What is Solving a Statics Problem: Finding the Resultant Force?

[findley]

Homework Statement

Here is a link to the problem that I am having issues with.

http://tinypic.com/r/8zfb5h/7

Homework Equations

F x perpendicular distance

The Attempt at a Solution

http://tinypic.com/r/1tamma/7


I have no idea where to go from there, or if I am even on the right path.

I separated the force into components but I have no idea what to do next.

I tried to for the equation 500 Nm = -P_y (1.3164m) - P_x (1.2276m) but got stuck.

Any help would be greatly appreciated!

Thanks

 

[PhanthomJay]

I separated the force into components but I have no idea what to do next.

I tried to for the equation 500 Nm = -P_y (1.3164m) - P_x (1.2276m) but got stuck.

You can do it that way, just note that Px, Py, and P are trigonometrically related, so you can solve for P using that relationship.. Usually, using the definition of M = F times perpendicular distance works well, but in this case, try the equivalent definition of a moment using the vector cross product M = r X F

 

[findley]

Unfortunately I have not done the vector cross product. Is it something I could teach myself in a few days?

Thanks!

 

[fizzynoob]

You can teach yourself the cross product in less than 5mins, it kinda like a cofactor expansion to find a determinant

 

[PhanthomJay]

Unfortunately I have not done the vector cross product. Is it something I could teach myself in a few days?

Thanks!

The vector cross product r X F is (r)(F)(sintheta), where r is the magnitude of the position vector from the point about which you are summing moments to the point of application of the Force , F is the magnitude of the Force itself, and theta is the smallest of the angle between r and F. If M acts clockwise about the point you are summing moments about, then the moment is negative, and if it is ccw, the moment is positive (or vice versa, depending on sign convention you use). The moment has a direction, cw or ccw, or, mathematically speaking, a direction perpendicular to the plane in which the moment acts. Depending on the problem, one method is easier than the other, but they both lead to the same result.

 

[findley]

Thank you both very much. I have found the answer to the problem.