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

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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:
• Mechanical Engineering...

[@xtract7335]

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

[@yakuzauhm7093]

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.8159]

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.

[@MichelvanBiezen]

Not sure why you are asking that question. You need an F1 force downward to balance out the forces.

[@octco.ltd.8159]

@@MichelvanBiezen If the question does not specify to find F1 and F2, but to find the reaction force to achieve equilibrium, are there only horizontal (Fx) and vertical (Fy) reaction forces?

[@MichelvanBiezen]

Ah, now I understand the question. No there will be only one force required for the reaction force. But since the reaction force is a vector with both magnitude and direction, you can only determine the force by finding the x and y components of the force separately.

[@heinmt3178]

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

[@MichelvanBiezen]

It is possible to pick a different point on the bracket to place the 2 forces, F1, and F2, and find their magnitudes such that the sum of the forces in the x and y directions are zero, but also note that in this problem, the lines of action of each force all meet at a single point. If they didn't, chances are that the net moment would not be zero, and the bracket would start rotating until the net moment was zero.

[@genieyas5186]

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

[@iffah31]

The bow looks good on you, sir :--)

[@jonathaneugenio-y8v]

It looks like you mixed up F1x and F1y

[@jonathaneugenio-y8v]

it must be 700 cos 55 for Fx and 700 sin 55 for Fy

[@jonathaneugenio-y8v]

or 700 sin 35 for Fx and 700 cos 35 for Fy

[@jonathaneugenio-y8v]

just kidding, you're correct

[@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.

[@MichelvanBiezen]

Too bad that it didn't help.