Adding Forces Resultant and Equilibrium, Resolving Forces Component - Copy

UNIVERSITI TEKNOLOGI MARA FAKULTI KEJURUTERAAN KIMIA

CHEMISTRY LABORATORY (CHE 175)

Remarks :

No Title Allocated Marks (%) Marks (%)1. Abstract/Summary 52. Introduction 53. Aims/Objective 54. Theory 55. Procedure 36. Apparatus 57. Results 208. Calculations 109. Discussions 2010. Conclusions 1011. Recommendations 512. References 5

13. Appendices 2Total 100

Checked by : Rechecked by :

NAME : ABDUL HALIM BIN NORDIN (2008293172) MUHAMMAD RAZI BIN ZAHARI (2008424824) NOOR SYAFIQAH AMERAH BT AHMAD TARMIZI (2008293072) SITI NOR SAMRAH BT A.RAHIM (2008291992)

NURUL ADILAH BT NASARUDDIN (2008292022)

GROUP : 2

EXPERIMENT : (7) – 1A : ADDING FORCES RESULTANT AND EQUILIBRIUM 1B : RESOLVING FORCES COMPONENT

DATE PERFORMED : 18 MARCH 2009

SEMESTER : DIS 2008 – APRIL 2009

PROGRAMME/CODE : DIPLOMA IN CHEMICAL ENGINEERING / EH 110

SUMMARY

The resultant (R) of two or more vector is the single vector which produce the same effect (in

both magnitude and direction). Consider two forces, P and Q acting at a point O. The resultant

will be vector R

R is a diagonal of the parallelogram of which P and Q are two sides. This provide a useful rule.

The parallelogram rule of vector addition. If two vectors OA and OB are represented in

magnitude and direction by the sides OA and OB of a parallelogram OABC, then OC represent

their resultant. In experiment 2, concurrent forces vector ally is added to determine the

magnitude and the direction of the combined force. In this experiment if will do the opposite; it

will find two forces which then added together have the same effect as the original force. Any

force vector in the x-y plane can be expressed as a sym of vector in the x direction and a vector

in y direction. The magnitude of the force independent of the length of the two ropes. The angle

are measure from the positive x-axis in the anti-clockwise direction, the angle have a positive

(+ve) sign. If the measurement is in the clockwise direction, the angle has a negative (-ve) sign,

the angle are measure from the (+ve) y-axis in the clockwise direction.

INTRODUCTION

The force that causes an object to fall due to gravity is called the weight of object. On the

surface of the earth, the acceleration of an object due to the gravity is about 9.8m/s². This is

usually referred to as gravity, so the equation can stated before in a little different format to

specify weight:

Fw = mg

The subscript ʷ is used to donate that is it the weight force. Notice that it also replace with y.

So when an object is placed on a table, it s not falling because the table is holding it up. The

object is said to be at equilibrium because it is not being accelerated. The net force on the table is

0 N. There is the weight force pulling the object down so something must be pushing up to make

the net force also 0 N. This force is called the normal force. The normal force (F N) is always

exerted perpendicular to the surface that is holding up the object, in this case it represent the

table.For a level surface, the magnitude of the normal force is equal to the magnitude of the

weight force, except that is acts in the opposite direction.Notice how the normal force is still

perpendicular to the slanted surface. It cancels out the component of the weight force

perpendicular to the slanted surface (shown by the dotted line perpendicular to the slanted

surface). But it does not cancel out the weight force entirely. So what is left is the component of

the weight force parallel to the slanted surface (shown by dotted line parallel to the slanted

surface).This is called the net force (Fnet). The net force is directed towards the bottom of the

ramp. This is why the block will slide down the ramp.The magnitude of the normal force is equal

to the magnitude of the component of the weight force perpendicular to the slanted surface.

