Sum of Forces in X and Y Directions - axis set up normally

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The discussion centers on calculating the sum of forces in the X and Y directions using standard axis orientation. Participants clarify the use of equations Vx = Vcosθ and Vy = Vsinθ for determining vector components, specifically for a force F1 at a 60° angle from the x-axis. There is confusion regarding the signs of the components, with consensus that the x-component is negative and the y-component is also negative due to the direction of F1. A suggestion is made to redraw the vector for clarity on the signs of the components. Overall, the focus is on accurately applying vector decomposition to solve the problem.
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Being asked for the sum of forces in the X and Y directions and the axis are set up 'normally' so like a plus sign ( + ).

I attempted the first one and I think I did it correctly, but I'm not so sure about the second one.
 

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The first one is fine. Where's your work on the second one?

Note: While a diagram is useful for reference, please type up your work right here in the thread. That helps people give you better help.

Please see: How to Ask for Help
 
So I was looking through my notes, and I saw the equation Vx = Vcosθ and Vy = Vsinθ.

I'm not sure though that this is where i can put these equations to use, since i don't know the value of V. In looking at the angles, F1 is 60° from the x-axis going counterclockwise. I'm lost in more of how to put the information I do know together.
 
cdornz said:
So I was looking through my notes, and I saw the equation Vx = Vcosθ and Vy = Vsinθ.
That's how you'd find the x and y components of some vector V, where θ is the angle made with the +x axis.

I'm not sure though that this is where i can put these equations to use, since i don't know the value of V.
According to the diagram, it would be F1.

In looking at the angles, F1 is 60° from the x-axis going counterclockwise.
F1 is 60° below the negative x axis. So will the x-component of F1 be negative or positive?
 
If the x-component is below the negative x-axis i would think that that component would be negative. But using those equations I think I've determined the answer.

F1x=F1cos60°
F1y=F1sin30°

I broke this down by drawing out the picture of the triangle somewhat given and by finding the f1x and f1y components and plugging those into the equations I had.
 
cdornz said:
If the x-component is below the negative x-axis i would think that that component would be negative. But using those equations I think I've determined the answer.

F1x=F1cos60°
F1y=F1sin30°

I broke this down by drawing out the picture of the triangle somewhat given and by finding the f1x and f1y components and plugging those into the equations I had.
Looks OK except for the signs.
 
In question with the signs, would the x-component be negative because it is technically below the F1 and the y-component positive because it is technically above the F1?
 
cdornz said:
In question with the signs, would the x-component be negative because it is technically below the F1 and the y-component positive because it is technically above the F1?
Hint: Redraw the vector F1 so its tail is at the point of application. Or at least imagine a set of axes with the origin at the tail of F1. That will make it easier to see the signs of the components.

The x-component of F1 is negative since F1 points to the left of the vertical axis.

The y-component of F1 is negative since F1 points below the horizontal axis.
 
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