LAB 1: EQUILIBRIUM OF FORCES FOR 3 POINT LOAD.

INTRODUCTION

When all the forces that act upon an object are balanced, then the object is said to be in a state of

equilibrium. The forces are considered to be balanced if the rightward forces are balanced by the

leftward forces and the upward forces are balanced by the downward forces. This however does not

necessarily mean that all the forces are equal to each other. Note that the two objects are at

equilibrium because the forces that act upon them are balanced; however, the individual forces are

not equal to each other. If an object is at rest and is in a state of equilibrium, then we would say that

the object is at "static equilibrium." "Static" means stationary or at rest. A common physics lab is to

hang an object by two or more strings and to measure the forces that are exerted at angles upon the

object to support its weight. The state of the object is analyzed in terms of the forces acting upon

the object. The object is a point on a string upon which three forces were acting. If the object is at

equilibrium, then the net force acting upon the object should be 0 Newton. Thus, if all the forces are

added together as vectors, then the resultant force (the vector sum) should be 0 Newton. (Recall

that the net force is "the vector sum of all the forces" or the resultant of adding all the individual

forces head-to-tail.) Thus, an accurately drawn vector addition diagram can be constructed to

determine the resultant. In this lab work, we will prove and compared the calculated angle (theory)

with measured angle (experiment).

OBJECTIVE.

The objectives that we’ll get are we’ve investigating the conditions required to achieve static

equilibrium. We also have proved the force calculation method using geometrical method and

resolution of force method. We also compared experimentally obtain result with theoretical

calculation.

Part A: THEORETICAL WORK

TASK 1: FIND THE THEORIES.

1. Calculation of equilibrium of force using geometrical methods which are triangle method

and parallelogram law.

PARALLELOGRAM LAW

The parallelogram law is put the tails (end without the arrow) of the two vectors at the same point,

(just the vector A and the vector B on the left of the diagram) then close the parallelogram by

drawing the same two vectors again (the vector B and vector A to the right of the diagram). The

parallelogram rule is just the Triangle method used twice at the same time.

TRIANGLE METHOD

The triangle method is one of the rules of adding vectors that conveniently by geometric methods.

Draw vector A with its magnitude represented by a convenient scale and then draw vector B to the

same scale with its tails starting from the tips of A. The resultant vector R= A+B is the vector drawn

from the tails of A to the tips of B.

2. Calculation of equilibrium of force using resolution of forces method.

3. Calculation of random error and systematic error.

¿CALCULATION VALUE−MEASUREDVALUE∨ ¿CALCULATEVALUE

×100¿

TASK 2: DRAW THE FREE BODY DIAGRAM.

FREE BODY DIAGRAM

F₁ = 0.10 X (9.81) = 0.981 N

F₂ = 0.15 (9.81) = 1.4715 N

F₃ = 0.05 (9.81) = 0.4905 N

∑F = F₃ = F₁ + F₂

m₂ = 0.15kgm₂ = 0.1kg

TASK 3: DERIVE THE EQUATIONS AND CALCULATE THE ANGLES

1. Based on your FBD above, derive the equation to calculate angle Ө1 and angle Ө2.

2. Report your data using table below. Start the load 3, m3 with 50g. Calculate for five different

weights in increment order.

T1=m*g

T1=(0.25)*(9.81)

T1=2.45

T2=m*g

T2=(0.25)*(9.81)

T2=2.45

ΣFx = - T1 cosӨ + T2 cosӨ =0

ΣFy = T2 sinӨ +T1 sinӨ – mg=0

0 = 2.45 sinӨ +2.45 sinӨ –mg=0

mg = 4.9 sinӨ

sinӨ = mg/4.9

Ө =sin-1 (mg/4.9)

LOAD 50g

LOAD 100g

LOAD 150g

LOAD 200g

LOAD 250g

mg= 0.25*9.81=2.45

Ө= sin-1 (2.45/4.9)

Ө= 30ᵒ

mg= 0.2*9.81=1.96

Ө= sin-1 (1.96/4.9)

Ө= 23.6ᵒ

mg= 0.15*9.81=1.47

Ө= sin-1 (1.47/4.9)

Ө= 17.48ᵒ

mg= 0.1*9.81=0.98

Ө= sin-1 (0.98/4.9)

Ө= 11.54ᵒ

mg= 0.05*9.81=0.49

Ө= sin-1 (0.49/4.9)

Ө= 5.74ᵒ

TASK 3: RESULT FOR THE CALCULATION.

