Determining the support reactions on a frame

[JJones_86]

Homework Statement

Determine the SUPPORT REACTIONS acting on the frame (See Picture Below)


You can see I have drawn the FBD, but I am lost on applying static equilibrium about point "F" because I don't know the distance from point "F" to any other point. In order for me to sum the moment about point "F" I'd need to know the distance from point "F" to ATLEAST point "B". So I'm not sure I have the support reactions set up correctly, any ideas or advice would be greatly appreciated. Thanks!

 

[PhanthomJay]

There is only one external support and one externally applied load in this system. You don't need a dimension from F to B to solve for the support reaction. What are the applied forces in the x and y directions?

 

[JJones_86]

How would I solve for the moment about point F then, because I need the Force and distance to calculate the torque.. The applied force in the X-Direction is 600# in the negative direction, and the Y-Direction is 600 # in the negative direction, correct?

 

[JJones_86]

Or are you trying to say that I don't need the Couple "CF"?, which would eliminate the need for summing moments..

 

[PhanthomJay]

How would I solve for the moment about point F then, because I need the Force and distance to calculate the torque.. The applied force in the X-Direction is 600# in the negative direction, and the Y-Direction is 600 # in the negative direction, correct?

No, that is incorrect. There is an applied force in the y direction, from the objects weight, but what applied external force is there in the x direction?

Or are you trying to say that I don't need the Couple "CF"?, which would elimante the need for summing moments..

The frame would tip over if there were no couple at support F.

 

[JJones_86]

The force in the Y direction would be 2668.93 N, and the X direction would be tension, is that what your asking?

 

[PhanthomJay]

The force in the Y direction would be 2668.93 N, and the X direction would be tension, is that what your asking?

The problem is given in units of pounds and feet, so don't convert the force units to Newtons, as it is unnecessary. The externally applied force in the y direction is 600 lbs. The tension force in the cable is internal to the system. When solving for support reactions, one must look at the externally applied forces only. Is there any externally applied force in the x direction??

 

[JJones_86]

Ohhh, so there is no X direction force, so in that case I just need the perpendicular distance from point F to point A?

 

[PhanthomJay]

Ohhh, so there is no X direction force, so in that case I just need the perpendicular distance from point F to point A?

Yes, corrrect,{see edit. this is incorrect}; to solve for the couple. You also need the sum of Fx = 0 and sum of Fy = 0 equations to solve for the x and y support reactions at F.

Edit: In solving for the couple at F, don't forget to factor in the pulley radius, sorry.

 

[JJones_86]

Aww, i see now, thanks man! Much appreciated, this has been buggin the crap out of me all day, haha.

 

[JJones_86]

Ok, so I got CF = 3600#ft Clockwise Direction, Fx = 0, Fy = 600# Positive direction, sound right?

 

[PhanthomJay]

Ok, so I got CF = 3600#ft Clockwise Direction, Fx = 0, Fy = 600# Positive direction, sound right?

Your magnitudes are correct, and direction for Fy at F is correct. But your couple direction is not correct. you are looking for the couple of the support on the frame, not the couple of the applied load about the support.

 

[JJones_86]

Oh, I meant counter clockwise, my bad

 

[PhanthomJay]

Oh, I meant counter clockwise, my bad

OK, good.