Through simple trigonometry the magnitude of the normal force is:

FN = FW cos ᶿ

By the same reasoning the magnitude of the net force is:

F net = Fᵂ sin ᶿ

Notice that happen when the ᶿ equal to 0.(a level surface) since cos 0 =1 and sin 0 = 0 the

equation rarely simplify to:

Fᴺ=Fᵂ and Fnet=0

Addition of vector can be done graphically by placing the tail of the successive error at the tipoff

the previous one. The sum or the resultant vector is the arrow down from the tail of the first

vector to the last vector. Two vectors also can be added using parallel method. Vector can be

added more accurately by adding their component along chosen axis with the aid of trigonometry

functions. A vector of magnitude V making an angle ᶿ with the axis has component :

Vᵪ= V cos ᶿ Vᵧ= V sin ᶿ

Given the component, a vector magnitude and direction can be find from:

V=√V²x + V²y tan ᶿ =Vy/Vx

RESULTANT VECTORS

The resultant is the vector sum of two or more vector. It is the result of adding two or more

vector together. If displacement vector, A,B and C is added together, the result will be vector R.

As shown in the diagram scaled vector addition diagram.

To say that vector R is the resultant displacement of displacement vector A,B and C is to says

that a person who walked with displacement by the same amount as a person who walked with

displacement R. Displacement vector R gives the same result as displacement vector A+B+C=R

that is why it can be said that

A+B+C=R

The above discussion pertains to the result of adding displacement vectors. When displacement

vector are added, the result is a resultant displacement. But any two vectors can be added as long

as they are the same vector quantity. If two or more velocity vector is added then the result is a

resultant force.

In all such cases, the resultant vector (whether a displacement vector, force vector, velocity

vector,etc) is the result of adding the individual vector. It is the same thing as adding A+B+C+...

In summary, the resultant is the vector sum the entire individual vectors together. The resultant

can be determined by adding the individual forces together by using vector addition methods.

EXPERIMENT 1A.

OBJECTIVE

To find the equilibrant by adding the forces resultant experiment

THEORY

In figure 2.1 spaceship x and y are pulling on a asteroid with forces indicated by vectors Fᵪ and

Fᵧ. Since these forces are acting in the same point of the asteroid, they are called current forces.

As any vector quantity, each is defined both by its magnitude, which is proportional to length of

the arrow. (The magnitude of the force is independent of the length of the rope)

The total force on the asteroid can be determined by adding vectors Fᵪ and Fᵧ. In the illustration,

the parallelogram method is used. The diagonal of the parallelogram defined by Fᵪ and Fᵧ is Fᵣ

the vector indicating the magnitude and direction of the total force acting on the asteroid. Fᵣ is

called the resultant of Fᵪ and Fᵧ.

Another useful vector is Fᵉ, the equilibrant of Fᵪ and Fᵧ. Fᵉ is the forces needed to exactly offset

the combined pull of the two ship. Fᵉ has the same magnitude as Fᵣ but is in the opposite

direction. It will see in the following experiment the equilibrant provides a useful experimental

method for finding the resultant of two or more forces.

APPARATUS

-Experiment board

-Degree scale

-Three pulleys

-Masses

-Spring balanced

-Force ring

-Three mass hanger

-String

PROCEDURES

The equipment was set up as shown in figure 2.2. Mass hanger and mass provide a gravitational

force of F=mg downward. However, since the force ring is not accelerated, the downward force

must be exactly balanced by an equal and opposite, or equilibrant force Fᵉ is of course providing

by the spring balanced.

1. The magnitude and direction of F, the gravitational force provide by the mass and mass

of the hanger (F=mg) is

F: Magnitude = 0.49N

Direction = Downward

2. The spring balance and the degree plate was used to determine the magnitude and the

direction of Fe

Fᵉ: Magnitude = 0.49N

Direction = Downward

The pulley and hanging masses was used as shown in figure 2.3 and the equipment was set up so

that two to know forces, F1 and F2 are pulling on the Force Ring. The Holding Pin was used to

prevent the ring from being accelerated. The Holding Pin was provide a force, F ᵉ that is exactly

opposite to the resultant of F1 and F2.

The spring balance was adjusted to determined to magnitude of Fᵉ. The spring balance was kept

vertically and a pulley was used to direct the force from the spring in the desired direction. The

spring balanced was moved towards or away from the pulley to vary the magnitude of the force.

The pulley and spring balanced was adjusted so that the holding pin is centred in the forced ring.

3. The magnitude in Newton’s of F1,F2 and F3 was record the value of the hanging masses

M1,M2, and M3 (include the mass of the hanger) and also ᶿ1,ᶿ2,and ᶿ3 was recorded, the

angle each vectors makes with respect to the zero-degree line on the degree scale.