Load 3, m₃ (g) Angle 1, Φ₁ (◦) Angle 2, Φ₂ (◦)50 g 5.74ᵒ 5.74ᵒ

100 g 11.54ᵒ 11.54ᵒ150 g 17.48ᵒ 17.48ᵒ200 g 23.6ᵒ 23.6ᵒ250 g 30ᵒ 30ᵒ

PART B: EXPERIMENTAL WORK.

TASK 4: FIND THE EQUIPMENT.

List of Equipment:

1. Ls-13101 equilibrium of forces Apparatus

2. Magnetic pulley x2

3. Magnetic protractor x1

4. Weight hanger and set of weights (500g) x3

FIGURE 2: LS 13101 EQUILIBRIUM OF FORCES APPARATUS.

LEGEND

A – Mounting Board

B – Protractor

C – Pulley with Magnetic Base

D – Weigh

TASK 5: PROCEDURE

1. Two magnetic pulleys had been placed 200mm away from each other in a horizontal line.

Confirm that the pulleys are perfectly aligned on the grid.

2. After that take 400mm cord and tie both end of the cord with 50g load hooks.

3. Place the cord on the pulleys and make the two 50g weight are balance.

4. We’ve put the weight to both hooks until the both weight become 250g. We put 100g on m₁ and

150g m₂ and recorded it as m₁ and m₂

5. After making sure it’s in a balanced condition, we placed the 50g weight hook on the middle of

the hook. We also make sure it is in between the two pulleys (100mm) and we’ve recorded the

load as m₃.

6. After that, we’ve measure the angle Φ₁ and Φ₂ at the pulleys of m₁ and m₂ using the protractor

respectively.

7. Increase the load m₃ to 100g and measure the angle Φ₁ and Φ₂ respectively using the protractor.

8. Keep repeating the step 7 until the load m₃ reached 250g and then record the data.

9.

TASK 6: RECORD THE MEASURED ANGLE

Load 1, m₁ = 100g

Load 2, m₂ = 150g

Load 3, M₃ (G) Angle 1, Φ₁ (◦) Angle 2, Φ₂(◦)

50 g 8ᵒ 10ᵒ

100 g 14ᵒ 17ᵒ

150 g 22ᵒ 23ᵒ

200 g 30ᵒ 29ᵒ

250 g 35ᵒ 32ᵒ

TASK 7: SHOW YOUR CALCULATION SAMPLE OF ERROR

¿CALCULATION VALUE−MEASUREDVALUE∨ ¿CALCULATEVALUE

×100¿

LOAD 50g

LOAD 100g

LOAD 150g

LOAD 200g

LOAD 250g

θ1=¿5.74−8∨ ¿5.74

×100¿

Ө1=39.4%

Ө1 =| 11.54 – 14 |X 100

Ө1=21.3%

11.54

Ө2 = |11.54 – 17 |X 100

Ө2=47.3%

11.54

5.74

Ө1 = |17.48 – 22 |X 100

Ө1=25.9%

17.48

Ө2 =| 17.48– 17 |X 100

Ө2=31.6%

17.48

Ө1 = |23.6 – 30 |X 100

Ө1=27.1%

Ө2 = |23.6– 29 |X 100

Ө2=22.9%

23.623.6

Ө1 =| 30 – 35 |X 100

Ө1=16.7%

Ө2 = |30 – 32 |X 100

Ө2=6.7%

30 30

θ₂=¿5.74−10∨ ¿5.74

×100¿

θ2=74.2%

TASK 7: RESULT OF ERROR ANALYSIS.

