FE Review force balance Question

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SUMMARY

The discussion revolves around solving a force balance problem presented in the 2010 Lindeburg FE Review Manual, specifically Problem #2 in Chapter 10. The problem involves two intersecting lines, P and Q, at a 70-degree angle, with a force F applied at 25 degrees from P and 45 degrees from Q. The correct resolution of forces Fp and Fq requires understanding vector addition and the Law of Sines, rather than simply using trigonometric projections. The provided solution indicates that Fq is derived from the sine of the angles, demonstrating the interdependence of the forces rather than treating them as independent projections.

PREREQUISITES
  • Understanding of vector addition in physics
  • Familiarity with trigonometric functions, specifically sine and cosine
  • Knowledge of the Law of Sines for solving triangles
  • Experience with force equilibrium problems in engineering mechanics
NEXT STEPS
  • Study the Law of Sines and its application in vector resolution
  • Review trigonometric identities and their use in force calculations
  • Practice problems involving force balance and vector addition
  • Explore the concept of projections versus resultant forces in physics
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Engineering students, particularly those preparing for the FE exam, and professionals in mechanics or physics who need to understand force balance and vector resolution techniques.

ratman720
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I am working out of the 2010 lindeburg book which can be found here. The solution is in the book so I am actually looking for an explanation rather than a solutution.

http://www.scribd.com/doc/113765067/FE-Review-Manual-Lindeburg-2010

this is chapter 10 pg 8 in the chapter, reviewing the PDF its pg 163 on scribd. Problem # 2

Homework Statement



Given Line P and Q which intersect at a 70 degree angle. A force F is applied between them at 25 degrees from p and 45 degree from Q.

Find Fp and Fq

Homework Equations


Trig related


The Attempt at a Solution



My approach was simply to orient the system with line P being the x axis. Thus Fp should be 300cos(25) and because we know the angle between F and Q, Fq should also be easy to resolve as 300cos(45).

Both answers are wrong
Lindeburg provides the solution as Fy=Fsin(25)=Fqsin(70) thus Fq=Fsin(25)/sin(70), Which I can see and understand. However for his Fp he initially uses the Fcos(25) then subtracts Fqcos(70).

I am curious to know the following

1. Why simply using the force multiplied by the cosine of the angle between the vector and direction doesn't work. This approach does work for x,y force components.

2. If P and Q are simply directional lines why Fq has any relevance on the magnitude of Fp


Thank you for looking.
 
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If you draw the (x,y) components of F in terms of 25 degrees and 70 degrees (on an x-y plane!) you should see the problem. What you're basically doing is giving a value for Fq relative to one position and a value for Fp from another.
 
I can kind of see that, at least as an explanation for Fq. But if I orient the system such that Fp is the x-axis then Fp=Fx=Fcos25
 
ratman720 said:
I can kind of see that, at least as an explanation for Fq. But if I orient the system such that Fp is the x-axis then Fp=Fx=Fcos25
Well that is the 'projection' onto the x (or p) axis, but you are not looking for projections. You are looking for 2 vectors, Fp and Fq, such that Fp + Fq = F , using the laws of vector addition. Draw a sketch. I'd use the Law of Sines to solve.
 

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