FE Review force balance Question

[ratman720]

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.

 

[S.E.]

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.

 

[ratman720]

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

 

[PhanthomJay]

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.

 

[rezeew]

Hello,

Thank you for bringing this question to my attention. After reviewing the problem and solution provided in the Lindeburg book, I have a few thoughts to share with you.

Firstly, I agree with your approach of orienting the system with line P as the x-axis. This is a valid starting point for solving the problem.

As for your first question, I believe the issue lies in the use of the cosine function. When using trigonometric functions to resolve vectors, it is important to consider the direction of the vector. In this case, the vector F is not in the same direction as line P, so simply using the cosine of the angle between them will not give an accurate result. Instead, we must use the component of F that is parallel to line P, which is given by Fcos(25). However, this is not the full magnitude of Fp, as there is also a component of F that is perpendicular to line P, which is given by Fqcos(70). Therefore, the full magnitude of Fp is given by Fcos(25) - Fqcos(70), as shown in the solution provided in the book.

For your second question, I believe the relevance of Fq on the magnitude of Fp lies in the fact that both forces are acting on the same point (the intersection of lines P and Q). Therefore, the total force in the x-direction must be balanced, meaning that the sum of the x-components of Fp and Fq must be equal to the x-component of F. This is why the solution uses Fqcos(70) in the calculation for Fp.

I hope this explanation helps to clarify the solution provided in the book. Trigonometry can be tricky when dealing with vectors, so it's important to carefully consider the direction and components of each force. Keep up the good work on your FE review and good luck on your exam!