An improvised car lift can be made with a piece of lumber as illustrated in the figure above. With distance load to fulcrum 0.2 m, distance effort force to fulcrum 2 m and a load on the lever - half the weight of the car 2000 kg - the effort force can be calculated
Fe = (0.5 2000 kg) (9.81 m/s2) (0.2 m) / (2 m)
= 981 N
≈ 100 kg
The effort force at a distance of 2 ft from the fulcrum can be calculated as
Fe = (1 lb) (1 ft) / (2 ft)
= 0.5 (lb)
The formula (1) can be modified to express required load if you know the effort, or required distance from fulcrum if load and effort forces are known and so on.
The level above where the fulcrum located between the load and effort force is often characterized as a first-class level mechanism.
A level where the load and effort force are located on the same side of the fulcrum is often characterized as a second-class level mechanism.
A force (weight) of 1 pound is exerted at a distance of 1 ft from the fulcrum.
The effort force at a distance of 2 ft from the fulcrum can be calculated as
Fe = (1 lb) (1 (ft) / (2 ft)
= 0.5 (lb)
The effort force at a distance of 2 m from the fulcrum can be calculated as
Fe = (1 kg) (9.81 m/s2) (1 m) / (2 m)
= 4.9 N
A lever mechanism where the input effort is higher than than the output load is often characterized as a third-class lever mechanism.
A force (weight) of 1 pound is exerted at a distance of 2 ft from the fulcrum.
The effort force at a distance of 1 ft from the fulcrum can be calculated as
Fe = Fl dl / de
= (1 lb) (2 ft) / (1 ft)
= 2 (lb)
A lever with two acting load forces and one effort force is indicated in the sketch below:
The generic equation for one effort force with one or more acting load forces can be expressed as
Fe = (FlA dlA + FlB dlB +.. + FlN dlN ) / de (2)
This equation is modified for three acting loads below.
A weight A of 1 pound is exerted at a distance of 1 ft from the fulcrum. A weight B of 2 pound is exerted at a distance of 2 ft from the fulcrum, and a weight C of 3 pound is exerted at a distance of 3 ft from the fulcrum.
The effort force at a distance of 2 ft from the fulcrum can be calculated as
Fe = (FlA dlA + FlB dlB +FlC dlC ) / de
= ((1 lb) (1 ft) + (2 lb) (2 ft) + (3 lb) (3 ft)) / (2 ft)
= 7 (lb)
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