Resolving forces in the direction perpendicular to a line

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SUMMARY

The discussion focuses on resolving forces F1 and F2 in the direction perpendicular to a line AB using trigonometric principles. The user successfully determined the horizontal and vertical components of force F1 by sketching triangles and applying basic trigonometry. The sine and cosine rules were utilized to find the necessary components, leading to the correct solution. This method emphasizes the importance of visual representation in solving force resolution problems.

PREREQUISITES
  • Understanding of basic trigonometry, including sine and cosine rules
  • Familiarity with force components in physics
  • Knowledge of free body diagrams and their applications
  • Concept of levers and fulcrums in mechanics
NEXT STEPS
  • Study the application of the sine and cosine rules in resolving forces
  • Learn how to create and interpret free body diagrams
  • Explore the principles of levers and their mechanical advantages
  • Practice problems involving force resolution in two dimensions
USEFUL FOR

Students in physics, engineers dealing with mechanics, and anyone looking to improve their skills in resolving forces and understanding trigonometric applications in real-world scenarios.

Bolter
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Homework Statement
See below
Relevant Equations
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Hi I'm very stuck on what to do for these 2 questions I got wrong

Screenshot 2020-10-16 at 12.17.44.png

Screenshot 2020-10-16 at 12.17.56.png


Can someone please help me on what triangle I need to sketch out in order to find the 2 components of forces for F1 and F2. I'm assuming you have to make use of the sine or cosine rule here

I'd be grateful for any help given! Thanks
 
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What have you done so far?
Remember that what counts here are the actual forces that are perpendicular to the levers, all respect to the fulcrum A.
 
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What are the horizontal and vertical components of force F1?
 
I actually managed to find the right answer at the end by drawing triangles and labelling the forces that are perpendicular to the line AB. Then with some basic trig I got the force needed
 
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