Download Determining the Net Force

NEWTON’S LESSON 5
Determining the Net Force
In order to continue, Newton's first law of motion must be thoroughly understood:
An object at rest tends to stay at rest and an object in
motion tends to stay in motion with the same speed and in
the same direction unless acted upon by an unbalanced
force.
In the statement of Newton's first law, the unbalanced force refers to that force which
does not become completely balanced (or canceled) by the other individual forces.
If either all the vertical forces (up and down) do not cancel each other and/or
all horizontal forces do not cancel each other, then an unbalanced force
exists.
A free body diagram indicates if there is an unbalanced force.
What are the unbalanced forces in these free body diagrams?
The net force is the sum of all the forces.
A force is a vector and two forces of equal magnitude and opposite direction will cancel
each other out.
Observe the following examples of summing two forces:
Observe in the diagram above that a downward vector will provide a partial or full
cancellation of an upward vector. And a leftward vector will provide a partial or full
cancellation of a rightward vector. The addition of force vectors can be done in the same
manner in order to determine the net force (i.e., the vector sum of all the individual
forces). Consider the three situations below in which the net force is determined by
summing the individual force vectors which are acting upon the objects.
Recall, a net force (i.e., an unbalanced force) causes an acceleration.
Use this knowledge to determine whether or not a net force exists in the following
situations.
Description of Motion
Net Force: Yes or No?
Yes! There is a net force
since there is an acceleration.
No! There is no net force
since there is not an
acceleration.
Yes! There is a net force
since there is an acceleration.
No! There is a no net force
since there is not an
acceleration (zero slope on a
v-t graph means zero
acceleration).
Yes! There is a net force
since there is an acceleration
(the slope on a v-t graph
means acceleration).
Yes! There is a net force
since there is an acceleration
(the slope on a v-t graph
means acceleration).
The Big Misconception
What is so terribly difficult about remembering that F = ma (force causes acceleration),
or a body in motion stays in motion in the absence of a force?
The big deal however is not the ability to recite the first law nor to use the second law to
solve problems; but rather the ability to understand their meaning and to believe their
implications.
PHYSICS CONCEPTS CAN NOT JUST BE MEMORIZED, THEY MUST BE
UNDERSTOOD!
When you entered this classroom, you already had oodles of experience with physics.
Through this you developed a set of beliefs which sometimes are not easy to change.
To change misconceptions involves self-reflection (to ponder your own belief systems),
critical thinking (to analyze the reasonableness of two competing ideas), and evaluation
(to select the most reasonable and harmonious model which explains the world of
motion).
The most common misconception is one which dates back for ages; it is the idea that
sustaining motion requires a continued force.
Newton's laws declare loudly that a net force (an unbalanced force) causes an
acceleration; the acceleration is in the same direction as the net force. To test your own
belief system, consider the following question and its answer.
MISCONCEPTION TESTING QUESTION:
Two students are discussing their physics homework prior to class. They are discussing
an object which is being acted upon by two individual forces (both in a vertical
direction); the free-body diagram is:
Anna Litical suggests to Noah Formula that the object under discussion could be
moving.
Anna’s argument: If friction and air resistance could be ignored (because of their
negligible size), the object could be moving in a horizontal direction.
According to Anna, an object experiencing forces as described at the right could be
experiencing a horizontal motion as described by the ticker tape diagram:
Noah’s argument: Noah states the object could not have any horizontal motion if there
are only vertical forces acting upon it. Noah claims that the object must be at rest,
perhaps on a table or floor. After all, says Noah, an object experiencing a balance of
forces will be at rest.
WHO DO YOU AGREE WITH? WHY?
Remember as a kid when you went sledding down a hill and across the level surface?
Imagine a the moment that there was no friction along the level surface from point B to
point C and that there was no air resistance to impede your motion. How far would your
sled travel? And what would its motion be like?
Most students I've talked to quickly answer: the sled would travel forever at constant
speed. Without friction or air resistance to slow it down, the sled would continue in
motion with the same speed and in the same direction.
The forces acting upon the sled from point B to point C would be the normal force (the
snow pushes up on the sled) and the gravity force:
These forces are balanced and since the sled is already in motion at point B it will
continue in motion with the same speed and direction.
So, as in the case of the sled and as in the case of the object which Noah and Anna are
discussing, an object can be moving to the right even if the only forces acting upon the
object are vertical forces. Forces do not cause motion; forces cause accelerations.
Newton's first law of motion declares that a force is not needed to keep an object in
motion. Slide a book across a table and watch it slide to a rest position. The book in
motion on the table top does not come to a rest position because of the absence of a force;
rather it is the presence of a force - that force being the force of friction - which brings
the book to a rest position. In the absence of a force of friction, the book would continue
in motion with the same speed and direction - forever (or at least to the end of the table
top)!
