Forces in Equilibrium (Vectors)

[Oblivion77]

Hey guys, I am stuck on this question. It is first year engineering vector statics. Here it is.

Thanks for any help!

 

[tiny-tim]

Hey guys, I am stuck on this question. It is first year engineering vector statics. Here it is.

Thanks for any help!

Hi Oblivion77!

Show us what you've tried, and where you're stuck, and then we'll know how to help.

 

[Oblivion77]

Well, i resolved the vectors into the components and added the sums and equaled them to zero (since its in equilibrium)

$$\Sigma Fx = -118.65 -F3Cos[\alpha] + F1 =0$$

$$\Sigma Fy = 68.5 - F3Sin[\alpha] = 0$$

I don't know how to go from here, with the "minimum"

 

[tiny-tim]

Hi Oblivion77!

… just woken up … :zzz:

$$\Sigma Fx = -118.65 -F3Cos[\alpha] + F1 =0$$

$$\Sigma Fy = 68.5 - F3Sin[\alpha] = 0$$

I don't know how to go from here, with the "minimum"

Yes, that's right.

Tip: when you have two equations with cos and sin,

put the cos and sin on their own on the left, then square and add (using cos² + sin² = 1, of course)

That will give you F3 in terms of F1, and then … ?

 

[MrEnergy]

...after you solve a little bit, you get;


137/2=F3*Sin(a)

so for a=>90 degrees sin is max and F3 is min.

Min F3 is 137/2...And a is 90 degrees...


and magnitude of F1 is;

F1=Cos30*137

 

[tiny-tim]

Hi MrEnergy!

137/2=F3*Sin(a)

so for a=>90 degrees sin is max and F3 is min.

Min F3 is 137/2...And a is 90 degrees...


and magnitude of F1 is;

F1=Cos30*137

oops! I over-complicated it!

Yes, you're absolutely right.

(hmm … not sure how I helped there … )