Mechanical Engineering: Particle Equilibrium (9 of 19) Forces on a Bracket

In this video I will calculate the forces acting on the anchor of a bracket.

Next video in the Particle Equilibrium series can be seen at:

[xtract]

Hii sir I am from india.and because of you my concept about any chapter of physics is very clear..thanks.

[yakuzauhm]

At 2:10, you can to find the degree of the forces of f2, since it passed through the 3rd quadrant which is 270 degrees and plus the 35 degrees which is equal to 305 degrees.
Then calculate f2x cos and sin 305 degrees has the same value of sin and cos 35 degrees.
For f2x cos 305 = 0.5736 and for f2y sin 305 = -0.8192
I owe this knowledge to Prof Michel Van Biezen. You're the best. :)

[octco.ltd.]

If the F1 were not downward, it was at horizontal.
F1=400N
F2=700N
F2x=401.5N
F2y=573.4N
Is it right? And why.thank you.

[genieyas]

i would rather prefer if you appaly tow forces one horizontal and the second vertical wich are the reaction in hing support...

[heinmt]

F1 and F2 can be at another position to be stable? how do we decide the position?

[jonathaneugenio-yv]

It looks like you mixed up F1x and F1y

[nononnomonohjghdgdshrsrhsjgd]

your drawing in this example is really bad. I still don't understand how exactly this bracket example can be connected to a real bridge.