Load 3, m₃ (g) Angle 1, Φ₁ error (%) Angle 2, Φ₂ error (%)50 g 39.4% 74.2%

100 g 21.3% 47.3%150 g 25.9% 31.6%200 g 27.1% 22.9%250 g 16.7% 6.7%

DISCUSSION.

After doing the experiment, we noted that the forces acting upon an object in equilibrium is

commonly used to analyze situations involving objects at static equilibrium. The most common

application involves the analysis of the forces acting upon a sign that is at rest. For example, consider

the picture at the right that hangs on a wall. The picture is in a state of equilibrium, and thus all the

forces acting upon the picture must be balanced. That is, all horizontal components must add to 0

Newton and all vertical components must add to 0 Newton.

If an object is at equilibrium, then the forces are balanced. Balanced is the key word that is used to

describe equilibrium situations. Thus, the net force is zero and the acceleration is 0 m/s/s. Objects at

equilibrium must have an acceleration of 0 m/s/s. This extends from Newton's first law of motion.

But having an acceleration of 0 m/s/s does not mean the object is at rest. An object at equilibrium is

either at rest and staying at rest, or in motion and continuing in motion with the same speed and

direction.

What we get in this lab work are the different between the theoretical and experiment values. What

we can conclude is there got slightly different value between the theoretical and experiment values.

The reason why got the different was got some error that causing it. There are lot of things that can

be the reason in error. Firstly was the random error or called human error, it’s caused by inherently

unpredictable fluctuations in the reading of measurement apparatus or experiment. The different

between the theoretical and experiment values maybe because of the human error which is take the

reading wrongly or wrongly put the apparatus. Other thank that the observational error and

systematic error may caused the different between the theoretical and experiment values.

Discussion 2

What we have experience in this lab work was we know the conditions required to achieve static

equilibrium. We also have proved the force calculation method using geometrical method and

resolution of force method. We also compared experimentally obtain result with theoretical

calculation. Other than that we together find the solution and try to solve it by asking lecturer and

technician there. Besides that, we have compared about the theoretical and experiment values.

CONCLUSION

In conclusion, equilibrium is the state of an object in which all the forces acting upon it are balanced.

In such cases, the net force is 0 Newton. Knowing the forces acting upon an object, trigonometric

functions can be utilized to determine the horizontal and vertical components of each force. If at

equilibrium, then all the vertical components must balance and all the horizontal components must

balance.

We also know the Calculation equilibrium of force using resolution of force method

Force is a vector, therefore it is adds according to the parallelogram law. In the above section we

discovered that if a set of given forces acting over a body is unable to produce any displacement of

motion in the body, this means the force are in equilibrium, and the result may be associated with

only some internal stress of the body.

If two force components acting on a point, follow either the parallelogram law or triangle rule for

vector to draw its resultant force if we have in figure 1.

Figure 1

When it is required to resolve a force into two components in two specific directions, we start to

draw parallel lines from the head of the vector to the desire directions until they intersect the

directions.

In case of determining the magnitude and direction of the resultant force it is recommended to

apply the law of cosines or law of sines to the triangle.

F1

F2

Solution

F1F2

FR = F1+F2

FV

FR

FU

U

Cosine law:

√B2+C2−2 BC∗cosa

Sine law:

Asina

= Bsinb

= Csin c

Calculation for random error and systematic error.

i) Random Errors

Take more data. Random errors can be evaluated through statistical analysis and can be

reduced by averaging over a large number of observations.

ii) Systematic Errors

Systematic errors are difficult to detect and cannot be analyzed statistically, because all

of the data is off in the same direction either too high or too low. Spotting and

correcting for systematic error takes a lot of care.

C B

A

a

b c