both the condition and the
conclusion, using vectors. Like “If and only if (such and such formula) then (such and such other
formula)”. Write it on the white board and copy it here.
If the vector sum of forces acting on an object is zero
 F  0,
then both x- and y-component are separately zero. So the above equation can be written alternatively
F
x
 0,
F
y
 0.
That is, you can add all the x-components together, and then add all the y-components together, then
check separately to see whether both sums are zero or not. This is how you check the equilibrium
condition.
In the exercise below you will see that the above condition for equilibrium is in fact the case for several
situations and in particular to see what the phrase “vector sum of all forces acting on an object” means.
st
Note that the 1 law begins with the statement of an if-and-only-if condition. Be able to remember this
when your TA asks.
PRE-LAB EXERCISE (Do these problems before you come to lab.)
a and b shown on the right
sketch to the right their difference c  b  a .
1. Given the vectors
a
b
a
b
Label the vectors.
a and b shown on the
right, sketch the combination c  2a  b .
2. Given the vectors
Label vectors.
4-2
Force Table
a with the x-y axes shown only has a y-component. The x-component of a is zero. Draw
another set of x-y axes near the vector b such that b only has a y-component in this new coordinate
3. Vector
system.
y
a
b
x
c is shown on the right. Show the xcomponent, c x on the diagram. Is the x-
4. The vector
y
component positive or negative?
Show the y-component,
c y on the diagram. Is the
c
y-component positive or negative?
In terms of the magnitude c and

:
x
Write an expression for c x :
Write an expression for c y :
5. The vector D is shown on the right. Show the xcomponent Dx on the diagram. Is the x-
y
x
component positive or negative?
Show the y-component,
Dy on the diagram. Is
the y-component positive or negative?
In terms of the magnitude
D and  ,
Write an expression for Dx :
D
Similarly, write an expression for Dy :
4-3

Force Table
6. Three forces as shown in the figure right
are applied at an object A.
Are these forces equal? If not why not?
F2  15N
F3  15N
F1  15N
150
30
A
What is the net force on A?
To make the net force on A be 0 , what
force (magnitude and direction) should be
added to the three existing forces?
Equipment: Force Table, weigh hangers, pulleys (4), mass set. Your computer has a calculator.
Pulley
Central peg
Force table from top showing ring
(not quite centered) center peg
and pulleys
Ring
4-4
Force Table
Problem 1 Forces in equilibrium
When the vector sum of all forces acting on a certain object is zero, one expects that the object does not
accelerate. Thus its velocity vector should remain constant (not necessarily zero, so the object need not
be at rest). This special case of Newton’s second law is also covered by Newton’s first law (Galileo’s “law
of inertia”).
Suppose you run a high-priority package
delivery service that will deliver, at the fastest
It’ll be better if it
possible rate of speed, night or day, and
doesn’t get
reasonably small package by pickup truck to oiljiggled too much,
if
you
know what
company and well logging-company crews
I mean.
working out on the prairies of Wyoming. A
company will pay you a great deal of money to
transport a dangerous, fragile cargo over rough
terrain at high speed. You figure you can
suspend the cargo in a sealed container inside
a barrel of oil to damp vibrations. But to do so,
you need to use a system of stabilizing wires to
keep the cargo in equilibrium at the center of
the damping fluid. It isn’t good enough to trust
what you think you know from your physics
book, although you have to start there. This leads you to consider the following problem experimentally:
Prediction Questions: How do the forces acting in a plane on an object balance one another? What does
it mean for them to “balance”? First, what is the definition of balancing and then, further, what is the
result of the fact that they balance? Why are these two questions separate?
(Questions continued) Discuss these questions with your group, report the answers on the white board
and also in this space.
From the picture, you can see that there must be forces acting vertically on the ring; for example gravity,
or the weight force, must act downward. The forces due to tension in the four attached strings are nearly
horizontal, but they must not be exactly horizontal, since they must provide some upward, vertical force
component to balance the weight. In this experiment, we will ignore the small upward component of the
string forces, since we will be concerned with the horizontal components. The horizontal components
must cancel one another, and (independently) the vertical components must cancel one another, in order
that the ring remains at rest. Moreover, each string exerts a horizontal force on the ring (along the string
directed away from the ring) that has a magnitude nearly equal to the weight of the mass plus mass
hanger attached to its end.
Recall that force, including weight, is a push or pull measured in Newtons. Mass (measured in kilograms)
is the resistance of an object to acceleration. When you place masses on the hangers they cause
increased tension in the strings. If the total mass of mass hanger plus slotted mass is 150 grams, then
2
the tension in the string is the same magnitude as the weight, or mg = (.150 Kg)(9.80 m/s ) = 1.47
Newtons. Thus, to get the magnitudes of the forces, you should multiply the mass values by g.
4-5
Force Table
Exercise 1: Be sure the central peg is in place to hold the ring in place when the tension forces do not add
o
to zero. Mount a pulley at the 20 mark and suspend a total of 100 grams from the string. Mount another
o
pulley at 120 and hang a total of 200 grams from the string.
(a) Using either components or a scale diagram, determine the magnitude and direction of the sum of the
two forces. Record your prediction here.
(b) Now put another pulley in place to provide a third force. Place the pulley and adjust the masses you
hang on the string so that the ring remains in place at the center without the central peg. Notice that
this additional force will have to be equal in magnitude and opposite in direction to the other two in
order to make the sum of the three forces equal zero. Before removing the center peg, be sure the
strings attach to the ring such that the line of each force passes through the center of the ring.
Otherwise the angles will be off.
(c) Explain why the result either does or does not support your calculation in part (a).
Exercise 2: Mount the first two pulleys the same as in the previous exercise, and place the same
o
corresponding masses on the strings. Then place the third pulley at 220 and suspend a total of 150
grams from the string.
(a) Determine the magnitude and direction of the sum of three forces. (This can be done by adding the
third vector to the sum of the first two, which you already found in the first exercise.)
4-6
Force Table
(b) Again use the force table to check to see whether your calculation is correct. As before, you can
check by adding a vector that should be equal in magnitude and opposite in direction to the sum.
(c) Assess the extent of the agreement between your computation and the measurement on the force
table.
Exercise 3: After finishing the second exercise, replace the peg and add 20 grams to the mass hanging
on the string over pulley four.
Question 1: How can you tell by looking at the force table that the tension forces on the ring do not add to
zero now?
Now remove the peg.
Question 2: After half a second, the ring does not move. Since it stays not moving, its acceleration must
be zero, so that the vector sum of the horizontal forces acting on it must still be zero. How can this be
when you just changed one of the forces? Look closely at the apparatus.
4-7
Force Table
Conclusion: At the end of class there will be a discussion. Record the most important points from the
discussion. Also, record your own group’s conclusions on the following: How do these laboratory
exercises relate to Newton’s second law? (Yup. The second.) State Newton’s second law and explain
the connection. How about Newton’s first law? What are two main points of the laboratory exercise?
Reference
1. Pre-lab problems are adapted from David R. Sokoloff and Ronald K. Thornton, Interactive lecture
Demonstrations: Active Learning in Introductory Physics (John Wiley & Sons Inc. 2004).
4-8