# Physics balancing problem?

It says "A 2kg mass is 5m away from the fulcrum of a see-saw. Where must a 3kg mass be placed to balance the see-saw?

Update:

can you guys explain how you did it please?

Relevance

lets balance the see-saw. So it'll rotate neither left nor right, right?

2*5=3*d;

d=10/3 ;

righto!

But what weighs 2 Kg and is found 5m from a seesaw fulcrum? A large seesaw too , if u ask me ;-)

• To balance, you have to have the fulcrum point be the "equal sign" in your equation, so that you have the (mass1 x distance1 from fulcrum) = (second mass2 x distance2 from fulcrum)

2kg * 5m = 3 kg * X

solve for X:

X = 2kg* 5m / 3kg

X = 3.33m

So this means if the 2kg is 5m from the pivot point, then the 3kg on the other seat of the see saw will be 3.33m from the pivot

• there must be a balance of torque. torque = force * distance. so on one side there is a 2kg mass 5m away. 2kg * 5m is 10 and that is the torque on that side. now on the other side u need the same torque. therefore u need 10= 3kg* d where d is the distance. you solve it and divide 10/3 to get 3 and a third. not 3.5 lol

• weight of 2kg = 2g N

weight of 3kg = 3g N

moment of 2g N about fulcrum = moment of 3g N about fulcrum

2g * 5 = 3g * x

x = 10/3 = 3.333 m ---answer

Source(s): my brain (Prof TBT)