O.K. seems like I was wrong about the 2kN/m force. It can NOT be represented by a point load of 0.8N with distance 2.5m.
The magnitude of the force is the area of the triangle and where it acts is somewhere called the "centroid" (of a right triangle) and acts at two third's of its base...
I don't know what this one is asking for, really... is it the resultant force on point A? Or the reaction force?
What I tried doing is represent the 2 kNm as 0.8N acting 2.5m away from A and normally calculate the moment by splitting the forces into components.
I'm not getting the right...
I have figured it out.
Since we know where the line of action acts, we must find the magnitude of the moment (M) that makes it act at that location. If we take moments about point A, then (M) is equivalent to the actions of the two forces about point A.
So I broke the two forces into their...
Homework Statement
Homework Equations
ƩMoments = Moment + Fx*d + Fy*d
R = (Fx)i + (Fy)j
The Attempt at a Solution
I'm only interested in part (ii)
I calculated the components of the resultant force and they are
|R| = 1.15(i) + 9.36(j)
I know that the x-intercept of line of action...
OK... it turns out I have to reverse the sign for the horizontal components so it becomes like this:
-(3 * cos 25) * 5.4
+(2 * cos 30) * 4
But I don't understand why... the 3 kNm force is pulling to the right hence its horizontal component is positive.
Can someone explain?
Homework Statement
Homework Equations
M = F*r
The Attempt at a Solution
I tried splitting each force into its xy components and solve but the answer is wrong. Here is my attempt:
For point A:
Horizontal moments:
(3 * cos 25) * 5.4
- (2 * cos 30) * 4
Vertical moments:
- 4 * 1.5
- (3...