Answer:
(a) Picture Attached
(b)
i. 30 Nm
ii. Distance between A and M = 0.375 m
Explanation:
(a)
The other two forces are as follows:
1. The weight of wheel barrows acting downward at Point m
2. The Normal Reaction acting upward on wheel
These are show in the picture attached in solution.
(b)
i-
Now, for the moment of 20 N force about the point A:
[tex]Moment = Fd\\\\where,\\\\F = Force = 20\ N\\d = perpendicular\ distance between\ force\ and point\ A = 1.5\ m\\Therefore,\\Moment = (20\ N)(1.5\ m)[/tex]
Moment = 30 Nm
ii.
We take the sum of all moments about point A. Taking counter-clockwise direction as positive:
[tex](Weight)(AM) - Moment\ due\ to\ 20\ N Force = 0\\(80\ N)(AM) = (30\ Nm)\\AM = Distance\ between\ A\ and\ M = \frac{30\ Nm}{80\ N}\\\\[/tex]
Distance between A and M = 0.375 m
