Question

A 60 N weight is suspended from one end of a 10 m wooden beam which weighs 30 N. Where should the beam be picked up so that it balances?

Answer 1

The beam must be picked up at 8.33m respect to the opposite corner from the weight (to 1.66 m from the weight).

Step-by-step explanation:

As we want the beam to be balanced, we will define the moment of the beam and the weight respect to an arbitrary point x from the opposite corner to the weight. The beam weight will be placed at the half of the beam:

`M(x)=R*0+Wb*(lb/2-x)+W(lb-x)`

(See the ilustration below to guide)

Where Wb and lb are the weight and the length of the beam.

Note that the zero distance from the picked-up force is due the force is located at the arbitrary point so no moment is doing.

Now due the beam has to be balanced the total moment must be zero:

`0=0+Wb*(lb/2-x)+W(lb-x)`

Then is possible to find the x position point by solving the equation above:

`0=0+Wb*lb/2-Wb*x+W*lb-W*x`

`Wb*x+Wb*x=Wb*lb/2-+W*lb`

`x(Wb+W)=Wb*lb/2-+W*lb`

`x=(Wb*lb/2-+W*lb)/(Wb+W)`

For the respecting values:

`x=(30 N*10/2 m+6 N* 10 m)/(30 N+60 N)`

`x=(45 Nm+60 Nm)/(90 N)`

`x=(105 Nm)/(90 N) = 8.33 m`