Understanding Forces in 3D Space: Solving Problems and Formulas Explained

AI Thread Summary
The discussion centers on understanding how to apply trigonometric formulas to resolve forces in 3D space, specifically in problem 2.61. The formula Fx=Fcosθ is used to find the x-component of a force vector, with the calculation involving projections onto the x-y plane and further onto the x-axis. Participants clarify that basic 2-D trigonometry principles apply, despite the problem being in 3D, emphasizing the importance of the accompanying diagram. The use of Fz=80*sin(30) is explained as a projection onto the z-axis, reinforcing the relationship between sine and cosine in vector resolution. Overall, the discussion highlights the application of fundamental trigonometric concepts in three-dimensional force analysis.
jdawg
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I'm having trouble number 2.61. I have the answers, but I don't understand why they are doing what they're doing.

They have this for (Fx)1:(Fx)1=(80 lb)*cos(30)*cos(40)=53 lb
I'm familiar with the formula Fx=Fcosθ , but I'm confused about using it in 3D. Is there a formula that I need to memorize specifically for 3D?
 
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jdawg said:
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I'm having trouble number 2.61. I have the answers, but I don't understand why they are doing what they're doing.

They have this for (Fx)1:(Fx)1=(80 lb)*cos(30)*cos(40)=53 lb
I'm familiar with the formula Fx=Fcosθ , but I'm confused about using it in 3D. Is there a formula that I need to memorize specifically for 3D?
Not really. You should study carefully the diagram which is included with the problem.

Calculating (80 lb) * cos 30° gives the projection of the force vector in the x-y plane. Multiplying this projection of the force vector in the x-y plane by cos 40° gives the x-component of the force. Similar calculations can give the components along the y and z axes.

It's just a matter of applying 2-D trigonometry and not being distracted because the force vector is 3-D.
 
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Ohh ok, that makes sense! I'm still a little confused about finding Fz. It says Fz=80*sin(30). I kind of have an idea of why, but I'm not super confident about it. Do you use sin(30) in this situation because the the line on the far left of the triangle is parallel to the z axis?
 
This is just basic trigonometry. Look at this right triangle:
Snapshot.jpg

From trigonometry you know that ##sin\alpha =F_z/F## and ##cos\alpha = F_x/F##.
You can rearrange these to get:
$$F_z=F sin\alpha$$
$$F_x=F cos\alpha$$
These have the intuitive meaning of projections: a vector times cosα means a projection of that vector onto the plane from which the angle alpha is measured, and sinα means a projection onto a plane perpendicular to the previous one.
 
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Perfect! Thanks so much, its been a minute since I had trig! :)
 
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