(a) It is second class lever
(b) Given,
AB=1m,AF=0.4mand
BF=1−0.4=0.6m
Mechanical advantage = load/effort =BFAF=0.60.4=1.5
(c) Moment at point F is zero
Load×AF=Effort×BF
Effort=Load×AFBF=15×0.40.6=10kgf
Value of E=10kgf
In fig below shows a lever in use.
(a) To which class of lever does it belong?
(b) If FA = 80 cm, AB = 20 cm, find its mechanical advantage.
(c) Calculate the value of E.
In the figure below shows a lever in use.
(b) If AB = 1m, AF = 0.4 m, find its mechanical advantage.
The figure below shows a lever in use.
In fig below shows the use of a lever.
(a) State the principle of moments as applied to the above lever.
(b) To which class of lever does it belong? Give an example of this class of lever.
(c) If FA = 10 cm , AB= 490 cm, calculate : (i) the mechanical advantage, and (ii) the minimum effort required to lift the load (= 50 N).