Solve 3D Vector Forces: F3, Alpha, Beta, Gamma

[ur5pointos2sl]

I have attached a copy of the problem and an attempt at a solution. Any help would be greatly appreciated. I know the answers I need to get to but I cannot seem to figure out what I am doing wrong. Thanks

btw the answers are:

F3 = 9.58 kN
Alpha= 15.5 degrees
Beta = 143 degrees
Gamma = 53.1 degrees

I just noticed at the bottom of my work when I was computing the magnitude I put -10 instead of -1. Either way it does not affect the answers that I get.

 

[ur5pointos2sl]

Ok so I figured out what my problem was..I had forgot that one or two of the directions were negative. Thanks anyways.

 

[thomate1]

I can help you with solving 3D vector forces. To begin, let's define our variables. F3 represents the magnitude of the third force, while Alpha, Beta, and Gamma represent the angles between the force and the x, y, and z axes, respectively.

To solve for F3, we can use the Pythagorean theorem in 3D space, which states that the magnitude of a vector is equal to the square root of the sum of the squares of its components. In this case, we have three components: Fx, Fy, and Fz.

To find Fx, we can use the formula Fx = F3cos(Alpha). Similarly, Fy = F3cos(Beta) and Fz = F3cos(Gamma).

Now, we can plug in our values for Fx, Fy, and Fz into the Pythagorean theorem formula:

F3 = √(Fx^2 + Fy^2 + Fz^2)

Substituting in our values, we get:

F3 = √(9^2 + 6^2 + (-10)^2)

F3 = √(81 + 36 + 100)

F3 = √217

F3 = 9.58 kN

To find the angles Alpha, Beta, and Gamma, we can use the inverse trigonometric functions.

Alpha = cos^-1(Fx/F3) = cos^-1(9/9.58) = 15.5 degrees

Beta = cos^-1(Fy/F3) = cos^-1(6/9.58) = 53.1 degrees

Gamma = cos^-1(Fz/F3) = cos^-1(-10/9.58) = 143 degrees

I hope this helps you understand the problem better. Remember to always double check your calculations and units to ensure accuracy.