Finding equivalent force and couple for 120 kN force at point A

[sean777]

Homework Statement

The question reads:
F1 is 120kn at A. B is located 300mm from A. f1 is inclined at 40 degrees to the x axis. Force one may be replaced by a statically equicilant force system consisting of a single force f2 and a couple c. Let force f2 pass through b


It asks you to
B) determine the plane, magnitude and direction and inclination of force f2
C) determine the plane of the couple c, it's sense of rotation and the magnitude of it's resultant

No idea where to start with this. Help would be greatly appreciated

 

[tiny-tim]

welcome to pf!

hi sean777! welcome to pf!

show us what you've tried, and where you're stuck, and then we'll know how to help!

 

[sean777]

I understand that a couple is made of 2 forces acting in opposite directions but how to replace this force with a couple and another force is what is confusing me.

 

[tiny-tim]

well, forget the couple for the moment …

what do you think the magnitude and the direction of the second force are?​

 

[sean777]

I imagine it would be 120kn to the left inclined at 40 degrees to the x-axis with the line of action passing through B

 

[tiny-tim]

I imagine it would be 120kn to the left inclined at 40 degrees to the x-axis with the line of action passing through B

ah, no …

it should be parallel to the original force (down and to the right), not opposite to it

ok, now subtract the orignal force from that to get the couple ​

 

[sean777]

ah, no …

it should be parallel to the original force (down and to the right), not opposite to it

ok, now subtract the orignal force from that to get the couple ​

And now I'm completely confused and don't know what to do

 

[pongo38]

The hint is: you have to add a force to the system, and then add to it an equal and opposite force. You then have three forces. You're neally there.

 

[NascentOxygen]

Imagine a heavy steel bridge strut lying in the middle of the welding factory floor. To slide it out of the way, you push against it. But unless you push at the exact right spot, not only will it slide along the floor, but it will also rotate. But it might happen that this optimum spot to push against (point A) is just where the paint hasn't dried yet. So you move along a bit and push with the same force[/color][/color] (i.e., same magnitude & direction) but where the paint is safely dry (point B). This gets the beam moving all right, but to stop it rotating you have to arrange for someone to exert a torque to cancel the turning movement you are now causing. How to counteract the torque you are causing by pushing at point B instead of point A? How to describe what must be done to cancel it out? One way is to express it as a pair of forces about point B.