Using Polar Plot Graph Paper to Resolve Vector Addition By Jack Graham March 2013 Key Idea to Learning Graphic Vector Addition: 1. All vectors acting on the same point can be represented by an “x-component” and a “y-component”. “y-component” Y “x-component” X Special Vector Analysis Graph Paper This is a polar plot graph. Each Ring is 1.0 Newton or the weight of a 100g mass. 1 2 3 4 5 6 7 8 9 10 Suppose we want to add two vectors, A at 0⁰ of 5 N To B at 120⁰ of 7 N We can draw the two vectors starting at the origin and extending the arrow head out to the 5 N ring and 7 N ring. Special Vector Analysis Graph Paper Y Suppose we want to add two vectors, A at 0⁰ of 5 N To B at 120⁰ of 7 N Adding a normal x-y axis to it allows us to easily find x & y component vectors. 10 9 8 7 6 5 4 3 2 1 1 2 3 4 5 6 7 8 9 10 -10 -9 -8 -7 -6 -5 -4 -3 -2 -1-1 -2 -3 -4 -5 -6 -7 -8 -9 -10 X Special Vector Analysis Graph Paper Y Suppose we want to add two vectors, A at 0⁰ of 5 N 3”x 5” Card To B at 120⁰ of 7 N Adding a normal x-y axis to it allows us to easily find x & y component vectors. Use the edge of a card to drop a dashed line from the head of each vector to the X and Y axis. 10 9 8 7 6 5 4 3 2 1 1 2 3 4 5 6 7 8 9 10 -10 -9 -8 -7 -6 -5 -4 -3 -2 -1-1 -2 -3 -4 -5 -6 -7 -8 -9 -10 X Special Vector3”x Analysis Graph Paper 5” Card Y Suppose we want to add two vectors, A at 0⁰ of 5 N To B at 120⁰ of 7 N Adding a normal x-y axis to it allows us to easily find x & y component vectors. Use the edge of a card to drop a dashed line from the head of each vector to the X and Y axis. 10 9 8 7 6 5 4 3 2 1 1 2 3 4 5 6 7 8 9 10 -10 -9 -8 -7 -6 -5 -4 -3 -2 -1-1 -2 -3 -4 -5 -6 -7 -8 -9 -10 X Special Vector Analysis Graph Paper Y Suppose we want to add two vectors, A at 0⁰ of 5 N To B at 120⁰ of 7 N The only Y component is drawn in from the origin to the dashed line from B The A vector has no Y component. The X component of the B vector is – 3.5 10 9 8 7 6 5 4 3 2 1 1 2 3 4 5 6 7 8 9 10 -10 -9 -8 -7 -6 -5 -4 -3 -2 -1-1 -2 -3 -4 -5 -6 -7 -8 -9 -10 The X component of the A vector is the whole A vector X Special Vector Analysis Graph Paper Y Suppose we want to add two vectors, A at 0⁰ of 5 N To B at 120⁰ of 7 N To find the X-component of R the Resultant Vector we combine both of the X-components adding as signed numbers: +5 – 3.5 = +1.5 10 9 8 7 6 5 4 3 2 1 1 2 3 4 5 6 7 8 9 10 -10 -9 -8 -7 -6 -5 -4 -3 -2 -1-1 -2 -3 -4 -5 -6 -7 -8 -9 -10 We draw the Rx-component. We recognize the By-component as the Ry-component (changed to R color here) X Special Vector Analysis Graph Paper 3”x 5” Car Y 10 Suppose we want to add two 9 8 vectors, A at 0⁰ of 5 N 7 6 To B at 120⁰ of 7 N 5 4 Using a 3” by 5” card as a 3 2 guide we draw 1 1 -10 -9 -8 -7 -6 -5 -4 -3 -2 -1 -1 perpendicular lines from the -2 -3 ends of the R-components. -4 -5 -6 -7 -8 -9 -10 2 3 4 5 6 7 8 9 10 X Special Vector Analysis Graph Paper Y 10 Suppose we want to add two 9 8 vectors, A at 0⁰ of 5 N 7 6 To B at 120⁰ of 7 N 5 4 Using a 3” by 5” card as a 3 2 guide we draw 1 1 -10 -9 -8 -7 -6 -5 -4 -3 -2 -1 -1 perpendicular lines from the -2 -3 ends of the R-components. -4 -5 -6 -7 -8 -9 -10 3”x 5” C 2 3 4 5 6 7 8 9 10 X Special Vector Analysis Graph Paper Y 10 Suppose we want to add 9 8 two vectors, A at 0⁰ of 5 N 7 6 To B at 120⁰ of 7 N 5 4 Using a 3” by 5” card as a 3 2 guide we draw 1 1 -10 -9 -8 -7 -6 -5 -4 -3 -2 -1 -1 perpendicular lines from the -2 -3 ends of the R-components. Where these perpendiculars intersect is the head of the resultant which we draw. 2 3 4 5 6 7 8 9 10 -4 -5 -6 -7 -8 -9 -10 We can see from the graph the R vector is 6.2 N at 76⁰. X Key Ideas to Learning Graphic Vector Addition: 1. All vectors acting on the same point can be represented by an “x-component” and a “y-component”. 