A force is not required to keep a moving book in motion; and a force is not required to
keep a moving sled in motion; and a force is not required to keep any object horizontally
moving object in motion.
NEWTON’S LESSON 5 HOMEWORK
1. What is the maximum resultant force resulting from a 3-N force and an 8-N force?
______
2. What is the minimum resultant force resulting from a 3-N force and an 8-N force?
______
3. Consider the following statements in answering the next three questions.
A.
B.
C.
D.
E.
F.
G.
H.
I.
J.
The position of the object remains constant.
The velocity of the object remains constant.
The position of the object is changing.
The velocity of the object is changing.
The velocity of the object is 0 m/s.
The acceleration of the object is 0 m/s2.
The resultant of all the individual forces acting upon the object is 0 N.
The net force acting upon the object is 0 N.
All individual forces acting upon the object are equal.
The individual forces acting upon the object balance each other.
a. If an object is known to be at equilibrium, then which of the following statements
MUST be true?
b. If an object is known to be at equilibrium, then which of the following statements
MUST NOT be true?
c. If an object is known to be at equilibrium, then which of the following statements
COULD be true?
4. Free-body diagrams for four situations are shown below. For each situation, determine
the net force acting upon the object.
5. Free-body diagrams for four situations are shown below. The net force is known for
each situation. However, the magnitudes of a few of the individual forces are not known.
Analyze each situation individually and determine the magnitude of the unknown forces.
For the next several questions, consider the velocity-time plot below for the motion of an
object along a horizontal surface. The motion is divided into several time intervals, each
labeled with a letter.
6. During which time interval(s), if any, are there no forces acting upon the object? List
all that apply.
7. During which time interval(s), if any, are the forces acting upon the object balanced.?
List all that apply.
8. During which time interval(s), if any, is there a net force acting upon the object? List
all that apply.
9. During which time interval(s), if any, is the net force acting upon the object directed
toward the right? List all that apply.
10. During which time interval(s), if any, is the net force acting upon the object directed
toward the left? List all that apply.
For the next several questions, consider the dot diagram below for the motion of an object
along a horizontal surface. The motion is divided into several time intervals, each labeled
with a letter.
11. During which time interval(s), if any, are there no forces acting upon the object? List
all that apply.
12. During which time interval(s), if any, are the forces acting upon the object balanced.?
List all that apply.
13. During which time interval(s), if any, is there a net force acting upon the object? List
all that apply.
14. During which time interval(s), if any, is the net force acting upon the object directed
toward the right? List all that apply.
15. During which time interval(s), if any, is the net force acting upon the object directed
toward the left? List all that apply.
16. The launch velocity and angle is given for three different projectiles. Use
trigonometric functions to resolve the velocity vectors into horizontal and vertical
velocity components. Then use kinematic equations to determine the time that the
projectile is in the air, the height to which it travels (when it is at its peak), and the
horizontal distance that it travels. Don’t round, numbers need to be used in following
calculations.
a.
b.
c.
Given:
Given:
Given:
Launch Vel. = 30 m/s
Launch Vel. = 30 m/s
Launch Vel. = 30 m/s
Launch angle = 30 degrees above
horizontal
Launch angle = 45
degrees
Launch angle = 50
degrees
Calculate:
Calculate:
Calculate:
vix = __________
vix = __________
vix = __________
viy = ___________
viy = ___________
viy = ___________
tup = ___________
tup = ___________
tup = ___________
ttotal = ___________
ttotal = ___________
ttotal = ___________
y at peak = ___________
y at peak =
___________
y at peak =
___________
x = ___________
x = ___________
x = ___________
17. What angle above gives the projectile the maximum range (i.e., maximum horizontal
displacement)?
HOMEWORK KEY
1. 11 N
2. 5 N
3. a. B, F, G, H, and J, b. D, c. A,C,E, and I
4. a. 0N, b. 5N left, c. 0N, d. 15N up
5. A=50N, B=200N, C=1100N, D=20N, E=300N, F=H, G=50N
6. none
7. BDFH
8. ACEG
9. AE
10. CG
11. none
12. ACEGI
13. BDFH
14. BF
15. DH
16. a. ttotal = 3.06 s, dyf at peak = 11.5 m, dx = 79.5 m
b. ttotal = 4.33 s, dyf at peak = 23.0 m, dx = 91.8 m
c. ttotal = 4.69 s, dyf at peak = 26.9 m, dx = 90.4 m
17. 45 degrees
NEWTON’S LESSON 5 HOMEWORK
1. What is the maximum resultant force resulting from a 3-N force and an 8-N force?
______
Answer: 11 N
2. What is the minimum resultant force resulting from a 3-N force and an 8-N force?
______
Answer: 5 N
3. Consider the following statements in answering the next three questions.
K.
