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Hi everbody ,
If F= 2i-3j , and if the moment vector of F about B (1,4) equals 16 k
find the equation of the action line of F .
Sorry can't write in LaTex
Help appreciated
Last edited by Avva (2011-03-25 20:37:20)
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hi Avva,
Only just read this and I'm busy catching up on loads of mail after being away from home.
I'll post an answer (I hope ) later today,
Bob
Children are not defined by school ...........The Fonz
You cannot teach a man anything; you can only help him find it within himself..........Galileo Galilei
Sometimes I deliberately make mistakes, just to test you! …………….Bob
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hi Avva,
See below for a diagram. I've put F with the right size vector components and (1,4) in the right place but the line of action for F will not be right as I have not worked it out yet.
The point (x,y) is where a line from (1,4) meets the line of action of F at 90 degrees.
so
using the 'cross or vector product'
and also
using the dot or scalar product as the vectors are perpendicular.
Are you able to construct the two equations for x and y from this?
Once you have (x,y) you can construct the equation of the line going through this point with gradient -3/2.
Post again if you need more details.
Bob
Last edited by Bob (2011-03-28 21:33:38)
Children are not defined by school ...........The Fonz
You cannot teach a man anything; you can only help him find it within himself..........Galileo Galilei
Sometimes I deliberately make mistakes, just to test you! …………….Bob
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thank u Bob I was about 2 lose hope but why did u impose that The point (x,y) is where a line from (1,4) meets the line of action of F at 90 degrees? Sorry for bothering u
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am sorry if it seems lyk a stupid question
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you make it on the perpendicular ??!
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hi Avva,
Definitely not a stupid question. Mathematicians should be asking these questions all the time.
The easy answer is because the torque is defined this way, but that doesn't help you much so here's the underlying reason.
Take a look at the diagram below.
B is the turning point; F is the force.
Let's suppose we take a point Y on the line of action of F and calculate the torque using the line BY.
Also let the acute angle between BY and the line of action be θ
Split F into two components, one acting along the line BY, the other at right angles to BY. I have shown these components as black lines.
The total moment will be the sum of the moments of these two components.
But the component acting along BY has no turning effect about B because it acts through B
So the total moment is just the turning effect of the component that is at right angles to BY T = (F.sinθ) x BY)
Now let's choose X instead of Y.
X is the point on the line of action of the force so that BX is perpendicular to the line of action.
BX = BY sinθ
So T = F.sinθ x BY = F x BY.sinθ = F x BX
So you can get the correct torque two ways.
(i) Choose any Y on the line of action and calculate (the component of the force that is at right angles to BY) x BY
or
(ii) Fix X so that BX is perpendicular to the line of action and compute (the magnitude of the force) x BX
Both give the same result.
Bob
Last edited by Bob (2011-04-01 06:48:39)
Children are not defined by school ...........The Fonz
You cannot teach a man anything; you can only help him find it within himself..........Galileo Galilei
Sometimes I deliberately make mistakes, just to test you! …………….Bob
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many thanks for your great illustration
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You're welcome.
Bob
Children are not defined by school ...........The Fonz
You cannot teach a man anything; you can only help him find it within himself..........Galileo Galilei
Sometimes I deliberately make mistakes, just to test you! …………….Bob
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