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Mindmap von Thomas Marshall, aktualisiert more than 1 year ago
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Erstellt von Thomas Marshall
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Mechanics-Chapter
5-Moments
- Equations
- Force of
Moment=Perpendicualr
distance from Pivot X Force
- P=FD
- Remember this must
include a direction
(Clockwise or
Anticlockwise)
- If not perpendicular the use the equation
- P=F X Dsinθ
- P=F X Dsinθ
- If not perpendicular the use the equation
- Remember this must
include a direction
(Clockwise or
Anticlockwise)
- P=FD
- Force of
Moment=Perpendicualr
distance from Pivot X Force
- We can find the sum of multiple moments
around a point by taking one from the other. EG
A force 3N acting clockwise with a perpendicular
distance of 5m and a force 10N with a
perpendicular distance of 2m acting anticlockwise are acting on the
point P. Find the resultant force.
- So we know that one is 5X3 and the other is
2X10. So if we take one and mius it from
the other = (2X10)-(5X3)= we get a force of
5NM. We know this will be anticlockwise as
that was the number we took away from.
- So the answer is 5NM anticlockwise.
- If you are unsure what way
it will be moving then that's
easy. First, what was the
first force you used. Is it
clockwise or anticlockwise?
If it's positive it will be that
one negative the opposite
one.
- If it is not perpendicular the
just use Dsinθ X F for each
moment like stated in the
equations section.
- If it is not perpendicular the
just use Dsinθ X F for each
moment like stated in the
equations section.
- If you are unsure what way
it will be moving then that's
easy. First, what was the
first force you used. Is it
clockwise or anticlockwise?
If it's positive it will be that
one negative the opposite
one.
- So the answer is 5NM anticlockwise.
- Still draw a diagram
for these as it will
help. I have not as it
is a simple diagram
and you should be
able to draw it
without my help on
that
- So we know that one is 5X3 and the other is
2X10. So if we take one and mius it from
the other = (2X10)-(5X3)= we get a force of
5NM. We know this will be anticlockwise as
that was the number we took away from.
- When something is in equilibrium its resultant
force =0. This means we can solve the reaction
force at any point by finding just one point
- For example- A uniform
rod AB of length 5m and
weight 20N is resting
on two points C and D.
Such that C is 1m from A
and 3m from D. Find the
reaction at C and D.
- We know RC + RD = 20. So if we
consider the moments about
point C we get 20 X 1.5 = RC X 3.
We know this as the clockwise
moments = the anticlockwise
moments. 30=3RC. RC=10
- If more
forces are
involved just
add them to
each side.
- When we have a
non-uniform rod
it is the same bu
the point of mass
is not in the
middle.
- This therefore only
changes your diagram
and just move the
Weight/mass to where it
should and change the
distances. Everything
will stay the same.
- If you don't have
the mass then just
rearrange the
formula to help
you find it.
- If you don't have
the mass then just
rearrange the
formula to help
you find it.
- This therefore only
changes your diagram
and just move the
Weight/mass to where it
should and change the
distances. Everything
will stay the same.
- When we have a
non-uniform rod
it is the same bu
the point of mass
is not in the
middle.
- If more
forces are
involved just
add them to
each side.
- We know RC + RD = 20. So if we
consider the moments about
point C we get 20 X 1.5 = RC X 3.
We know this as the clockwise
moments = the anticlockwise
moments. 30=3RC. RC=10
- Start by drawing a diagram like the one below. This will help a lot.
- For example- A uniform
rod AB of length 5m and
weight 20N is resting
on two points C and D.
Such that C is 1m from A
and 3m from D. Find the
reaction at C and D.
- Draw a diagram everytime it will help a lot.
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