Download Unit_2_Part_2---Forces_in_2

Unit 2, Part 2----Forces in 2-D
Vector Resolution
In the previous section, you learned how to take perpendicular vectors and “add” them to find one vector (the
resultant) that could cause the same action as the original two. In order to do some other types of physics’ problems,
you will need to do the exact reverse of finding the resultant. You’ll need to do this when the problems give you
some vector (could be a force vector, a velocity vector, etc) that is acting at some degree away from the horizontal.
Then it will say find the “_______”, and you’ll have to use one of the 4 funky equations or the force equations to
find whatever it is that the problem asked you to find. You need to remember that all the equations in this class
are specific for direction. All your variables in each of those equations must be directed in the same
dimension (either all horizontal or all vertical) in order to get the right answer. So, if you’re looking for some
vertical “distance” something went up in the air, you would have to use vertical velocities and a vertical
acceleration, not a horizontal velocity to calculate it.
For example: If someone is pulling a sled forward with a rope, but the rope they’re pulling it with isn’t exactly
parallel with the ground (it’s at some angle), some of the force that’s in the rope is directed vertical (the vertical
component of the force) and some of the force is directed horizontal (the horizontal component of the force). To
calculate the rate the sled was accelerating along the horizontal ground, you would need ONLY the amount of force
that was directed horizontally. The vertical component of the force wouldn’t have anything directly to do with the
forward motion of the sled.
10N
2kg

Let’s say in the above example a 2-kg sled is being pulled along the ground (assume there is NO friction for this
example) by a rope making a 300 angle with the ground. The sled is being pulled with an applied 10N force in the
rope at that angle. If you wanted to calculate the acceleration the sled was undergoing on the ground, you could use
any equation that has “a” (acceleration) in it. The only one that would work with the information you have been
given is “Fnet = ma”. However, the “a” in that equation is a horizontal vector (it’s being pulled along the
ground….”horizontally”). Therefore all other vectors (i.e., the “FA and the Ff”) in the equation must also be
horizontal vectors. You can’t just use the “10N at 300” because that’s not the amount of force that’s directed
horizontally. So, you make a triangle and find the “horizontal applied force component” ( FAx) of the triangle (how
much of the 10N is directed horizontally) using trig. It will look like this:
FA = 10N
FAy

FAx
To find the horizontal component of the applied force (FAx), it is the adjacent side of the angle in the right triangle.
You could use the cos function to calculate it.
Cos 300 =
Fx / 10 N

Fx = (cos 300) (10N)

8.7N
The amount of the 10N force on the sled that has anything at all to do with its horizontal motion is only 8.7N and
it’s directed to the right (look at the way the “x-axis” vector points).
So, to calculate the acceleration of the sled along the ground , use
Fnet (x) = max
Fnet(x) = 8.7N
m=
2-kg
a=
?
(FA = Fnet since Ff was stated to be 0N early in the problem)
Unit 2 Notes Part 2 ---Forces in 2-D
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*Only the amount of the force directed horizontally pulls the sled forward. The amount force directed vertically (Fy)
pulls the sled “up”. This will affect the normal force (the N), which is the amount of force the ground pushes back
up on the sled with, which in turn will affect the friction.
Adding non-perpendicular vectors (vectors that don’t make a 900 angle to each other)
If you need to add two vectors that aren’t perpendicular to each other, you have to first break each vector down into
its horizontal component (x-direction) and its vertical component (y-direction) and then add/subtract all the vertical
components together and add/subtract all the horizontal components together. The easiest way to visualize this is by
the “log problem” (this is when physics begins to get fun!!!). Picture a log with two ropes pulling up on it as in the
diagram below. What is the net force on the log?
8N
12N
“a”
“b”
600
100
log
The above drawing is a log that has two ropes attached to it. Two people are pulling on the ropes. Rope “a” has a
force of 8N in it and the rope is 120 0 away from 00 (so angle “a” is making a 600 angle with the closest horizontal).
Rope “b” has a force of 12N in it and it’s 10 0 from zero (so angle “b” is making a 100 angle with the closest
horizontal).
Question: What is the resultant force acting on the log?
Since the two vectors aren’t perpendicular to each other (they don’t make a 90 degree angle with each other), you
can’t just do vector addition (adding the two vectors head-to-tail) to find the resultant. But, if you found the x and y
components of each of the vectors, you could then add those together easily and begin to solve the problem. So,
first find the vertical (y) and horizontal (x) components of each vector. Here’s how: In the work below, Fx(a)
stands for the horizontal component of the force for vector “a”, etc…..
“Fx (a)”
“Fx (b)”


cos 600 = Fx (a) / 8N
cos 100 = Fx (b) / 12N
Fx (total) = Fx (b) - Fx (a)


