Solving 2D moments about point O

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To solve for the magnitude of force F that creates a 15 N-m clockwise moment about point O, the setup involves determining the perpendicular distance from point O to the line of action of the force. The discussion emphasizes the importance of measuring distances perpendicular to the force's line of action rather than along the y-axis. For part (b), with F set at 100 N, the goal is to find the angle θ that results in a 20 N-m clockwise moment, requiring a similar approach to calculating moments. Participants are encouraged to clarify their diagrams and calculations to ensure accurate results. Understanding the relationship between force, angle, and moment is crucial for solving these problems effectively.
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(a) The moment the force, F, makes about O is 15 N-m clockwise, and θ = 30˚, find the magnitude of F.
(b) If the force F = 100 N, determine the angle θ so the force develops a clockwise moment about point O
of 20 N-m. Limit your answers to a range of 0º ≤ θ ≤ 90º.

I'm a little lost can someone please help me with how to set up equations so i can figure out the answers. Thanx
 

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welcome to pf!

hi majunee! welcome to pf! :smile:

start by completing the diagram …

continue the line of force backwards, and find its perpendicular distance from O :wink:​
 
thanx so is this correct so far
-15 KN-m = F[-(sin30°)-(50mm)-(300sin60°)]
 
hmm … you're going to need another hint:

call that corner "C"

draw the line G through C parallel to F

draw the two perpendicular lines from O to G and from C to F …

how long are they? :wink:​
 
Sorry I am really having trouble understanding the concept...
wouldnt 0-G be 300sin60 and c-f be 50 + xsin30
 
ahhh, you seem to be measuring vertical distances

no, to find the moment of a force, you need to measure distance(s) perpendicular to the line of the force :wink:

start again :smile:​
 
how? if force has continuing line of action shouldn't we measure along the yaxis which would be like i did?
 
why the y axis? :confused:

(are we using the same diagram? … the force is at 30°)
 
yes we are
so would it be -300sin60 + 300 cos 60 - 50 + xsin30 - x cos 30
 
  • #10
why the y axis? :confused:

and what is the "x" in your equation?
 

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