e101 - resolution of forces

7/21/2019 E101 - Resolution of Forces

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ANALYSIS

Physics is not all about the study of motion, including accelerations, which

are changes in velocities. Physics is also about what can cause an object to

accelerate. That cause is a forcewhich is a pull or push upon an object resulting

from the interaction with another object. The force is said to act on the object to

change its velocity. Whenever there is an interaction between two objects, there is

a force upon each of the objects. When the interaction comes to an end, the two

objects no longer experience the force. Force only exist as a result of an

interaction. For example, when a car slams into a telephone pole, a force on the

car from the pole causes the car to stop.

Force is a quantity that is measured using the standard metric unit nown as

theNewton, which is abbreviated by an !"#. $ne "ewton is the amount of force

required to give a % g mass an acceleration of % m&s&s. ' force is a vector quantity,

which is a quantity that has both magnitude and direction. To fully describe the

force acting upon an object, you must describe both the magnitude, si(e or

numerical value, and the direction.

' single vector can be broen down into two components, namely its

hori(ontal and vertical vectors. The tas of determining the amount of influence of

a single vector in a given direction involves the use of trigonometric functions, that

is through the component method. To determine the vertical component of a vector,

cosine function is used while sine function is used in a hori(ontal component.

'fterwards, the Pythagorean Theorem is used in order to come up with the

magnitude and direction of the resultant. 'nother one would be to through the

parallelogram method, a graphical representation of the vectors. The single vector

which effect is the same as that of vectors when added is called a resultant, which

is the sum of vectors. The vector that balances a resultant is called the equilibrant,


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which is equal in magnitude with the resultant but oppositely directed. )quilibrium

is a state of balance. When an equilibrium is at rest, it is a static equilibrium. The

first condition of equilibrium states that the sum of all forces acting on a body or

system is (ero.

*n the performed experiment, concurrent forces act on a ring pulled by +

strings with loads attached to its ends. The objectives of this experiments are to

determine the resultant force of concurrent forces using the raphical and

'nalytical method, to determine the first condition of equilibrium and its

implication, and to differentiate scalar from vector quantities and compare resultant

from equilibrant. The materials that were used are force table, super pulley with

clam, mass hangers, slotted mass, and a protractor. We are given some instructions

for equipment care such as extra care should be given on the super pulleys to avoid

damages and use reasonable mass on the hanger. The goal is to find the equilibrant,

in order to now what the resultant of the three vectors is. This was done in a two-

dimensional plane. We need to find the equilibrant E . *t is equal to the

resultant/s magnitude but in opposite direction.

R=E

%0

*n the graphical method, accuracy in drawing and measuring length of lines

is needed. ' ruler and a protractor are needed as the resultant will depend on the

length of the arrow obtained. *n addition, nowledge in scaling is important. This

method includes the parallelogram and polygon methods.


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The parallelogram method is only used for adding two forces. *n this, the tail

of the arrow is in the same location of the two vectors or origin and a

parallelogram is drawn with the diagonal line representing the resultant force.

Figure % shows two forces, F% and F1,added with the resultant 2.

F1 R

F%Figure 1

$n the other hand, the polygon method is used if there are more than two

given forces. The arrows in this method are connected head to tail. The arrow that

closes the polygon will be the resultant 2. The tail of the resultant is located at the

tail of the first vector while its head is pointed toward the head of the last vector.

Figure 1 shows the resultant 2 is drawn from the tail of the first vector F% to last

vector F3.

R F%

F3 F1

Figure 2

For analytical method, a more reliable way of finding a resultant vector is by

the component method. ' component of a vector is the projection of the vector on

an axis. *n this method, the resultant may be determined by adding first all the


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components of forces along the x-axis, Fx and along the y-axis, Fy and

then using the following equations to get the resultant and its direction where Fn

is the last force being added.

Fx=F1x+F2x+F3x +Fnx 10

Fy=F1y+F2y+F3y +Fny 30

R=(Fx)

2

+(Fy )

2

+0

tan1(FyFx

) 40

Force is a vector quantity that, when applied to a rigid body, has a tendency

to produce translation, which means a movement in a straight line. When the

equilibrant was identified, the ring in the middle did not experience a change in

motion anymore. This means that it reached its state of equilibrium.


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To start with )xperiment

%5%6 2esolution of Force, we are

given 7 procedures to help us

solve this experiment. Without

these procedures, we will not have

any idea what to do or what to

expect. The first procedure is

assembling the system using four

pulleys of the force table. 's

shown in Figure 3, a force board

or force table0 that has three or

more0 chains or cables attached to a

centre ring. The chains or cables

exert forces upon the centre ring in

three different directions. Typically

the experimenter adjusts the direction of the three forces, maes measurements of

the amount of force in each direction, and determines the vector sum of three

forces. Forces perpendicular to the plane of the force board are typically ignored in

the analysis and it is use with a ring, bench pulley, and string.


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The second procedure is to attach a hanger at the end of each string that

passes over a frictionless pulley and arbitrarily suspend a mass on each hanger.

These strings are used to place the slotted mass. 's shown in figure +,

the slotted mass is placed on the

hanger.