Fx (a) = (cos 600) (8N)

Fx (a) = 4N, left

Fx (b) = (cos 100) (12N)

Fx (b) = 11.8N, right
11.8N – 4 N = 7.8N, right
So the TOTAL horizontal force from BOTH ropes together is 7.8N to the right. Since they are in the same plane
(both directed horizontal) you can just add/subtract them without doing any complicated math process.
Unit 2 Notes Part 2 ---Forces in 2-D
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To find the TOTAL vertical force (the Fy (total)), you do the same thing that we did above, but solving for the “y”
direction.
Fy (a)
Fy (b)
sin 600 = Fy (a) / 8N 
=
=
sin 100 = Fy (b) / 12 N 
Fy (total) = Fy (a) + Fy (b)

Fy (a) = (sin 600) (8N)
Fy (b) = (sin100) (12N)


6.9N, up
2.1N, up
(6.9N) + (2.1N) = 9N, up
Now what you have is a TOTAL horizontal force and a TOTAL vertical force. These are now perpendicular to each
other so you can do vector addition to find the resultant.
R
Fy (total)

Fx (total)
(9N)2 + (7.8N)2 = R2
tan = 9 N
11.9N
 = 490
tan = 1.15
7 .8 N
So, the two original vectors pulling on the log can be replaced with a single force of 11.9N acting at 49 0 and the
same thing will happen to the log.
Equilibrium and the Equilibrant Force
If an object is in equilibrium, the overall net force acting on the object is zero (and the acceleration is zero…).
If there are a bunch of different forces acting on an object, you could find the resultant force by doing everything
that you just learned (adding the force vectors together using vector addition). If the object is in equilibrium, there
must be a single force (called the equilibrant, “E”) that is the exact same size/magnitude as the resultant, but in the
exact opposite direction (1800 away from the resultant).
“R”
A
“E”
B
Picture the above diagram with two vectors (A and B) acting on a box. The resultant (R) is some value at some
angle. The object is in equilibrium. If the value of the resultant happens to be 12N at 30 0, the value of the
equilibrant is 12N at (300 + 1800) 2100.
Unit 2 Notes Part 2 ---Forces in 2-D
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A perfect example of forces in equilibrium is the “sign problem”. Picture the below diagram as a sign that weighs
168 N hanging from two ropes (a and b) that each make a 22.5 degree angle with the closest horizontal. The sign is
not moving. What forces does each rope exert on the sign?
a
b
mg = 168N
The question is really asking:
What is the amount of force in each rope (the tension, T, in each of the ropes)?
Before you begin this problem, you need to realize that the sign is in equilibrium, it’s not moving up or down or
side-to-side (hanging signs usually are in equilibrium). Therefore, the TOTAL downward force must equal the
TOTAL upward force and the total force in the left direction must equal the total force in the right direction. The
Net force acting on it is zero…the sign is in equilibrium, it’s not moving.
The total vertical downward force (as shown by the downward vector, mg) is just the weight of the sign
(mg = 168N). So, since the sign isn’t moving there MUST be a total upward force of 168N. Each of the ropes
make an equal angle with the closest horizontal (22.50), so the amount of force in each rope (T) must also be equal
(right triangles with equal angles have equal lengths of their sides, therefore T is the same for both ). So, the total
upward force (168N) must be equally divided between the two ropes. However, that value isn’t the tension in the
ropes, it’s the vertical component (Fy) of the tension in the ropes. Each rope has a horizontal (Tx) component and a
vertical (Ty) component. I’ve drawn the vertical components for you in the diagram below:
Ty (a)
Ty (b)


