HY 607: Finding the Angle for Resultant Force of Two Forces

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Homework Help Overview

The discussion revolves around determining the angle at which an 800N force must be applied so that the resultant force of two forces equals 2000N. The problem involves vector components and the use of the i,j method for force representation.

Discussion Character

  • Exploratory, Mathematical reasoning, Assumption checking

Approaches and Questions Raised

  • The original poster attempts to set up the problem using trigonometric functions to express the forces in terms of their components. They express confusion regarding the application of the Pythagorean theorem in this context.

Discussion Status

Some participants have provided guidance on expanding the equation and suggested steps for isolating the cosine of the angle. There is ongoing exploration of the mathematical relationships involved, with no explicit consensus reached yet.

Contextual Notes

Participants note the use of a diagram to illustrate the forces, but there is a mention of an attachment that is not visible to others. The problem is framed within the constraints of a homework assignment, which may limit the type of assistance that can be provided.

lektor
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with reference to the diagram, at what angle must the 800N force be applied in order that the resultant R of the two forces has a magnitude of 2000N.

We were instructed to use the i,j method.

So, the forces for Fi = 800sin(-) and Fj = 1400 + 800 cos(-) and the resultant is 2000N.

This is where i get stuck, a^2 + b^2 = c^2
so 2000^2 = (800 sin(-))^2 + (1400+800 cos(-))^2

any help would be great.
 

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bump^

please help :p
 
lektor said:
bump^

please help :p

We can't see the attachment yet...
 
it is simply a diagram showing

|.../
|.../
|(-)./
|.../
|./

with the vertical line saying 1400N and the diagonal line saying 800N
 
It appears correct, expand

2000^2= 800^2 sin^2(@) + 1400^2 +2*1400*800*cos@ + 800^2*cos^2@

2000^2 - 800^2 - 1400^2 = 2*1400*800*cos@

solve for cos@


MP
 

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