The third procedure is by trial

and error. 's shown in Figure 4, the

ring must be placed at the center by

adjusting the angle of the strings or

vary the load on the hanger. This

procedure requires patience in

getting the right angle and loaded

mass.

The fourth procedure is to

pull the ring slightly to one side and

then release. Then afterwards,

observe if the ring returns to the center. 'nd if not, going bac to the third

procedure is a must and adjust the position or load of one string if balance is

difficult to obtain.

The fifth procedure

is to record the mass on

each string and its angle

as

F1

, F2

, F3

, F4

, 1

,2

, 3

4

respectively once the ring

MASS ANGLE

F1= 14g 1=15

F2= 14g 2=65

F3= 14g 3=280


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is at the center and

balanced just lie in

Table %.

F4= 50g 4=151

Table %

The sixth procedure is determine the resultant ofF

1, F

2, F

3

, using the

polygon method and the component method. Then compare the result of F4.

The last procedure is to perform another trial by repeating procedures % to 8 but

with a different mass on the hangers or different angles for the strings.

*n trial %, as shown in Figure 8, the value of the fourth mass is 45 grams and

its angle is %4% degrees. 's shown in Table 1.%, using the Polygon 9ethod, the

2esultant is equal to 45g while the angle is 314o. The Polygon 9ethod/s

percentage error for the 2esultant is 5.55: while for the 'ngle is %.;%:. 's shown

in Table 1.1, using the


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error of F=|505050 |x 100=0.00

error of =|331325331 |x 100=1.81

=Fx=F1x+F2x+F3x

Fx=25cos15 +25cos65 +55 cos 280 Fx=44.26 g

=Fy=F1y+F2y+F3y

Fy=25sin15 +25sin65 +55sin 280 Fy=25.04 g

R=( Fx )2+ ( Fy)2 R=( 44.26 )

2+ ( 25.04 )2 R=50.85

( Fx

Fy)=29.5o Sof E330.5o

=tan1

error of F=|5050.8550 |x100=1.7

error of =

|

331330.5

331

|x 100=0.15

Table 1.% Table 1.1

While in trial 1, as shown in Figure 7, the value of the fourth mass is %14

grams and its angle is %7? degrees. 's shown in Table 3.%, using the Polygon

9ethod, the 2esultant is equal to %17g while the angle is 38% o. The Polygon

9ethod/s percentage error for the 2esultant is %.85: while for the 'ngle is 5.48:.

's shown in Table 3.1, using the


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FIGURE 7

%1+.4g while the angle is 385o. The


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error of F=|125127125 |x100=1.60 error of =|359361359 |x 100=0.56

Table 3.%

=Fx=F1x+F2x+F3x Fx=75cos0 +35sin 45+35sin315 Fx=124.5 g

=Fy=F1y+F2y+F3y Fy=75sin 0+35sin 45+35sin315 Fy=0 g

R=( Fx )2

+ ( Fy)2

R=( 124.5 )2

+( 0 )2 R=124.5 g

( Fx

Fy)=0o360o

=tan1

error of F=|125124.5125 |x 100=0.4 0 error of =|359360

359 |x100=0.28

Table 3.1

The biggest percentage error that we have acquired is %.;%: which was

from the first trial at the polygon method. This means that the ring was near at the

equilibrium state but not entirely at the center. There are numerous factors that can

be found. $ne of these factors could be the uneven flooring of the room. We

noticed how the table is shaing too much. *t could either be the floor or the table

itself.

Based on the frequent adjustments in the trial and error part, we can say that

F+is entirely dependent on F%, F1, and F3. )verything on the force table must be

balanced in order to get the ring to its equilibrium state.


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For every trial, we somehow get the correct load at the first to fifth try. 's

shown in Table +.% and Table +.1, there is somehow a clue in order to get the right

be closed to F3/s value to mae it balance. @iewise, in trial 1, the loaded mass in

F1and F3is 34g while F%is 74g, F+must be closed or higher than F3.

MASS ANGLE

F1= 14g 1=15

F2= 14g 2=65

F3= 14g 3=280

F4= 50g 4=151

Table +.% C Trial %

MASS ANGLE

F1= 74g 1=0

F2= 34g 2=45

F3= 34g 3=315

F4= 125g 4=179

Table +.1 C Trial 1

CONCLUSION

Forces are vector quantities. Their magnitudes are defined in terms of the

acceleration they would give the standard mass.


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adjusting the fourth mass and angle such that while plastic ring aligns with the

blac ring printed on the force table. Dince there is no resultant and is balanced,

then the sum of all forces acting on a body or a system is equal to (ero. The

positioning of the ring is important because it tells whether the force table

apparatus is at equilibrium or not. The mass of the hangers cannot be disregarded

because its weight can still affect the equilibrium even when there is no slotted

mass on it. When a pull is applied on the ring and then released, it sometimes fail

to return to the center might because that the other three pulls still have rings on it

or the string fail to retract because of the other three pulls that are connected to it.

The significance of the resultant of F%, F1, and F3 to the remaining force F+0 is

somehow or equal, in some cases, to the magnitude of F+. The efficient, accurate

and practical to use for me is the analytical method because it only requires

calculations which is more precise than the polygon method. The polygon method

is more time consuming and prone to error.

e101 - resolution of forces

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