# Thread: Statics-Need help with resultant moment problem

1. ## Statics-Need help with resultant moment problem

I have no idea if I am doing this right or exactly where to start. Moments keep confusing the hell out of me, its so frustrating. So far for part A I got the

resultant moment = -1200(sin x + cos x)

To get part A i took the cross product of the vector components of the distance between each force and A and the components of the force itself. Like this

M1 = r x F = (-3i + 2j) x (400cosx i + 400sinx j) = (-1200sinx - 800cosx) k
M2 = r x F = (0i + 1j) x (400cosx i + 400sinx j) = (-400cosx) k

Meq = (-1200sinx - 800cosx - 400cosx) k = -1200(sinx + cosx)

Umm so am i allowed to do this?

Is this correct and can someone guide me with the other parts of the problem?

Also one part that was left out of the drawing is that the distance from A to B is 1m

2. The force at D is in the opposite direction, so you should have:

M1 = r x F = (-3i + 2j) x (-400cosx i - 400sinx j) = (1200sinx + 800cosx) k

and so the total moment is

Meq = (1200sinx + 800cosx - 400cosx) k = 400(3sinx + cosx) k

(and you'll write $\theta$ instead of x in the solution, of course)

Part (d) should be easy - just substitute $90^{\circ}$ for x.

For (b), you can take the derivative and set it to zero. (I don't see how to do it without calculus - if you haven't had calculus, post again)

For (c), notice that if x is between 180 and 270 degrees, the forces will both be in the opposite direction, so the moment will have the opposite sign. So it must have passed through zero somewhere - just set Meq to zero and solve for x.

Post again if you're still having trouble.

3. Thank you. I have a question for part B though. Why do you take the derivative and set it to zero to solve for the maximum moment? I know how to do that, but I don't understand the reasoning behind it I guess.

4. If the derivative is greater than zero, you can increase x slightly to get a higher moment. If the derivative is less than zero, you can decrease x slightly to get a higher moment. So the derivative must be zero.

If you try to graph a function that has a maximum and is "smooth" enough, you'll have to draw it "flat" through the maximum. (Making a "corner" would not be "smooth" enough). All of these concepts can be defined formally, of course.

5. That makes sense, it has been a while since I have had to remember things like that. My formal math skills are subpar I'd say.

Thank you so much for your help.