Mechanical advantage of inclined plane(derivation)
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Suppose, inclination AB=l
Vertical height, AC=b
Angle of elevation, Angle ABC= θ
Suppose a load be raised from the bottom A to the top C of the inclined plane by applying an effort E along the inclined plane.
If there is no friction on the inclined plane,
When the body is in equilibrium, the forces acting on the load balances each other.
From the diagram,
mgcos θ=R
mgsin θ=E
Therefore, the minimum effort needed to pull the load along the inclined plane is mgsin θ.
The distance moved by effort E is given by,
d=AB=l
Distance moved by the load is d'=CA=b
We know that,
Mechanical advantage=Load/Effort
=mg/mgsin θ
=1/sin θ
From the triangle ABC,
sin θ=AC/AB=b/l
Hence,
M.A.=1/sin θ=l/b
Mechanical advantage of an inclined plane is its length divided by the vertical height through which the load moves.
Vertical height, AC=b
Angle of elevation, Angle ABC= θ
Suppose a load be raised from the bottom A to the top C of the inclined plane by applying an effort E along the inclined plane.
If there is no friction on the inclined plane,
When the body is in equilibrium, the forces acting on the load balances each other.
From the diagram,
mgcos θ=R
mgsin θ=E
Therefore, the minimum effort needed to pull the load along the inclined plane is mgsin θ.
The distance moved by effort E is given by,
d=AB=l
Distance moved by the load is d'=CA=b
We know that,
Mechanical advantage=Load/Effort
=mg/mgsin θ
=1/sin θ
From the triangle ABC,
sin θ=AC/AB=b/l
Hence,
M.A.=1/sin θ=l/b
Mechanical advantage of an inclined plane is its length divided by the vertical height through which the load moves.
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