L.
M.
N.
O.
P.
Q.
R.
S.
T.
The position of the object remains constant.
The velocity of the object remains constant.
The position of the object is changing.
The velocity of the object is changing.
The velocity of the object is 0 m/s.
The acceleration of the object is 0 m/s2.
The resultant of all the individual forces acting upon the object is 0 N.
The net force acting upon the object is 0 N.
All individual forces acting upon the object are equal.
The individual forces acting upon the object balance each other.
a. If an object is known to be at equilibrium, then which of the following statements
MUST be true?
b. If an object is known to be at equilibrium, then which of the following statements
MUST NOT be true?
c. If an object is known to be at equilibrium, then which of the following statements
COULD be true?
Answer: If an object is at equilibrium, then all the forces acting upon the object would
balance each other. That is, the net force acting upon the object would be 0 N. If the
forces were added together as vectors in a head-to-tail fashion, then the resultant would
be 0 N. If the net force is 0 N, then the acceleration of the object would be 0 m/s/s. That
means the velocity of the object (whether it be 0 m/s or not 0 m/s) must be a constant
value. So an object which is at rest (v = 0 m/s) would stay at rest in the same position; but
an object which is moving would remain moving with that very velocity. So an object at
equilibrium could be moving with a constant velocity (and change its position) or could
be at rest (and remain at a constant position). For these reasons, the following answers
exist for questions #24-26.
a. B, F, G, H, and J
b. D
c. A, C, E, and I
4. Free-body diagrams for four situations are shown below. For each situation, determine
the net force acting upon the object.
Answer:
Situation A The net force is zero Newtons. All the individual forces balance each other
(i.e., cancel each other out).
Situation B The net force is 5 Newtons, left. The vertical forces balance each other (i.e.,
cancel each other out). The leftward force (friction) remains unbalanced.
Situation C The net force is zero Newtons. All the individual forces balance each other
(i.e., cancel each other out).
Situation D The net force is 15 Newtons, up. The upward force of air resistance is only
partially balanced by the downward force of gravity - 15 N of upward force remains
unbalanced.
5. Free-body diagrams for four situations are shown below. The net force is known for
each situation. However, the magnitudes of a few of the individual forces are not known.
Analyze each situation individually and determine the magnitude of the unknown forces.
Answer:
A = 50 N (the horizontal forces must be balanced)
B = 200 N (the vertical forces must be balanced)
C = 1100 N (in order to have a net force of 200 N, up)
D = 20 N (in order to have a net force of 60 N, left)
E = 300 N (the vertical forces must be balanced)
F = H = any number you wish (as long as F equals H)
G = 50 N (in order to have a net force of 30 N, right)
For the next several questions, consider the velocity-time plot below for the motion of an
object along a horizontal surface. The motion is divided into several time intervals, each
labeled with a letter.
6. During which time interval(s), if any, are there no forces acting upon the object? List
all that apply.
7. During which time interval(s), if any, are the forces acting upon the object balanced.?
List all that apply.
8. During which time interval(s), if any, is there a net force acting upon the object? List
all that apply.
9. During which time interval(s), if any, is the net force acting upon the object directed
toward the right? List all that apply.
10. During which time interval(s), if any, is the net force acting upon the object directed
toward the left? List all that apply.
Answer:
6. None - If an object is on a surface, one can be guaranteed of at least two forces gravity and normal force.
7. BDFH - If the forces are balanced, then an object is moving with a constant velocity.
This is represented by a horizontal line on a velocity-time plot.
8. ACEG - If an object has a net force upon it, then it is accelerating. Acceleration is
represented by a sloped line on a velocity-time plot.
9. AE - If the net force is directed to the right, then the acceleration is to the right (in the
+ direction). This is represented by a line with a + slope (i.e., upward slope).
10. CG - If the net force is directed to the left, then the acceleration is to the left (in the direction). This is represented by a line with a - slope (i.e., downward slope).
For the next several questions, consider the dot diagram below for the motion of an object
along a horizontal surface. The motion is divided into several time intervals, each labeled
with a letter.
11. During which time interval(s), if any, are there no forces acting upon the object? List
all that apply.
12. During which time interval(s), if any, are the forces acting upon the object balanced.?
List all that apply.
13. During which time interval(s), if any, is there a net force acting upon the object? List
all that apply.
14. During which time interval(s), if any, is the net force acting upon the object directed
toward the right? List all that apply.