2. X-components to the right are positive numbers. 3. X-components to the left are negative numbers. 4. All the X-components can be algebraically added (like signed numbers) to get the X-component of the resultant. 5. Y-components going up are positive numbers. 6. Y-components going down are negative numbers. 7. All the Y-components can be algebraically added (like signed numbers) to get the Y-component of the resultant. 8. Resultant vector’s magnitude and direction angle can be found graphically from its “x-component” and “ycomponent”. Practice Vector Addition • Distribute the “Polar Plot Graphs Vector Addition Using Component Vectors Practice Worksheet 1” • Distribute 3”x5” cards • Do the six problems. When a student finishes, check their work. If it is correct have them help others and check the next students finished. Continue until all have completed the solutions. • Collect the papers and put them in the wire basket on Mr. Graham’s desk. Finding an Equilibrant Y 10 Suppose we want to add 9 8 two vectors, A at 0⁰ of 5 N 7 6 To B at 120⁰ of 7 N 5 4 Using a 3” by 5” card as a 3 2 guide we draw 1 1 -10 -9 -8 -7 -6 -5 -4 -3 -2 -1 -1 perpendicular lines from the -2 -3 ends of the R-components. Where these perpendiculars intersect is the head of the resultant which we draw. 2 3 4 5 6 7 8 9 10 -4 -5 -6 -7 -8 -9 -10 We can see from the graph the R vector is 6.2 N at 76⁰. X Finding an Equilibrant An equilibrant force is a force vector that balances a force vector or group of force vectors causing all the forces acting at a point to sum to zero giving no change in motion. Y Given forces A, B, and C, how can we determine the size and direction of their equilibrant E? R B A C By finding the sum of the given forces R and reversing its direction we find E. E X Finding an Equilibrant Given Force A at 0⁰ 4 N, B at 90⁰ 4 N, and C at 180⁰ 4 N, determine their equilibrant. Hints: Use the method for adding two or three vectors to find their resultant using special polar plot graphing methods, to add the given vectors. Then reverse the resultants components To find the equilibrium force vector. Use the force table to check your answer. Practice Finding an Equilibrant Distribute the blank Polar Plot Graphs for Vector Addition Using Component Vectors Find the equilibrant for each of the following problems. Check your answers using the force table. The first students done set up the force table solution for all to see. 1. Given Force A at 0⁰ 4.5 N, B at 60⁰ 4.5 N, and C at 200⁰ 4.5 N, determine their equilibrant. 2. Given Force A at 0⁰ 4.5 N, B at 110⁰ 4.5 N, and C at 220⁰ 4.5 N, determine their equilibrant. 3. Given Force A at 0⁰ 4.5 N, B at 60⁰ 8.5 N, and C at 200⁰ 4.5 N determine their equilibrant. 4. Given Force A at 0⁰ 4.5 N, B at 110⁰ 8.5 N, and C at 220⁰ 4.5 N, determine their equilibrant. 5. Given Force A at 0⁰ 8.5 N, B at 60⁰ 8.5 N, and C at 200⁰ 8.5 N, determine their equilibrant. 6. Given Force A at 0⁰ 8.5 N, B at 110⁰ 8.5 N, and C at 220⁰ 8.5 N, determine their equilibrant.