• Support PF! Buy your school textbooks, materials and every day products Here!

Force Required to be a minimum

  • Thread starter Emilio
  • Start date
  • #1
10
0

Homework Statement


If the resultant force acting on the bracket is required to be a minimum, determine the magnitude of F1. Set Φ=31°
XllYL1D.png


Homework Equations


Fx=F(cos(θ))
Fy=F(sin(θ))

The Attempt at a Solution


First I tried to find ∑Fx and ∑Fy
ΣFx=0=200+cos(67.38)260+cos(59)F1x
ΣFy=0=sin(67.38)260+sin(59)F1y

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."
 
Last edited:

Answers and Replies

  • #2
NascentOxygen
Staff Emeritus
Science Advisor
9,244
1,072
Set ϕ = 31
Something important missing here?
 
  • #3
10
0
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° ."
 
  • #4
NascentOxygen
Staff Emeritus
Science Advisor
9,244
1,072
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°?
 
  • #5
10
0
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.

nAUgaKZ.png


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).
 
  • #6
NascentOxygen
Staff Emeritus
Science Advisor
9,244
1,072
The phi isn't showing up on my screen, it's just a blank space.
 
  • #7
10
0
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.
 
  • #8
NascentOxygen
Staff Emeritus
Science Advisor
9,244
1,072
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.
 
  • #9
10
0
What does it mean to find a minimum? A minimum force that would keep it in equilibrium? The smallest possible resultant force?
 
  • #10
NascentOxygen
Staff Emeritus
Science Advisor
9,244
1,072
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.
 
  • #11
10
0
The resultant sum of those 3 external forces is to be its smallest possible value.
So instead of ∑Fx,y being equal to 0, they should equal the smallest possible value?

Wouldn't that be Fx=0 and Fy=0?
 
  • #12
NascentOxygen
Staff Emeritus
Science Advisor
9,244
1,072
So instead of ∑Fx,y being equal to 0, they should equal the smallest possible value?

Wouldn't that be Fx=0 and Fy=0?
The character set you are using seems problematical. I saw Fy= -∞ 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.
 
  • #13
10
0
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 Fx=0, but I want to get this value as small as possible. Do I set up a limit as Fx approaches 0?
 
  • #14
NascentOxygen
Staff Emeritus
Science Advisor
9,244
1,072
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 Fx=0, but I want to get this value as small as possible. Do I set up a limit as Fx 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 F1x, it's just F1 because the cos converts it to the horizontal component.
 

Related Threads on Force Required to be a minimum

Replies
4
Views
18K
Replies
3
Views
1K
  • Last Post
Replies
4
Views
6K
Replies
9
Views
2K
  • Last Post
Replies
3
Views
5K
Replies
2
Views
3K
  • Last Post
Replies
1
Views
2K
Replies
1
Views
2K
Replies
4
Views
8K
Replies
8
Views
769
Top