15. During which time interval(s), if any, is the net force acting upon the object directed
toward the left? List all that apply.
Answer:
11. None - If an object is on a surface, one can be guaranteed of at least two forces gravity and normal force.
12. ACEGI - If the forces are balanced, then an object is moving with a constant velocity
or at rest. This is represented by a section of a dot diagram where the dots are equally
spaced apart (moving with a constant velocity) or not even spaced apart at all (at rest).
13. BDFH - If an object has a net force upon it, then it is accelerating. Acceleration is
represented by a section of a dot diagram in which the spacing between consecutive dots
is either increasing or decreasing.
14. BF - If the net force is directed to the right, then the acceleration is to the right (in the
+ direction). This is represented by a dot diagram in which the dots are increasing their
separation distance as the object moves from left to right.
15. DH - If the net force is directed to the left, then the acceleration is to the left (in the direction). This is represented by a dot diagram in which the dots are decreasing their
separation distance as the object moves from left to right.
16. The launch velocity and angle is given for three different projectiles. Use
trigonometric functions to resolve the velocity vectors into horizontal and vertical
velocity components. Then use kinematic equations to determine the time that the
projectile is in the air, the height to which it travels (when it is at its peak), and the
horizontal distance that it travels. Don’t round too early, numbers need to be used in
following calculations.
a.
b.
c.
Given:
Given:
Given:
Launch Vel. = 30.0 m/s
Launch Vel. = 30.0 m/s Launch Vel. = 30.0 m/s
Launch angle = 30.0 degrees above
horizontal
Launch angle = 45.0
degrees
Launch angle = 50.0
degrees
Calculate:
Calculate:
Calculate:
vix = __________
vix = __________
vix = __________
viy = ___________
viy = ___________
viy = ___________
tup = ___________
tup = ___________
tup = ___________
ttotal = ___________
ttotal = ___________
ttotal = ___________
dyf at peak =
dyf at peak =
dyf at peak = ___________
___________
___________
dx = ___________
dx = ___________
dx = ___________
Answer: The solutions to all three of these non-horizontally launched projectile problems
involve the use of kinematic equations and an appropriate problem-solving strategy. The
kinematic equations and their use in projectile problems are listed and discussed
elsewhere. In each of these problems, it is known that ax = 0 m/s/s and ay = -9.8 m/s2. The
values of vix and viy can be determined using trigonometric functions:
vix = vi * cos(theta)
viy = vi * sin(theta)
Once vix and viy are known, the other unknowns can be calculated. The time up to the
peak (tup) can be determined using the equation
vfy = viy + ay *t
where the vfy = 0 m/s (there is no vertical velocity for a projectile when its at its peak)
and ay = -10 m/s/s. Once tup is known, the ttotal (time to travel the entire trajectory -both up
and down) can be determined by doubling the tup. The horizontal displacement of the
projectile (x) can be computed in the usual way using the equation
Δdx = vix*t + 0.5*ax*t2
where t is the ttotal value, ax = 0 m/s/s and vix was the first value calculated (using the
trigonometric functions). Finally, the y at the peak (i.e., the peak height) can be
calculated using the equation
Δdy = viy*t + 0.5*ay*t2
where t is the tup value, ay = -9.8 m/s/s and viy was one of the first values calculated (using
the trigonometric function). The t value in the equation is tup because the peak height is
reached when the projectile has traveled for one-half of its total time; tup is that time. This
method will yield the following answers:
a.
b.
c.
Answers:
Answers:
Answers:
vix = 25.9807 m/s
vix = 21.2132 m/s
vix = 19.2836 m/s
viy = 15 m/s
viy = 21.2132 m/s
viy = 22.9813 m/s
tup = 1.5306 s
tup = 2.1646
tup = 2.3450 s
ttotal = 3.0612 s
ttotal = 4.3292 s
ttotal = 4.69 s
dyf at peak = 11.4795 m
dyf at peak = 22.9591 m
dyf at peak = 26.9459 m
dx = 79.5321 m
dx = 91.8361 m
dx = 90.4400 m
ROUNDED ANSWERS:
a. ttotal = 3.06 s, dyf at peak = 11.5 m, dx = 79.5 m
b. ttotal = 4.33 s, dyf at peak = 23.0 m, dx = 91.8 m
c. ttotal = 4.69 s, dyf at peak = 26.9 m, dx = 90.4 m
17. What angle above gives the projectile the maximum range (i.e., maximum horizontal
displacement)?
Answer: Observe that the maximum range (x) in the above three calculations results
when the angle of launch is 45-degrees. This is always the case for situations where air
resistance is negligible.