Question

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?

Answer 1

`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`