Moment at a point applied to a bracket

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SUMMARY

The discussion focuses on calculating the force supported by point A due to a 2.5-kN load applied to a bracket, with specific dimensions provided: AB = 0.17 m and the distance from B to the force is 0.16 m at an angle of 24 degrees from the vertical. The user attempted to find the x and y components of the force but encountered errors in their calculations. The solution involves creating a Free Body Diagram (FBD) of the bracket and applying equilibrium equations around pin A, considering that the reaction at point B is perpendicular to the slot surface due to the absence of friction.

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Homework Statement


Calculate the magnitude of the force supported by the point at A under the action of the 2.5-kN load applied to the bracket. Neglect friction in the slot.

AB = .17 m

B to force = .16 m

angle between vertical and BtoForce = 24

The Attempt at a Solution



I found the x and y distance of the force from B

y = .16 cos 24 = .1462

x = .16 sin 24 = .0651

I tried multiplying the force by the y distance but it was incorrect.

I also found the distance from A to the force which is:

√((.17 + .0651)2 + (.1462)2) = .2768

and multiplied that by the force but that was also incorrect. Any ideas?

Any help appreciated!
 

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I think you must make a FBD of the bracket. Because there is no friction in the slot, any reaction at B must be perpendicular to the surface of the slot. Write your equations of equilibrium about pin A and solve.
 

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