# Statics - Moment using both vector and scalar approaches

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1. Sep 28, 2015

### Alison A.

1. The problem statement, all variables and given/known data

2. Relevant equations
Mo=Fd
Mo=r x F

3. The attempt at a solution
Alright guys, I did the whole process but I'm pretty sure I just made a little bump somewhere in my calculations which screwed up my answers.

First I found everything I could find
OA = 350j, so the unit vector of OA is 1j

OB = ( 350sin(33°) )i + ( 350 + 350cos(33°) )j
= 191i + 644j

tan(θ) = (300/350), θ = 40.6°

AC = √(300)^2 + (350^2) = 461 mm
OC = 461 sin(33°+40.6°) + (350 + 461cos(33°+40.6°), = 442i + 480j

Then,
AB = OB - OA
= 191i + 644j -350j
= 191i + 294j
Unit vector of AB = 0.545i + 0.839j

Given force is F = 130k

AC = OC - OA
= 442i + 130j

Vector Approach:
MA = AC x F
= (442i + 130j) x (130k)
= 16900i -57460j
In the position of AB = MA * Unit vector AB
MAB = 39000 N * mm

MOA = MA ⋅ Unit vector OA
= (16900, -57460) ⋅ (0,1,0)
= 57500 N * mm

Scalar Approach:
MAB = F x d
= 130 x 300
= 39000 N * mm

h = AC sin (40.6° + 33°)
= 461 sin (73.6°)
= 442 mm

MOA = F x h
=130 x 442
=57500 N * mm

My answers match, so I'm not sure where I went wrong, any help would be greatly appreciated!

2. Sep 28, 2015

### Alison A.

Omg, all I didn't do what convert to N * m,
I'm dumb.

Check your units boys and girls.