4. The value that record were used to construct F1,F2 and Fᵉ on a separate sheet of paper. An

appropriate scale was chose (2.0cm/ Newton) the length of each vector was make

proportional to the magnitude of the force it represent was indicate.

5. The parallel method was used to draw the result of F1and F2. The resultant Fᵣ was

labelled. The length of Fᵣ was measured to determine the magnitude of the resultant force

and this magnitude was record on the diagram.

6. The equilibrant force vector Fᵉ exactly balances the resultant vector Fᵣ. The magnitude

and direction of F1 and F2 was varies and the experiment was repeated.

RESULT FOR EXPERIMENT 1 A

1 The magnitude and direction of F, the gravitational force provide by the mass and mass

of the hanger (F=mg) is

F: Magnitude = 0.49 N

Direction = Downward

2 The spring balance and the degree plate was used to determine the magnitude and the

direction of Fe

Fᵉ: Magnitude = 0.49N

Direction =Downward

3 F1 : M1 = 0.050 kg Magnitude = 0.49 N Angle = 159˚

F1 : M1 = 0.070 kg Magnitude = 0.68 N Angle = 34˚

Magnitude = 0.98 N Angle = 112˚

Calculation For Experiment 1A

1. F = mg

F : magnitude = 0.050 kg (9.81 m/s2)

= 0.49 N

Direction = Downward

2. Fᵉ: Magnitude = 0.49 N

Direction = Downward

3. F1 : M1 = 0.050 kg Magnitude = mg Angle = 159˚

= (0.050kg)(9.81 m/s2)

= 0.49N

F2 : M2 = 0.070 kg Magnitude = mg Angle = 34˚

= (0.070kg)(9.81m/s2)

= 0.68 N

Fe: F = kx Magnitude = kx Angle = 112˚

= (40.88)(0.012m)

= 0.98 N

EXPERIMENT 1B

THEORY

In this experiment, concurrent forces are added vertically to determine the magnitude and

direction of the combined force. In this experiment, opposite way are using which is when two

forces are added together it will have the same effect as the original force. Any vector in x-y

plane can be expressed as the sum of a vector in the x- direction in a vector in the y- direction.

PROCEDURES FOR EXPERIMENT 1B

1. The equipment as shown in figure 3.1 was set up. A force, F as shown was determined by

hanging a mass from the force over a pulley. The holding pin was used to hold the force ring in

place.

The spring balance and a pulley were set up so the strings from the balanced run horizontally

from the bottom of the pulley to the force ring A. A second ring hanger was hanged directly from

the force ring.

Now the spring balance was pulled towards or away from the pulley to adjust the horizontal or x-

component of the force. The mass was adjusted in this way until the holding pin is centred in the

force ring. (Notice that this x and y component are actually the x and y component of the

equilibrant of the F rather than of F itself)

1. The magnitude and angle of F were recorded. The angle as shown in figure 3.1 was measured.

Magnitude= 1.03N angle= 53˚

2. The magnitude of the x and y component of the equilibrant of F were recorded.

X-component= 1.03cos53˚ y-component= 1.03sin53˚ - (0.105x9.81)

3. What are the magnitude of Fᵪ and Fᵧ, the x and y components of F?

Fᵪ= 0.62N Fᵧ= -0.21N

4.Record the angle of F, and the magnitude of F, Fx , and Fy.

F: Magnitude = 1.03N Angle= 53˚

Fx= 0.62N Fy= -0.21N

5.Record the magnitude and angle of the force vector, F , that you have constructed.

Magnitude = 1.03N Angle= 53˚

6.Calculate Fx and Fy , the magnitude of the x and y components of

F(Fx = F cos θ; Fy = F sin θ )

Fx= 0.62N Fy = - 0.21N

RESULT FOR EXPERIMENT 1B

1. The magnitude and angle of F were recorded. The angle as shown in figure 3.1 was measured.

Magnitude= 1.03 N angle = 53˚

2. The magnitude of the x and y component of the equilibrant of F were recorded.

X-component= 1.03cos53˚ y-component= 1.03cos53˚C – (0.105x9.81)

3. What are the magnitude of Fᵪ and Fᵧ, the x and y components of

Fᵪ = 0.62N Fᵧ = -0.21N

4. F: Magnitude = 1.03N Angle = 53˚

Fᵪ : 0.62N Fᵧ = -0.21N

5.Record the magnitude and angle of the force vector, F , that you have constructed.