mg
So, Ty (a) and Ty (b) must both be equal, and added together they must be equal to 168N. Therefore, each of the
Ty’s must equal 84N (which is one half of 168N). In order to find the amount of force in each of the ropes (T), you
can now use the “sin” function since you have the value of the angle, the side opposite it (Ty) and you’re looking for
the hypotenuse.
Sin 22.50 =
84 N
hyp
hyp =
84N
sin 22.5
T = 219.5N
Here are some easy practice problems:
1.
A box is pulled across a wooden floor with a rope. The rope makes a 60 degree angle with the floor. A
force of 75 N is exerted on the rope. What are the components of the force parallel and perpendicular to
the floor?
Draw a picture. In your diagram, you’re looking for the values of F x and Fy . The 75 N is the tension in the rope
and the rope should be the hypotenuse in the right triangle. The 60 0 angle is the angle the rope makes with the
horizontal (Fx).
Answer: Fx= 37.5N, Fy = 64.95N
Unit 2 Notes Part 2 ---Forces in 2-D
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2.
A student exerts a force of 72 N along the handle of a lawn mower to push it across a lawn. Find the
horizontal and vertical components of the force when the handle is held at 60 degrees, 40 degrees and 30
degrees to the lawn.
Answer:
x
y
600 = 36N
62.35N
400= 55.16N
46.28N
300= 62.35N
36N
The 72N is the force in the handle (the FA and will be the hypotenuse in the diagram). You’ll be looking for Fx and
Fy for each of the three angles given.
3.
A 40-kg crate is pulled across the ice with a rope. A force of 100 N is applied at an angle of 30 o with the
horizontal. Neglecting friction, calculate
a. the upward force the ice exerts on the crate as it is pulled.
b. the acceleration of the crate
a. this question is asking for the normal force (N)
m = 40 Kg
mg = 392 Kg, down
FAy = 50N, up
N = 392N-50N = 342N
b. to find the acceleration of the crate, you first need to find the force that’s pushing it in the horizontal
dimension (FAx) .
FAx = 86.6N
Fnet = ma
FAx -Ff = ma
FAx = ma
(Ff is zero as stated in the problem)
FAX
a
m
a = 2.17 m/s/s
4.
Now find the acceleration of the crate in the previous problem if there is friction and the coefficient of
friction is 0.09.
FAx = 86.6N
m = 40 Kg
mg = 392 Kg
FAy = 50N
N = 392N-50N = 342N
μ = 0.09
Ff = μ N = (342N)(0.09) = 30.78N
FAx - Ff = ma
FAX  F f
m
a
Unit 2 Notes Part 2 ---Forces in 2-D
86.6 N  30.78 N
a
40 Kg
a = +1.39 m/s/s, speeding up
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Joe pushes on the handle of a 10-kg lawn mower. The handle makes a 45o angle with the horizontal. Joe
wishes to accelerate the mower from rest to 1.39 m/s in 1.5 s. What force must Joe apply to the handle?
Neglect friction
* You ‘re looking for the actual force in the handle. To do this, you’ll need to find the “horizontal component”
of the force by first finding the horizontal acceleration using the four funky equations. Once you find the
horizontal acceleration, you can then find the horizontal net force (which is equal to the horizontal applied force
in this particular problem since there is no friction) using the F net equation. Once you find the horizontal force,
you can use trig to find the value of the force in the handle, which is the “hypotenuse” in your diagram.
Answer: 13.15N
5.
6.
A net force of 55N acts due west on an object. What added single force on the object will produce
equilibrium?
Answer: 55N, East
7.
Two forces act on an object. One force is 6N, East. The other force is 8N, North. (a) What is the
resultant? (b) If the object is in equilibrium, find the equilibrant.
Answer: 10N at 53 degrees North of East, 10N at 223 degrees. (away from East)
8.
A 62N force acts at 30 degrees and a second 62N force acts at 60 degrees. (a) What is the resultant?
(b)What is the equilibrant?
Answer: 120N at 45 degrees, 120N at 225 degrees
9.
Two forces act on an object. A 36N force acts at 225 degrees and a 48N force acts at 315 degrees. What
would be the equilibrant?
Answer: 60N at 99 degrees
10. If the sign in the “sign problem” (earlier mentioned in the notes
) is now hung by ropes that each
make an angle of 42 degrees with the horizontal, what force will each rope exert now?
Answer: 126N
11. If the ropes in the “sign problem” were pulled more horizontal to raise it higher, and the tension in each
rope increases to 575N (the angles are still equal), what angle does each rope make with the horizontal?
Answer: 8.4 degrees
Unit 2 Notes Part 2 ---Forces in 2-D
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