Why Use 110 Degrees Instead of 70 for F2 Components?

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Homework Help Overview

The discussion revolves around the use of angles in resolving force components in a physics problem, specifically questioning the choice of using 110 degrees instead of 70 degrees for the angle related to force F2. The subject area includes vector resolution and trigonometry in physics.

Discussion Character

  • Conceptual clarification, Assumption checking

Approaches and Questions Raised

  • Participants explore the implications of using different angles for calculating the components of force F2, questioning whether using 70 degrees would yield a negative value for the x-component. There is also a discussion about the interpretation of the angle in relation to the diagram provided.

Discussion Status

Some participants have offered insights into the relationship between the angles and their effects on the components, while others express concerns about the accuracy of the diagram and its representation of angles. Multiple interpretations of the angle's reference point are being explored.

Contextual Notes

There is a noted confusion regarding the diagram's representation of angles, particularly the angular difference between forces F1 and F2, which some participants believe may be misleading. This has led to varying assumptions about how the angles should be interpreted in the context of the problem.

Benjamin_harsh
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Homework Statement
why taking angle 110 instead of 70 for finding the magnitude of F2 in X and Y components?
Relevant Equations
F2x = 150.cos 110 = - 51.30, F2y = 150.sin 110 = 140.95
243677


F1x = 120. Cos 300 = 103.92 ; F1y = 120.sin 300 = 60N
F2x = 150.cos 110 = - 51.30, F2y = 150.sin 110 = 140.95

why taking angle 110 instead of 70 for finding the magnitude of F2 in X and Y components?
 
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If you take angle ##70°## insteady of ##110°##, do you get negative value for ##F_{2x}## component (as it should be according to the diagram)?
 
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If you are interested in just the magnitude of ##F_{2}## it doesn't matter which angle you take. That's because
##F_2=\sqrt{F_2^2\sin^2(\theta)+F_2^2\cos^2(\theta)}.## Note that ##\cos(110^o)=-\cos(70^o)## and when you square, the negative sign drops out.
 
Last edited:
I can't let this go. The diagram is very misleading. ?:)

The diagram shows the angular difference between F_1 and F_2 as 110^o. But that doesn't make any sense. That would make the total angle from the positive x-axis to the negative x-axis as 210^o. But that's impossible, since the x-axis is a straight line, and all angles that span a straight line are 180^o. So first and foremost, that needs to be fixed.

From here on out, I'm assuming that the 110^o angle spans from the positive x-axis to F_2 (not F_1 to F_2 as is it shown in the original post).

From there, just follow @kuruman's and @lomidrevo's advice. If you use the 70^o angle, which is the angle with respect to the negative x-axis, then the \cos 70^o will be projected along the negative x-axis. That's a tidbit you need to keep in the back of your mind when interpreting the result. However, if you use the 110^o angle, which is the angle with respect to the positive x-axis, the \cos 110^oresult is already negative so it takes care of the negative sign automatically.
 
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collinsmark said:
From here on out, I'm assuming that the ##110^o## angle spans from the positive x-axis to ## F_2## (not ##F_1## to ##F_2## as is it shown in the original post).
That was my interpretation of the situation. My reading was that OP was questioning whether the difference between ##\cos(70^o)## and ##\cos(110^o)## is more than just an algebraic sign.
 

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