LR - The Addition and Resolution of Vectors- The Force Table PDF

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Steeven Imbaquingo 06/13/17 PHY 215 Prof. Kibrewossen Tesfagiorgis The Addition and Resolution of Vectors: The Force Table Introduction: Physical quantities are commonly categorized as scalar or vector quantities. A scalar m , which is quantity has only magnitude and units. For example: a scalar quantity will be 10 s telling us the speed. Thus, scalar quantities are temperature, energy, and time; they are always positive but it does not say the direction. However, vector quantity does tell the direction. For m example: 3 East , which is velocity indicating the direction is going. Consequently, vector s quantities are displacement, acceleration, force, and momentum; they tell the direction is going, magnitude, and they are either positive or negative. Since vector quantities tells the direction, then we can add them with special methods like the triangular method, the parallelogram method, or the component method. Using these methods will helps us find the resultant or the vector sum. Objective: Our objective was to use the force table to figure out the mass of the resultant and the degree of the resultant. Then, it was to add the vectors graphically to find the resultant. Next, add the vectors analytically to find the resultant. Finally, we are given the resultant and the degree, so our objective was to find the two vectors. Procedure: 1. First, we set up the table force with strings and suspended weights. Vector Addition I: 2. We are given two vector quantities with magnitudes F1= ( 0.200) g N , θ 1=30 ° and F2 =(0.200)g N , θ2=120 ° (the total mass for each one will be 200g). First, in order to find the sum of these two vectors, we find it experimentally, then graphically, and analytically. Experimentally: - On the force table, we set up the clamp pulleys at 30 °∧120 ° and we added enough weights to total 200g - Now, we have to determine the third clamp pulley by simultaneously adding weights and manipulating the magnitude and direction of the equilibrant force that sustains the central ring centered in steadiness around the center pin.

Then, when it’s centered, it will give us the resultant’s mass and angle, but it is in the opposite direction. Finally, record the data on the table. Graphically: - We are going to use the triangular method in order to draw these two vectors that were given to us. However, first we have to set up a scale (25 mm=100 g). - Then we are going to measure the magnitude and the direction of the resultant with a ruler and protractor. Finally, record it on the table. Analytical: R x and  R y by computing the two vectors that were given to us: - First, we find  x = A x + B x o R -

 R y = A y + Bx Then we are going to use the Pythagorean Theorem to find the resultant: 2 2 o R= √ R x + R y o

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Last, is to calculate the direction of the resultant; Ry tan−1 o Rx

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Vector Addition II: 3. We are going to repeat the step 2 (experimental, graphically, and analytical). But, we have to remember we have different magnitudes and angles. Vector Addition III: 4. We are going to repeat the step 2 (experimental, graphically, and analytical). But, we have to remember we have different magnitudes and angles. Vector Resolution: 5. We are given the magnitude and the angle of the resultant, so the objective is to find the other two vector components, which is F x ∧F y . First, let’s find the two vectors experimentally, then graphically, and finally analytical. Experimentally: - First let’s set out the clamp pulleys at 240 ° , 90 ° , and 0 ° on the force table. - Then, we are going to add the total mass, which is 300g, on the 240 ° . This is the same as the magnitude and direction we are given, F= ( 0.300) g N , θ =60 ° , because 60 ° +180 °=¿ 240 ° . - Next, we are going to add weights on the 0 ° and 90 ° pulleys until forces of the system is in balance. This will give us the F x ∧F y . Finally put on the table. Graphically: - We are going to draw the resultant vector along with the scale. - Then, we are going to measure F x ∧F y , which it will give us the magnitude.

