Equilibrium of a T-Shaped Bracket: Determining Reactions at A and C

[tacet777]

Homework Statement

A T-shaped bracket supports a 150-N load as shown. Determine the reactions at A and C when (a) \alpha=90^o, (b) \alpha=45^o

ans: (a)A=150N going down, C=167.7N,63.4degrees (b)A= 194.5N going down; C=253N, 77.9degrees
base on the book.

http://www.glowfoto.com/static_image/23-165339L/2124/jpg/07/2010/img6/glowfoto
[URL]http://www.glowfoto.com/static_image/23-165339L/2124/jpg/07/2010/img6/glowfoto[/URL]

Homework Equations

\SigmaFx=0 ;right +
\SigmaFy=0 ;up +
\SigmaM=0 ;clockwise +

The Attempt at a Solution

\SigmaFx=-Ax+Cx+Bx
-Asin(\alpha)+Cx+Bsin(\alpha)=0
\SigmaFy=Ay+Cy-By
Acos(\alpha)+Cy-Bcos(\alpha)=0
M_c=-(30)Asin(\alpha)-(15)Acos(\alpha)+(30)Bsin(\alpha)+(25)Bcos(\alpha)

substituting the \alpha=90degrees

A=150N (but i don't know how to check weather it is going down or up since i got it in the moment equation)

using A=150
Cx=0
Cy=0Thank you guyz...

 

[tvavanasd]

Homework Statement

A T-shaped bracket supports a 150-N load as shown. Determine the reactions at A and C when (a) \alpha=90^o, (b) \alpha=45^o

ans: (a)A=150N going down, C=167.7N,63.4degrees (b)A= 194.5N going down; C=253N, 77.9degrees
base on the book.

The Attempt at a Solution

\SigmaFx=-Ax+Cx+Bx
-Asin(\alpha)+Cx+Bsin(\alpha)=0
\SigmaFy=Ay+Cy-By
Acos(\alpha)+Cy-Bcos(\alpha)=0
M_c=-(30)Asin(\alpha)-(15)Acos(\alpha)+(30)Bsin(\alpha)+(25)Bcos(\alpha)

substituting the \alpha=90degrees

A=150N (but i don't know how to check weather it is going down or up since i got it in the moment equation)

using A=150
Cx=0
Cy=0

Does force A have a horizontal component? Can we not assume that Ax=0 since it is supported by a roller?

Even if force A had both horizontal and vertical components, do not use the same variable name for both A and B (don't use \alpha twice).

Once you assume the direction of your forces, you must keep to that assumption in all of your equations. Some problems will be obvious, others will not. If your answers are negative, you assumed the wrong direction (it doesn't matter if your initial assumption is correct). Just be sure to state the vector directon correctly in your final answer.

When you create your moment equation, the directions of the forces that cause the moments must be the same as those in your force equations.


Assuming the Ay pushs down (wouldn't work very well otherwise), Cx pushes to the left (only horizontal force available to oppose force B) and Cy pushes up, your first equations should be:

Ax=0 (assume due to roller)
\SigmaFx = 0
Bsin(\alpha) - Cx = 0

Try to create the others again yourself.


The answers you list above (from your book?) appear to incorrect (by my calcs). The directions are correct, but the magnititudes are wrong.

 

[pongo38]

A good way to determine force directions in this case is: (1) recognise that there are just 3 forces acting on the bracket (2) for equilibrium these forces must all meet at one point. You should be able to identify that point (3) draw the triangle of forces, knowing the magnitude of just one, but the directions of the others.(4) put arrows on the sides of the triangle for all the known forces i.e. 150 N (5) put on the other arrows, making sure the arrows go round the triangle cyclically. That is, all cw or acw. These arrows are then in the correct directions for the unknown forces. The magnitudes of the forces should also agree with the trig/algebraic approach, thus making your request for confirmation unnecessary.