VectorComponents Worksheet

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Embry-Riddle Aeronautical University *

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Mechanical Engineering

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Apr 3, 2024

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Vector Components Lab (Worksheet) Name: Skye Gauthier Section Number: Section 1: Vyomkesh Sangewar Tuesday (8:15am – 11am) To start the lab, refer to the lab manual sheet provided. The equipment is shown below: Part 3: ∑ 𝑭 ⃑⃑ = 0, 3 Forces in 2 Dimensions 1. Use the balance to measure the mass of each of the mass hangers and record them. Make sure you label them with a sticky note. (3 points) Mass Hangers Mass [g] Hanger 1 50g Hanger 2 50g Hanger 3 50g

2. Keep the first pulley at the 0 ° mark and balance it with the other two pulleys. What angle does each of the forces make with the force aligned with zero mark? (3 Points) Pulley Angle Angle Aligned with Zero Mark [°] Pulley 1 0* Pulley 2 120* Pulley 3 240* 3. What angle do the two forces make with each other? (2 Points) Pulley Angle made with each other[°] Pulley 1 to Pulley 2 120* Pulley 2 to Pulley 3 120* 4. Imagine the x-axis that cuts the center of the force table from 180 ◦ mark and the 0 ◦ mark. Let the y-axis be perpendicular to that, through the center of the table along the line from 90 ◦ to 270 ◦ . The force aligned with 0 ◦ mark is pointing in the positive x-direction. a. Do the other two forces (not at the 0-degree pulley) point in the positive or negative x-direction? (1 point) Pulley 1 would be negative x- direction. Pulley 2 would be negative x- direction. b. Do the forces (not at the 0-degree pulley) point in the positive or negative y- direction? (1 point) Pulley 1 would be positive y – direction. Pulley 2 would be negative y – direction.

Part 4: Vector Components Keep the vectors the way they were in Part 3. 5. What must be true about the y-components of the two off-axis forces for the ring to be in equilibrium? (1 point) The y – components of the two off- axis forces for the ring are balanced because one force is positive and the other force in the y is negative. 6. Add and remove masses to each hanger so that the total mass of each is about 250 g . (As long as all three forces are kept equal, the equilibrium should not be disturbed. You may want to replace the nail to lock the ring while you are adjusting. Don’t forget to include the mass of the hanger itself.) What is the total mass of each of your three hanging masses after the additional mass? (3 point) Hanging Masses Mass [g] Mass 1 250g Mass 2 250g Mass 3 250g 7. You will now alter the force table with an additional pulley. The first pulley will remain at the 0  degree mark and the second pulley will remain at an angle . The first and second pulleys should continue to have masses of 250 grams each; you are not adding or removing masses to the first and second pulleys for this part of the experiment. The third pulley should have been at an angle on the force table. Move the third pulley to the 180  mark ( - x-axis ). Use the fourth line to create a fourth pulley. Hang the fourth mass hanger from it. Place the fourth pulley at the 270  mark ( - y-axis ). The force table will be unbalanced. You will add and remove masses to ONLY the third pulley and the fourth pulley to achieve equilibrium on the force table. You can determine the masses that you need for the components by trial and error, or you can calculate them in advance using Equations 2 and 3 shown below. Confirm that the component replacements for the original vector are equivalent by checking that the ring remains in equilibrium with the nail removed. When you have accomplished this task, you will have split the force that was initially exerted onto the third pulley into its x- and y- component forces. N (1) (2) (3)

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a. What mass did you need for your x-component? (1 point) Pulley 1: 250g Pulley 2 250g Pulley 3: 490g Pulley 4: 110g b. What do you calculate for the mass of the x-component using Equation 2? (1 point) I calculated the mass for the x-component as Fx = -23.49g. c. What mass did you need for your y-component? (1 point) Pulley 1: 250g Pulley 2 250g Pulley 3: 490g Pulley 4: 110g d. What do you calculate for the mass of the y-component using Equation 3? (1 point) I calculated the mass of the y component as Fy = 15g.

Part 5: Using Components to find a Vector 8. Install the nail to hold the ring in the center of the force table. Put one pulley at the 0  mark and hang a mass hanger with an additional 100g mass over it (150g). Put a second pulley at the 30  mark in Quadrant I. Hang a mass hanger with an additional 200 g of mass over it (250g) . These are the two forces that you will try to balance by adding components, putting the first one on the x-axis and the other one on the y-axis. Make predictions for a force to balance the two forces that you have installed (a.k.a. the equilibrium force): a. Will its x-component be positive or negative for the predicted equilibrium force? (1 point) The x component will be negative for the predicted equilibrium force. Fx = -366.51g. b. Will its y-component be positive or negative for the predicted equilibrium force? (1 point) The y component will be negative or the predicted equilibrium force. Fy = - 125g. c. Which quadrant will the equilibrium force of these components be in? (1 point) Quadrant 3 is where the equilibrium force will be in.

