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

Answer:

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

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:

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

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

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