A lever is used to lift a heavy load. When a 52 N force pushes the left end of the lever down 0.71 m, the load rises 0.3m. If the lever is only 78% efficient, what would be the weight of the load it could lift with this output?

1 Answer
Nov 28, 2015

Answer:

#96"N"#

Explanation:

Applying the conservation of energy we can say that the work done on the load is equal to the work done by the effort.

Since work done = force x distance moved in the direction of the force we can write:

#Lxxd_l=Exxd_e#

#:.(Lxxd_l)/(Exxd_e)=1#

The term #L/E# is called the mechanical advantage #(MA)#.

The term #d_e/d_l# is called the velocity ratio #(VR)#.

So we can write:

#(MA)/(VR)=1#

This assumes that the machine is 100% efficient and that there are no energy losses due to friction etc. However, we do not live in an ideal world and machines are rarely 100% efficient.

The ratio MA/VR is used as a measure of efficiency and is usually expressed as a percentage:

Efficiency = #(MA)/(VR)xx100#

#:.#Efficiency#=((L/E)/(d_e/d_l))xx100#

#:.#Efficiency#=(Lxxd_l)/(Exxd_e)xx100#

Now we can apply this to this problem:

#78=(Lxx0.3)/(52xx0.71)xx100#

#:.L=(78xx52xx0.71)/(0.3xx100)#

#L=96"N"#

You don't need to go through this derivation every time you get a problem like this.

Just remember these 3 points:

1.

#MA=L/E#

2.

#VR=d_e/d_l#

3.

Efficiency =#(MA)/(VR)xx100#