d. Do you expect the magnitude of your needed predicted equilibrium force to be greater than, less than, or equal to the vector sum of the two forces already installed? (1 point) I expected the magnitude of the needed predicted equilibrium force to be equal to the vector sum of the two forces already installed (Pulley on 0* and Pulley on 30*). 9. Find the components (one vector aligned with the x-axis and one vector aligned with the y-axis) of the force necessary to put the ring in equilibrium, by putting a third pulley on the x-axis, a fourth pulley on the y-axis, and hanging the appropriate masses over each. You can determine the masses by trial and error, or you can make use of Equations 2 and 3 to calculate a predicted value for each. Confirm that the ring is in equilibrium by removing the nail. a. What mass did you need for your x- component? (1 point) Pulley 1: 250g Pulley 2: 150g Pulley 3: 80g Pulley 4: 170g b. What do you calculate for the mass of the x-component using Equation 2 with the force in Quadrant I summed with the force along the x-axis? (1 point) The calculated mass of the x- components using equation two is Fx = - 209.90g. (250cos(0) + 150cos(30) + 80cos(270) + 170cos(180) = Fx = - 209.90g) c. What mass did you need for your y-component? (1 point) Pulley 1: 250g Pulley 2: 150g Pulley 3: 80g Pulley 4: 170g d. What do you calculate for the mass of the y-component using Equation 3 with the force in Quadrant I? (1 point) I calculated the mass of the y – component to be Fy = -5.

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(250sin(0) + 150sin(30) + 80sin(270) + 170sin(80) = Fy = -5g. e. Would the magnitude of the x-component be larger or smaller if you had set the angle of the force in Quadrant I to 40  ?(1 point) The magnitude of the x component would be smaller if I had set the angle of the force in Quadrant 1 To be 40* instead of 30*. f. What about the y-component? (1 point) The magnitude of the y- component would be smaller if I had set the angle differently. g. Would the magnitude of the x-component be larger or smaller if you had set the angle of the force in Quadrant I to 20  ?(1 point) The magnitude of the x – component will be larger if the angle was 20* instead of 30*. h. What about the y-component? (1 point) The magnitude of the y – component will be larger if the angle was 20* instead of 30*. If you refer to figure above, viewing it as a right triangle, you can see that the Pythagorean theorem gives: (4) 10. Use Equation 4 to calculate the magnitude of the resultant force that will come from the combination of the components you have constructed in Question 9. ( Note : This is also the equilibrium force for the vectors that you created in Question 8). Then, use Equation 1 to calculate the angle.

Once you have determined the magnitude and direction of the resultant of your two components, remove the components and construct the resultant force. This new force should be perfectly balanced by the forces that are in Quadrant I. Remove the nail to confirm that the ring is in equilibrium. a. What is the magnitude of your resultant force? How does this compare to your predictions from Question 8? (1 point) - Magnitude of combination: 209.96 (root (209.90)^2 + (-5)^2 ) = 209.96. b. What angle does it make from the positive x-axis? (1 point) Theta is 181.36*. - (Tan-1 (fy/fx) = -5/-209.90 = 1.36 + 180 = 181.36 is theta c. What quadrant is it in? How does this compare to your predictions from Question 8? (1 point) The angle is in the Third Quadrant. This matches my initial predictions in question 8 because I said Quadrant 3 as well.

Part 6: Calculate a Vector from Components Now you will calculate the components of the resultant of two vectors and use that calculation to directly create the vector needed to put the ring in equilibrium. Install the nail to hold the ring in the center of the force table. Put one pulley at the 15  mark , in the first quadrant, and hang a mass hanger with an additional 100g mass over it . Put a second pulley at the 110  mark in Quadrant II. Hang a mass hanger with an additional 200g of mass over it. These are the two forces that you will try to balance by calculating their resultant's magnitude and angle , and then creating a third force vector to balance it and put the ring in equilibrium. 11. Which quadrant should the third vector be placed in to put the ring in equilibrium? (1 point) The second quadrant is where the third vector should be placed. 12. To add the two vectors at 15  and 110  , you will first need their components. Use Equations 2 and 3 to calculate each vector's x- and y-components. a. What is the 15  vector's x-component? Is it positive or negative? (1 point) 150cos(15) = 144.89*. This positive. b. What is the 15  vector's y-component? Is it positive or negative? (1 point) 150sin(15) = 38.82. This is positive. c. What is the 110  vector's x-component? Is it positive or negative? (1 point) 250cos(110) = -85.51. This is negative d. What is the 110  vector's y-component? Is it positive or negative? (1 point) 250sin(110) = 234.92. This is positive. 13. The resultant vector will have an x-component that is the sum of the two contributing vectors' x-components, and a y-component that is the sum of the two contributing vectors' y-components. a. What is the resultant vector's x-component? Is it positive or negative? (1 point) Resultant vector’s x component = 59.38*.

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b. What is the resultant vector's y-component? Is it positive or negative? (1 point) Resultant vec tor’s y component is = 273.34*. c. Which quadrant is the resultant vector in? (1 point) The resultant vector is quadrant 1. 14. Using Equations 4 and 1, calculate the magnitude and angle of the resultant of the two forces installed at 15  and 110  - from the components you have just calculated for it. a. What is the magnitude of your resultant force? (1 point) Magnitude of resultant force is 111.018. (root (110)^2 + (15)^2 ) = 11.018 b. What angle does it make from the positive x-axis? (1 point) 77.74* from the positive x – axis (tan-1 (273.34/59.38) = 77.74 15. Now construct a third vector that will balance the two vectors at 15  and 110  . It must have the same magnitude as the resultant that it is pulling against (which you just calculated). And it must be directed in the opposite direction as that you just calculated for the resultant. a. What is the angle that you will place the new vector at? (1 point) The angle that I placed my new vector is 270* b. Which quadrant is it in? (1 point) Quadrant 3 16. Place the third pulley at your calculated angle and hang the correct mass from the string. Confirm that the new force vector holds the ring in equilibrium by removing the nail. Take a picture and attach it here. (5 points)

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