Magnitude = 1.03N Angle= 53˚

6.Calculate Fx and Fy , the magnitude of the x and y components of

F(Fx = F cos θ; Fy = F sin θ )

Fx= 0.62N Fy = - 0.21N

Calculation For Experiment 1 B

1. F = mg Angle = 53˚

F : magnitude = 0.105 kg (9.81m/s2)

= 1.030 N

2. X-component= 1.03cos53˚ y-component= 1.03sin53-(0.105x9.81)

3. Fᵪ = 1.03cos53˚ Fᵧ = 1.03sin53-(0.105x9.81)

= 0.62N = -0.21N

4. F: Magnitude = (0.105 kg)(9.81m/s2) Angle = 53˚

=1.03N

Fᵪ = 1.03Ncos53˚ Fᵧ = 1.03Nsin53˚- (0.105x9.81)

= 1.03cos 53˚ = -0.21N

5.Record the magnitude and angle of the force vector, F , that you have constructed.

Magnitude = 1.03N Angle= 53˚

6.Calculate Fx and Fy , the magnitude of the x and y components of

F(Fx = F cos θ ; Fy = F sin θ )

Fx= 0.62N Fy = - 0.21N

DISCUSSION

In the experiment 1A, it is about adding forces resultant and it equilibrant. The magnitude of

the force is equal to mg and has the measured 0.49N and the direction is downward while the

equilibrant force is 40.88N and the direction is downward. The mass that used is constant that is

0.050 kg. This way will make the calculation be easier and more accurate. When the angle is set

up to 159˚ the measurement of the magnitude is 0.49N in the other hand when the angle is 34˚

the magnitude is 0.68N and lastly when the angle set to 112˚ the magnitude is 0.98 N.

The diagonal of the parallelogram defined by Fᵪ and Fᵧ is Fᵣ the vector indicating the

magnitude and direction of the total force acting on the asteroid. Fᵣ is called the resultant of Fᵪ

and Fᵧ.

For the experiment in 1B, the magnitude of force is also 1.030 N because has a constant mass

same with experiment 1A that 0.050 kg. The angle is set up to 53˚ . The x- component and Fᵪ has

the same value, that is 0.62N and the y-component also has the same value with Fᵧ that is -

0.21N. So this result is due to theory that said when two forces added together, they will have

the same effect as the original force.

CONCLUSION

Based on the experiment 1A, the mass that used is constant that is 0.050 kg to make

calculation more accurate. The measure of magnitude is 0.49 N and direction is downward. For

the equilibrium force the magnitude is 0.49N and the direction is opposite that is downward. Fᵣ is

called the resultant of Fᵪ and Fᵧ.

Based on the experiment 1 B, the value of x- component and Fx is equal that is 0.62 N same

also with y-component and Fy the value is -0.21 N. It shows that when two forces added together,

they will have the same effect as the original force.

After spend a lot of time on these two experiments which is about adding forces resultant and

equilibrant and also resolving forces component, the conclusion that can be said is the

measurement of the result is totally successful because it is due to the theory given.

→ The total force of Fᵪ and Fᵧ are equal to the Fᵉ.

RECOMMENDATION

1. While taking the reading it should be in the vertically to prevent parallax error.

2. Be more careful on the condition of the string could lead to an error to our reading.

3. Make sure the position of the board is on the stable position because it can cause the

force ring could not come to its true equilibrium.

4. The adjustment of pulley and spring balanced must be accurate so that the holding pin is

centre in the force ring.

REFERENCES

1. Manual lab physics book

2. Physics book by Giancoli –sixth edition

3. Tom Henderson, 1996- 2007, Addition of Forces (online available: 22 Sept 2008)

http://www.glenbrook.k12.il.us/gbssci/phys/Class/vectors/u3l3a.html

4. Static Equilibrium- Balancing Force (online available : 22 Sept 2008)

http://www.cdli.ca/courses/phys3204/unit01_org04_ilo01/b_activity.html

Appendices