Force Required to be a minimum

[Emilio]

Homework Statement

If the resultant force acting on the bracket is required to be a minimum, determine the magnitude of F1. Set Φ=31°

Homework Equations

F_x=F(cos(θ))
F_y=F(sin(θ))

The Attempt at a Solution

First I tried to find ∑Fx and ∑Fy
ΣF_x=0=200+cos(67.38)260+cos(59)F_1x
ΣF_y=0=sin(67.38)260+sin(59)F_1y

Solving for F₁, I got
x=-384.06
y=279.99

And then to find the magnitude, I found the root of the sum of their squares

√(-384.06² + 279.99²)=479.335N

which was wrong.

A few things: I know the x-component can't be negative, it makes no sense, but I don't know what to change to make it positive. I also don't know what is implied by "required to be a minimum."

 

[NascentOxygen]

Set ϕ = 31

Something important missing here?

 

[Emilio]

Something important missing here?

I'm not sure what you mean, am I missing a step or did I leave something out? All it says in the prompt is "Set ϕ = 31° ."

 

[NascentOxygen]

I'm not sure what you mean, am I missing a step or did I leave something out? All it says in the prompt is "Set ϕ = 31° ."

There's a degree symbol? Typesetting sometimes misses symbols. Do you think it says set F1's angle to be 31°?

 

[Emilio]

There's a degree symbol? So what did you set to be 31°?

Oh sorry, I see a degree symbol in my original post, not sure why it doesn't show up.

I did set ϕ = to 31°, but I used 59° in my equations just because it was a little easier to visualize for me. I didn't think it would make a difference because sin(31)=cos(59).

 

[NascentOxygen]

The phi isn't showing up on my screen, it's just a blank space.

 

[Emilio]

The phi isn't showing up on my screen, it's just a blank space.

Oh that's odd, I even see the phi in your quoted text. I rewrote it using the editor instead of copy-pasting symbols, so hopefully it looks a little more accurate.

 

[NascentOxygen]

Your mistake is in equating the sum of the components to zero. That would certainly be ideal, but with the angle fixed all you can aim for is a minimum, not zero.

 

[Emilio]

What does it mean to find a minimum? A minimum force that would keep it in equilibrium? The smallest possible resultant force?

 

[NascentOxygen]

What does it mean to find a minimum? A minimum force that would keep it in equilibrium? The smallest possible resultant force?

The resultant sum of those 3 external forces is to be its smallest possible value.

 

[Emilio]

The resultant sum of those 3 external forces is to be its smallest possible value.

So instead of ∑F_x,y being equal to 0, they should equal the smallest possible value?

Wouldn't that be F_x=0 and F_y=0?

 

[NascentOxygen]

So instead of ∑F_x,y being equal to 0, they should equal the smallest possible value?

Wouldn't that be F_x=0 and F_y=0?

The character set you are using seems problematical. I saw F_y= -∞ in your post, which seemed a puzzling value; yet when I quote that to respond, that -∞ has transformed into a 0. You should be using the characters that appear when you tap $\Sigma$ in the forum editor toolbar, or else use Latex.

You can't make the resultant horizontal force equal to zero, as all forces have a component to the right---there is not going to be any cancelling in the x-direction.

 

[Emilio]

I had typed -∞ at first, but after considering it, I realized it didn't make sense so I changed it to 0.

So I can't set F_x=0, but I want to get this value as small as possible. Do I set up a limit as F_x approaches 0?

 

[NascentOxygen]

I had typed -∞ at first, but after considering it, I realized it didn't make sense so I changed it to 0.

Just my caching problem, then.

So I can't set F_x=0, but I want to get this value as small as possible. Do I set up a limit as F_x approaches 0?

Probably not.

You had the right idea back in your opening post:

to find the magnitude, I found the root of the sum of their squares

So go back to your first post, and improve on that. If force up is +, then downwards will be -

200+cos(67.38)260+cos(59)F1x

That won't be F_1x, it's just F₁ because the cos converts it to the horizontal component.