Analytical: - We are going to compute the resultant that was given along with the angle: o F x =Rcos ( θ) ∧ F y = Rsin (θ ) Vector Addition IV: 6. We are given three different magnitudes with different angles, which they are F1= ( 0.100 ) g N , θ1=30 ° F 2=( 0.200 ) g N , θ 2=90 ° and F3 = ( 0.300) g N , θ 3=225 ° . We are going to compute the resultant the same we did on step two. However, since there are three vectors we are going to use the parallelogram method graphically, then we are going to compute the resultant by using the component method analytical, and experimentally we are going to use the force table and balance 7. Finally record the data on the table. Data Set: Forces (N) Vector Addition I Vector Addition II Vector Addition III Vector Resolutio n Vector Addition IV

F1= ( 0.200 ) g N , θ1=30 °

Resultant R (magnitude and direction) Graphical Analytical Experimental F= ( . 280) g N θ=77 ° F= ( .312) g N θ=45 °

F= ( .283) g N θ=75 ° F= ( .304) g N θ=45 °

F= ( .286) g N θ=75 ° F= ( .311 ) g N θ=41 °

F= ( .24 0) g N θ=35°

F= ( .25 0) g N θ=36.8°

F= ( .251) g N θ=35.2°

F x = ( .148) g N

F x = ( .150) g N

F x = ( .150) g N

F=( 0.200) g N , θ =6 0 °

F y = ( .260 ) g N

F y = ( .260 ) g N

F y = ( .267 ) g N

F1= ( 0.100 ) g N , θ1=30 °

F= ( .140) g N θ=160 °

F= ( .13 0) g N θ=163 °

F= ( .131) g N θ=162 °

F2 = ( 0.200 ) g N , θ2=120 ° F1= ( 0.200 ) g N , θ1=20 ° F2 = ( 0.150 ) g N , θ2=80 ° F1=F x = ( 0.200 ) g N , θ1=0 ° F2 =F y = ( 0.150 ) g N , θ2=90 °

F2 =( 0.200 ) g N , θ2=90 ° F3 = ( 0.300 ) g N , θ3=225 °

R= √ (A x +B x )2+( A y +B y )2

Results and Calculations: Vector Addition I: F1 :   A x =200 cos ( 30 ° )=173 F2 :   

 A y =200 sin ( 30 ° )=100

 B x =200 cos ( 120 ° )=−100

R= √ ( 173 −100) +( 100 + 173 ) =283 g 273 =75 ° θ=tan−1 73 2

( )

Vector Addition II: F1 :   A x =200 cos ( 20 ° )=188 F2 :   

 A y =200 sin ( 20 ° )=68

 B x =150 cos ( 80° )=26

 B y =150 sin ( 80 ° )=148

R= √ ( 188 + 26 ) + (68 + 148 ) =304 g −1 216 =45 ° θ=tan 214 2

2

( )

Vector Addition III: F1 :   A =200 cos ( 0 °) =200

 A y =200 sin ( 0° ) =0

x

F2 :   

 B y =200 sin ( 120 ° )=173

2

 B x =150 cos ( 90° )=0

 B y =150 sin ( 90 ° )=150

R= √ ( 200 + 0 ) +( 0+150) =250 g 150 θ=tan−1 =36.8 ° 200 2

2

( )

Vector Resolution: F : F ( 0.300 ) g N ,θ =60 °  F x =300 cos ( 60 ° )=150 g F y =300 sin ( 60° ) =260 g Vector Addition IV: F1 :    A x =100 cos ( 30 ° )=87 A y =100sin ( 30 ° )=50

θ=tan

−1

( ) Ry Rx

F2 :  F3 : 

 B x =200 cos ( 90° )=0

 B y =200 sin ( 90 ° )=200

 C x =300 cos( 225 ° )=−212

 C y =300 sin ( 225° )=−212

R= √ ( 87 + 0−212 ) +( 50 + 200 −212 ) =130 g 38  =163 ° θ=tan−1 −125 Percent error: 1. 1.1 % 2. 2.3 % 3. 0.4 % 4. F x =0 % F y =2.7 % 5. 0.8 % 

2

(

2

)

Conclusion: We can conclude that there were minimal errors results between experimental, graphical, and analytical. For example, when we were solving the resultant analytical, it gave a precise result because we were using vector formulas. However, when we did it graphically, it gave us a slightly different result because we drew the vectors according to scale. Therefore, these minimal errors tell us that we were close to our accurate measurements....