Homework Help: Vector problem

1. Mar 20, 2009

Dell

given 3 forces working on one object with dimentions of 6x2

F1=4500N (from the front at an anlge of 39)
F2=9000N (from the front at an angle of -α )
F3=2250N (grom the back at an angle of 180)

see diagram below:

what must the angle α be in order for the total force applied to be parallel to the line AB, ?
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F=F1+F2+F3 (vectors)

Fx=4500*cos(39) + 9000*cos(α) - 2250
=2250(2*cos(39) + 4*cos(α) - 1)

Fy=4500*sin(39) -9000*sin(α)
=2250(2*sin(39)-4*sin(α))

now in order for this vector to be parallel to AB, it must be at the same angle,-->

2/6=[2250(2*sin(39)-4*sin(α))] / [2250(2*cos(39) + 4*cos(α) - 1)]

(2*cos(39) + 4*cos(α) - 1) = (6*sin(39)-12*sin(α))
cos(α)+3*sin(α) = 0.8054

from here i am stuck, i need to find some trigonometrical identity to help me out, alternatively, another way to solve the problem

Last edited by a moderator: May 4, 2017
2. Mar 20, 2009

tiny-tim

Hi Dell!

(i haven't checked how you got there, but …)

Hint: to solve an equation like this, put it in the form cos(A)cos(α)+sin(A)sin(α) = …

3. Mar 20, 2009

Dell

i dont understand, could you continue a bit

4. Mar 20, 2009

tiny-tim

well, for example, if it was cos(α)+sin(α) = 0.8054, you could write that as

cos(α)/√2 + sin(α)/√2 = 0.8054/√2

ie cos(π/4)cos(α) + sin(π/4)sin(α) = 0.8054/√2

so cos(α - π/4) = 0.8054/√2