Understanding Torque Balance and Center of Gravity in a See-Saw

[Wasseem92]

A poorly constructed see-saw has a fulcrum 2/3 of the way along its length. a) If the see-saw weighs 30 kg, where would a 20 kg child have to sit in order to balance the see-saw? b) What is the least mass that a child must have in order to balance the see-saw?
well i tried solving this question and my teacher gave this as a solution:

The center of gravity of the see-saw is assumed to be in the center, assuming that it is uniform. The force diagram is therefore as shown below. mss = 30 kg is the mass of the see-saw, mc = 20 kg is the mass of the child, and L is the length of the see-saw.

Figure 8.3: Problem 8.1


In order to find the distance x from the child to the fulcrum we can do a torque balance about the fulcrum:
sum of torque = (L/6)mssg - xmcg = 0

so that
x = (L*mss)/(6*mc) = = (1/4) L.


My question is: why did he start of with a distance of (L/6). Please i need help, this really baffles me.
Thank you.

 

[tiny-tim]

Welcome to PF!

Hi Wasseem92! Welcome to PF!

My question is: why did he start of with a distance of (L/6).

Because L/6 is the distance from the fulcrum to the centre of mass of the seesaw …

you can regard the whole mass of the seesaw as concentrated at its centre of mass.

(in fact, that's exactly why the centre of mass is defined, and is called that! )

 

[Wasseem92]

alright man thank you a lot! that really helped!