WHAT factor affects the mechanical advantage of machine?
Answers
Answer:
length of the incline divided by the vertical rise, the so-called run-to-rise ratio.
The ideal mechanical advantage (IMA) - ignoring internal friction - of a lever depends on the ratio of the length of the lever arm where the force is applied divided by the length of the lever are that lifts the load. The IMA of a lever can be less than or greater than 1 depending on the class of the lever.
If on one end of a class 1 lever in equilibrium force
F
is applied on a distance
a
from a fulcrum and another force
f
is applied on the other end of a lever on distance
b
from a fulcrum, then
F
f
=
b
a
Explanation:
Consider a lever of the 1st class that consists of a rigid rod that can rotate around a fulcrum. When one end of a rod goes up, another goes down.
This lever can be used to lift up a heavy object with significantly weaker than its weight force. It all depends on the lengths of points of application of forces from the fulcrum of the lever.
Assume that a heavy load is positioned at a length
a
from the fulcrum, the force it pushes down on a rod is
F
.
On the opposite side of a rod at a distance
b
from the fulcrum we apply a force
f
down such that two a lever is in equilibrium.
The fact that a lever is in equilibrium means that the work performed by forces
F
and
f
when a lever is pushed on either side by a small distance
d
must be the same - whatever work we, using force
f
, perform to push down our end of a lever on a distance
b
from the fulcrum should be equal to work to lift a heavy object on a distance
a
on the other end of a lever.
Rigidity of a rod that serves as a lever means that the angle a lever turns around a fulcrum is the same on both ends of a lever.
Assume that a lever turned by a small angle
ϕ
around a fulcrum slightly lifting a heavy weight. Then this heavy weight that exhorts a force
F
on one end of a rod at a distance
a
from a fulcrum was lifted by
a
⋅
sin
(
ϕ
)
height. The work performed must be
W
=
F
⋅
a
⋅
sin
(
ϕ
)
On the other end of a rod, on distance
b
from the fulcrum, force
f
pushed the lever down by
b
⋅
sin
(
ϕ
)
. The work performed equals to
W
=
f
⋅
b
⋅
sin
(
ϕ
)
Both works must be the same, so
F
⋅
a
⋅
sin
(
ϕ
)
=
f
⋅
b
⋅
sin
(
ϕ
)
or
F
f
=
b
a
From the last formula we derive that the advantage of using a lever depends on a ratio between lever ends' distance from fulcrum. The more the ratio is - the more advantage we have and more weight we can lift.
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