# Determining the support reactions on a frame

1. Oct 14, 2008

### JJones_86

1. The problem statement, all variables and given/known data
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!

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2. Oct 14, 2008

### 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?

3. Oct 14, 2008

### 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?

4. Oct 14, 2008

### JJones_86

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

Last edited: Oct 14, 2008
5. Oct 14, 2008

### PhanthomJay

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

6. Oct 14, 2008

### 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?

7. Oct 14, 2008

### PhanthomJay

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??

8. Oct 14, 2008

### 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?

9. Oct 14, 2008

### PhanthomJay

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.

10. Oct 14, 2008

### JJones_86

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

11. Oct 14, 2008

### JJones_86

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

12. Oct 14, 2008

### PhanthomJay

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.

13. Oct 14, 2008

### JJones_86

Oh, I meant counter clockwise, my bad

14. Oct 14, 2008

OK, good.