(6) A force F is applied on bar BC at B through rod AB asshown. If the t-component of the force is F₁ = 50 lb,determine the n-, x-, and y-components of the force F. The t-axis is directed along bar BC, and the n-axis is perpendicularto the bar BC.Ans: Fn = 284 lb, Fx = -271 lb, Fy = -98.5 lb-30°4FA20°

Step 1: Calculate the component of the force F in the n-axis Notice that the force F and the t-axis…

Answered: (6) A force F is applied on bar BC at B…

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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Two rescue vessels are pulling a broken-down motorboat toward a boathouse with forces of 840 lbs. and 960 lbs. The angle between these two forces is 24.5°. Determine the angle between the resultant vector and the 840-lb force (x). Find the magnitude and direction ( e) of the equilibrant. r = r =₁ x=1 8=₁ -r :: cos(24.5°) (840)² + (960)² - 2(840)(960) :: 400 lbs :: 1790 lbs e :: 77 :: 169 O 840 lbs 960 lbs. :: cos(167.75°) :: 1759 lbs :: 11° :: 139 :: 103° :: 167 :: cos(155.5°) O
Find the magnitude of the resultant force and the angle it makes with the positive x-axis. y4 200 N Magnitude: 300 N 60° Angle:
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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
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