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2D Moments confusion, does d have to be perpendicular?

  • Thread starter Willjeezy
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  • #1
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Homework Statement


http://imageshack.us/a/img812/8836/5tfn.jpg [Broken]

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Find F such that the moment around A will not exceed 1000N*m for any value of θ.

Homework Equations


M = fd


The Attempt at a Solution



I got the right answer, I found the distance of a -> b
AB= √( (2^2) + (3^2) )
AB=3.6

and then I did
M = fd
1000 = f (3.6)
f = 277.8 N

This is correct according to the answer in the back, but I don't understand why.
1. I was under the impression for: M = fd , d must be the perpendicular distance. Clearly, the d we are using is not perpendicular to f. Does d not have to be perpendicular to the force?

2. I thought this was a trick question at first, and logically I thought it made most sense that the greatest moment around A would be caused by a force only in the vertical direction. So I tried doing it where Fx = 0 and Fy is the variable of interest.

So i tried, M = (Fy)d
1000 = (Fy)(3)
but the fy I get is clearly not the answer. Can someone explain this to me.
 
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Answers and Replies

  • #2
6,054
390
If the force is not perpendicular to the distance, what is its moment?
 
  • #3
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I don't understand what you mean, Voko.
 
  • #4
6,054
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You wrote: "I was under the impression for: M = fd , d must be the perpendicular distance. Clearly, the d we are using is not perpendicular to f."

Clearly F and d are not perpendicular for every possible angle. So how does one go about computing the moment in such a case?
 
  • #5
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Would I break it into components instead?

Mx = fcosθ * (d)
My = fsinθ * (d)
 
  • #6
6,054
390
In 2D, moments are not vectors, so they have no components.

Have you not been given a definition of the moment of force? It should be in your course material/textbook, find it.
 
  • #7
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right right right.

M = fcosθ(3) - fsinθ(2)
10 000 = fcosθ(3) - fsinθ(2)

I have no clue what comes next. I've tried reading it, clearly I am confused. Wait, I am confused even more, my solution in the initial post yielded the right answer, was that by complete fluke?

It says in my book "d is the perpendicular to the line of action of the force" I still can't see how d will ever be the perpendicular on the line of action .
 
  • #8
6,054
390
What that definition really means is that when you compute the moment of force at point B about point A, you first find the component of force perpendicular to AB. Then the moment is the product of that component with AB.
 
  • #9
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http://imageshack.us/a/img547/2850/nzd7.jpg [Broken]

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Your not given θ or F, maybe because its 1am, but I swear I can't see what you are trying to help me see.

I understand F can be broken down into Fc and Fc1. Fc1 isn't important because Fc1 goes through our moment pt A, thus causing no moment.

I have no clue why Fc = F.
 
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  • #10
6,054
390
So you do understand that ##M = F_c d ##, correct? Now, what is ##F_c## if ##F## is at angle ##\alpha## with AB (note I said ##\alpha##, not ##\theta##)?
 
  • #11
6,054
390
Or let me ask that differently: does ##F_c## ever exceed ##F##? What is the max value of ##F_c##? At what angle with AB?
 
  • #12
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alpha = 90
 
  • #13
6,054
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So, what is the magnitude of ##F_c## at 90 degrees with AB, if the total magnitude of the force is F?
 
  • #14
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so wait, is θ just thrown in there to mess with me. Should I have known right from the jump that the largest moment generated around A would be Fc (perpendicular to AB)
 
  • #15
6,054
390
You see, since you are required to consider all possible angles ##\theta##, you might as well consider all possible angles ##\alpha##, which is easier, right?
 
  • #16
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Oooooooooohhhhh, Voko, you're JEDI MASTER.
 
  • #17
6,054
390
So, do you understand now why your initial solution was correct?
 
  • #18
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Yah, I missed the obvious until you pointed it out.

I appreciate you takin the time, voko.
 

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