# Eng Mechanics: Resultants of force systems

1. Nov 22, 2007

### Edwardo_Elric

1. The problem statement, all variables and given/known data
In the figure , the moment of a certain force F is 180ft - lb clockwise about O and 90ft-lb counterclockwise about B. If its moment about A is zero, determine the force.

2. Relevant equations
Moment of a force
Mo = Fy(x)
Mo = Fx(y)

3. The attempt at a solution
Since its moment about A is zero then the force passes through A down to the right because the problem mentioned its directions:

Moment of F w/ respect to O = 180ft - lb = Fx(y)
180ft - lb = Fx (3)
Fx = 30lb

Moment of F w/ respect to B = 90ft - lb = Fy(x)
90ft - lb = Fy(6)
Fy = 15lb
using Summation of components:
F^2 = (30)^2 + (15)^2
F = 61.8lb

the answer at the book is 75lb down to the right at theta = 36.9 degrees
is the book a typo?

2. Nov 22, 2007

### PhanthomJay

your math is off.....Fx = 60

What about the conrtribution of Fx(y)??
No

3. Nov 24, 2007

### Edwardo_Elric

Moment of F w/ respect to O = 180ft - lb = Fx(3)
Fx = 60lb

I dont understand this part:
Since F passes through A... and you compute its moment w/ respect to O
Fx on this part is clockwise so it is a positive moment
Fy on this part is counterclockwise so it has a negative moment

Moment of F w/ respect to B = 90ft-lb = Fx(3) - Fy(6)
90 = (60)(3) - Fy(6)
Fy(6) = 90 - 180
Fy = 30lb

but when i put the counter clockwise as positive moment the answer would be 45 and the answer would be the same as of the back of the book....
is it because of the signs?

Last edited: Nov 24, 2007
4. Nov 26, 2007

### PhanthomJay

You have chosen clockwise as positive therfore counterclockwise is negative. Therefore, your equation should read, asuming Fy acts down,
-90 = 60(3) -Fy(6)
-270 = -Fy(6)
Fy = 45 downward