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On the graph to the right, draw a coordinate axis. Draw in the two force vectors. Show by adding the vectors. graphically that the net force on the central: ring is zero. Verify this i algebraically by showing that the sum of the components is also zero. L -491+491), = ON

On the graph to the right, draw a coordinate axis. Draw in the two force vectors. Show by adding the vectors. graphically that the net force on the central: ring is zero. Verify this i algebraically by showing that the sum of the components is also zero. L -491+491), = ON

Advanced Engineering Mathematics
Advanced Engineering Mathematics
10th Edition
ISBN:9780470458365
Author:Erwin Kreyszig
Publisher:Erwin Kreyszig
Chapter2: Second-order Linear Odes
Section: Chapter Questions
Problem 1RQ
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On the graph to the right, draw a coordinate axis. Draw in the two force vectors. Show by adding the vectors graphically that the net force on the central ring is zero. Verify this algebraically by showing that the sum of the. components is also zero. L +) = N the position. Now, notice that you can slide the ring back and forth along the r axis, placing it at any distance, and the ring will stay in place. This means that every point along the r axis is an equilibrium point. Explain why this is the case. The forces Cancel it does not matter FYEN Now we're going to change the axes. Remove the two pulleys, and place them at angles of 90° and 270°, respectively. Keeping the z axis along = 0, we've now placed the forces along the y axis. Repeat the earlier analysis with the new forces. 1D Trial 2 Results Force Magnitude (N) Direction (6) 0297 F₁ F₂ 0.48A/ 1 80 13 JEN Vector 0.294160 GrMB.NK 1800 F 11800
Transcribed Image Text:On the graph to the right, draw a coordinate axis. Draw in the two force vectors. Show by adding the vectors graphically that the net force on the central ring is zero. Verify this algebraically by showing that the sum of the. components is also zero. L +) = N the position. Now, notice that you can slide the ring back and forth along the r axis, placing it at any distance, and the ring will stay in place. This means that every point along the r axis is an equilibrium point. Explain why this is the case. The forces Cancel it does not matter FYEN Now we're going to change the axes. Remove the two pulleys, and place them at angles of 90° and 270°, respectively. Keeping the z axis along = 0, we've now placed the forces along the y axis. Repeat the earlier analysis with the new forces. 1D Trial 2 Results Force Magnitude (N) Direction (6) 0297 F₁ F₂ 0.48A/ 1 80 13 JEN Vector 0.294160 GrMB.NK 1800 F 11800
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