Physics Lab 2 Force Vector PDF

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Physics Lab 2 Force Vector...

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Objective: To study vectors and compare experimental results with graphical and analytical calculations by finding a resultant force that balances out the given force so that the system will be in equilibrium. Theory: Vectors A and B can be added graphically by drawing them to scale and aligning them head to tail. The vector that connects the tail of A to the head of B is the resultant vector R. Vector addition is both associative and commutative. The components (Ax and Ay) of a vector A can be calculated by projecting the length of A onto the coordinate axes as shown in figure 1. The components can be obtained by using the following equations: Ax = |A| cos θA

Ay = |A| sin θA

The sign of a component gives its direction along the x or y axis. Conversely, from the components, the magnitude |A| and direction θ of the vector can be calculated using the following: |A|=

√A

x

2

+ Ay

2

θ = tan−1 ( Ayx ) A

In order to add vectors analytically, they must be in component form. The components of a vector sum of two vectors A and B yields the components of a new vector, called a resultant vector and will be denoted by R. The components of R can be calculated by: R x = Ax + B x

R y = Ay + B y

In this experiment, the direction and magnitude of a force C that balances out the forces of A and B so that the system will be in equilibrium was found. In order for the system to be in equilibrium, the following must hold: A + B = -C

A+B+C=0

Apparatus: Force table, weight holders, sets of masses, rulers, protractors, spirit levels

Procedure: 1. The force table was placed on a flat surface. Using the spirit level, it was ensured that the force table was level. 2. Three pieces of string ~12 to 15 inches long were cut. A loop was tied at the end of each piece of string, the other end of the string was attached to the ring. 3.The ring was placed in the center of the force table so that it encircles the pin. The strings were put over the pulleys attached to the force table. It was ensured that the pulleys are fixed at the same height around the table. 4. Three mass holders were obtained. For vector A, mass was added to one mass holder until the entire setup (mass holder and added mass) was 25 g. The mass was placed on the end of one of the strings looped over a pulley and the pulley was set at an angle of 63°.

5. For vector B, to the second mass holder, mass was added until the entire setup was ~41 g. This mass was placed on the end of one of the available strings looped over a pulley and the pulley was set at an angle of 154°. 6.For vector C (the resultant), the last mass holder was attached to the last string looped over a pulley. Mass was added to the system and the angle was adjusted until the system is in equilibrium, that is, when the ring with the attached strings is parallel to and suspended above the ring painted on the force table, the pin was then removed. When equilibrium was reached, the entire mass for the setup of vector C was determined. 7.The values for mass and angle for vectors A, B, and C in was recorded in Table 1. Using the formulas given, the mass and x and y- components of vectors A and B was calculated, as well as the mass, force, components, and angle for vector C. 8. The vectors A, B, and C and their corresponding components were then drawn to scale. The complete system of vectors A, B, and C together was also drawn. The experimental results for mass and angle measure of vector C was compared with the analytical calculations and the percentage errors was determined.

Calculations: Ax = |A| cos θA Ax = |25|cos 63 Ax = 11.35g Ay = |A|sin θA Ay = |25|sin 63 Ay = 22.28g

B x = |B| cos θB B x = |41.1| cos 154 B x = -36.94g

B y = |B| sin θB B y = |41.1| sin 154 B y = 18.01 g

To find the components of resultant Vector C C x = Ax + B x C x = 11.35 + (− 36.94) C x =− 25.59 C y = Ay + B y C y = 22.28 + 18.01 C y = 40.29

√C + C |C | = √(− 25.59) |C | =

x

2

y

2

2

+ 40.292

|C | = 47.73

θc = tan−1

∑y ∑x

−1 40.70 θ c = tan −25.59

θc = 57.84° % Error =

Observed−Expected Expected

% Error =

47.0−47.73 47.73 X

100% Error = 1.52%...