In Step 2 of method A, why should the center weight be less than the sum of the left and right weights?

College Physics
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Author:Raymond A. Serway, Chris Vuille
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Chapter1: Units, Trigonometry. And Vectors
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In Step 2 of method A, why should the center weight be less than the sum of the left and right weights?

134°
118°
109°
0000
Transcribed Image Text:134° 118° 109° 0000
A. Graphical Method:
1. Set up the dynamic kit stand with clamps and pulley system as shown in Fig. 1.
14
2. Hang a number of weights on the left, center and right cords such that the center weight is less
than the sum of the left and right weights.
3. Give the system a little jerk and let it come to rest.
4. With the use of a hard cardboard, trace the lines of the cords on the bond paper. See Figure 2.
5. With a convenient scale of unit length for every weight, (2 cm = 1 weight), construct graphically
the vector sum of the left, right and center cords. Use the graphing paper provided.
6. Complete table A and reproduce the graphical vector sum of Step No. 5 just below table A for
your final report.
Transcribed Image Text:A. Graphical Method: 1. Set up the dynamic kit stand with clamps and pulley system as shown in Fig. 1. 14 2. Hang a number of weights on the left, center and right cords such that the center weight is less than the sum of the left and right weights. 3. Give the system a little jerk and let it come to rest. 4. With the use of a hard cardboard, trace the lines of the cords on the bond paper. See Figure 2. 5. With a convenient scale of unit length for every weight, (2 cm = 1 weight), construct graphically the vector sum of the left, right and center cords. Use the graphing paper provided. 6. Complete table A and reproduce the graphical vector sum of Step No. 5 just below table A for your final report.
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