static lab 1.docx

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−MEASURED VALUE ∨ ¿

¿ ×10 0 CALCULATE VALUE

TASK 2: DRAW THE FREE BODY DIAGRAM.

FREE BODY DIAGRAM

m₂ = 0.1kg

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 = 0.15kg

TASK 3: DERIVE THE EQUATIONS AND CALCULATE THE ANGLES 1. Based on your FBD above, derive the equation to calculate angle Ө 1 and angle Ө2. 2. Report your data using table below. Start the load 3, m 3 with 50g. Calculate for five different weights in increment order.

T1=m*g

T2=m*g

T1=(0.25)*(9. 81)

T2=(0.25)*(9. 81)

T1=2.45

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 mg= 0.05*9.81=0.49 Ө= sin-1 (0.49/4.9)

LOAD 100g mg= 0.1*9.81=0.98 Ө= sin-1 (0.98/4.9)

LOAD 150g mg= 0.15*9.81=1.47 Ө= sin-1 (1.47/4.9)

LOAD 200g mg= 0.2*9.81=1.96 Ө= sin-1 (1.96/4.9)

LOAD 250g mg= 0.25*9.81=2.45 Ө= sin-1 (2.45/4.9)

TASK 3: RESULT FOR THE CALCULATION. Load 3, m₃ (g) 50 g 100 g 150 g 200 g 250 g

Angle 1, Φ₁ (◦) 5.74ᵒ 11.54ᵒ 17.48ᵒ 23.6ᵒ 30ᵒ

Angle 2, Φ₂ (◦) 5.74ᵒ 11.54ᵒ 17.48ᵒ 23.6ᵒ 30ᵒ

PART B: EXPERIMENTAL WORK. TASK 4: FIND THE EQUIPMENT. List of Equipment: 1. 2. 3. 4.

Ls-13101 equilibrium of forces Apparatus Magnetic pulley x2 Magnetic protractor x1 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) 50 g 100 g 150 g 200 g 250 g

Angle 1, Φ₁ (◦) 8ᵒ 14ᵒ 22ᵒ 30ᵒ 35ᵒ

Angle 2, Φ₂(◦) 10ᵒ 17ᵒ 23ᵒ 29ᵒ 32ᵒ

TASK 7: SHOW YOUR CALCULATION SAMPLE OF ERROR

¿ CALCULATION VALUE−MEASURED VALUE ∨ ¿

¿ ×10 0 CALCULATE VALUE

LOAD 50g

¿ 5.74−8∨ ¿ ×100 5.74 1 θ =¿

¿ 5.74−10∨ ¿ × 100 5.74 θ ₂=¿ θ2=74.2

LOAD 100g Ө1 =| 11.54 – 14 |X 100 11.5

Ө2 = |11.54 – 17 |X 100 11.5

Ө1=21.3%

Ө2=47.3%

LOAD 150g Ө1 = |17.48 – 22 |X 100 17.4 5.74

Ө2 =| 17.48– 17 |X 100 17.4

LOAD 200g Ө1 = |23.6 – 30 |X 100 23.6

Ө2 = |23.6– 29 |X 100 23.6

Ө1=27.1%

Ө2=22.9%

LOAD 250g Ө1 =| 30 – 35 |X 100 30

Ө2 = |30 – 32 |X 100 30 Ө2=6.7%

TASK 7: RESULT OF ERROR ANALYSIS.

Load 3, m₃ (g) 50 g 100 g 150 g 200 g 250 g

Angle 1, Φ₁ error (%) 39.4% 21.3% 25.9% 27.1% 16.7%

Angle 2, Φ₂ error (%) 74.2% 47.3% 31.6% 22.9% 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. F1

F2

F1 Solution FR = F1+F2

F2

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.

FV

FR

FU U

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. C

a

b

B

c A

Cosine law:

√ B2 +C 2−2 BC∗cos a

Sine law:

A B C = = sin a sinb sin 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.