STATICS EQUILIBRIUM OF FORCE AT A POINT

Experiment outcomes

To show that the force acting at a point is in equilibrium.

Force in Equilibrium

When the force acting on an object is in equilibrium , the resultant force acting on the object equal to zero

If the forces resolved into horizontal and vertical component respectively , then :a) The sum of all horizontal components of the force is equal to zero

Σ Fx = 0

b) The sum of all vertical components of the forces is equal to zero.

Σ Fy = 0

The two objects are at equilibrium since the force ar balance . However , the forces are not equal

Resolution Force

The Resolution of a force is a single force that can be resolved into two perpendicular components ( Vertical and horizontal components )

Resultant Force

A resultant force is a single force that can represents the combined effect of two or more forces by taking into account both the magnitude and the direction of the forces .

F R=√((Σ Fx2)+(ΣFy 2¿))¿

F Resultant Force ( F at the angle θ )

Fx Horizontal Vector Component.

Fy Vertical Vector Component.

Vertical Component Fy = F sin θ

Horizontal Component Fx = F cos θ

tan θ = (FyFx) tan−1=(FyFx)

Theory

When an object is in static equilibrium, Newton’s First Law states that the vector sum (or

resultant) of all the external forces acting on the object must add up to zero: F =Σ 0

. In this lab, we will study the vector sum of three forces acting on a single point in static

equilibrium. One of these forces Fg is provided by a 2.00N weight suspended from a

vertical string. The other two forces are the tensions in the strings and T hanging

from spring scales attached to a backboard. The magnitudes of the forces are measured

using the spring scales.

A vector can be expressed graphically by an arrow, or numerically by its x and y

components. In Problem 1, we will verify that the resultant force has zero components in

both x and y directions, ΣFx

= 0,ΣFy

= 0 , by drawing a scale diagram and measuring the

components. In Problem 2, we will use a scale diagram to graphically add the vectors

using the head to tail method and verify that the resultant force is a vector with zero

length, or that the length is less than the uncertainty.

→ + Σ Fx = 0, F1cosθ – F2 cosβ = 0------- (1

↑ + Σ Fy = 0, F1sinθ – F2 sinβ = 0 -------- (2)

Apparatus/Equipments

A set of weight Protector Ruler 2 units of spring balances String

Safety Precautions

- Make sure the apparatus and its accessory are in good and save condition.

- Make sure the workplace environment is clean and tidy.

- Use an appropriate safety guideline of equipment handling.

- Place the apparatus to its initial place.

- Keep the workplace clean and tidy.

- Do a report to the lecturer if any problem occured.

Procedures

i. The apparatus have been set up as figure 2.

ii. A 10 N mass has been hung on spring balance and inditicate as F 3.

iii. The string was been tie horizontally through the knot point to act as reference line.

iv. The reading has been record in table A.

v. A protector has been used to measure angles between the horizontal string with spring 1

and spring 2.

vi. The position of the knot is change to three different distances along the string.

vii. Procedure 4 , 5 and 6 were repeated.

viii. At the end of experiment the discussion is stated.

Result/Data

EXPERIMENTDATA COLLECTION

F1 F2 F3 θ ß F1 Sin θ F2 Sin ß

POSITION 1 7 8 10 43 49 4.77 6.04

POSITION 2 8 4 10 66 31 7.31 2.06

POSITION 3 4 9 10 31 63 2.06 8.02

TABLE A

Discussion

I. In order for a system to be in static equilibrium it will need to follow equations:

Equilibrium is When the force acting on an object is in equilibrium , the resultant force acting on the object equal to zero

∑ Fx = 0 and ∑ Fy = 0

By using each of these condition equations, we will be able to calculate the unknown mass in a

system that incorporates three different places…

II. The force that has been generated by the force that allocated at three different location on the

string , the first position is put up at the center of the horizontal line,

ERRORS IN MEASUREMENT

A) It is important to note that it is not always possible to make a prefect measurement.B) There is always some uncertainty or error in any measurement.C) Experimental uncertainty is due to either random errors or systematic errors.

SYSTEMATIC ERRORS

a) Systematic errors are errors caused by the condition of the measuring instrument or the state or the state of the environment in which the measurement are taken.

b) Systematic errors usually cannot be avoided and often apparent throughout the entire experiment.

c) It is reproducible inaccuraries that are consistently in the same direction.

Conclusion

In conclusion, the experiment was very instructive and successful because it helped explain how a system

involving three different places that we put the weight can be at static equilibrium. After measuring

the angles and weighing the masses used in the system, we were able to calculate the theoretical

mass as well as the percent error present in each of the three trials occurring in the experiment. We

found that for each of our calculated masses for the third elevator were within the range of less than

five to the experimental mass for each mass. This led us to have a percent error of 1.9, .06, and 2.1

percent for each corresponding trial. If I had to add some criticism for the static equilibrium

experiment would be that I would have like to have done it alongside an experiment involving kinetic

equilibrium. In the end, I concluded that this experiment was one of the more enjoyable experiments

that we have done so far. It helped my understanding of what exactly the different is between static

and kinetic equilibrium.

References

Fakhru’l – Razi Ahmadun , Chuah Teong Guan , Mohd Halim Shah Ismail ( 2005) Safety : Principles & Practices in the Laboratory , UPM Press

Gunt Hamburg Catalogue

Lotus Scientific (M) Sdn. Bhd. Catalogue

BA101 Engineering Science 1 Book , JMSK

WikiPedia , Bernoulli’s Beam Theory