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

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To determine the angle for the 800N force so that the resultant force equals 2000N, the i,j method is applied. The equations used are Fi = 800sin(θ) and Fj = 1400 + 800cos(θ), leading to the relationship 2000^2 = (800sin(θ))^2 + (1400 + 800cos(θ))^2. The discussion highlights the need to simplify this equation to solve for cos(θ). Ultimately, the goal is to isolate cos(θ) to find the required angle for the resultant force.
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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 verticle 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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