# Torque and equilibrium

1. Nov 16, 2008

1. The problem statement, all variables and given/known data

On a seesaw, Mass A is 60 kg, mass B is 30 kg, and mass C is 10kg. Mass A is 1.0 m from the piot and mass C is 3.0 m from pivot. Calculate torques of mass A and C about the pivot. Where should you place mass B in relation to the pivot for the system to be in rotational equilibrium

The seesaw is at an angle of 30 degrees from the horizontal. Mass A is on the left side and Mass C is on the right side
2. Relevant equations

torque = rFsin $$\Theta$$

3. The attempt at a solution
I found the torques but i dont know if they are right

mass A (g=9.8m/s2)
$$\tau$$=(1)(60g)sin 30 = 294 N$$\cdot$$m

mass C
$$\tau$$=(3)(10g)sin30 = 147 N$$\cdot$$m

then I tried to find where to put mass B

$$\tau$$mass A+$$\tau$$massB+$$\tau$$mass C= 0

i dont know if i should use this equaiton of soemthing else to find where to place B

Last edited: Nov 16, 2008
2. Nov 16, 2008

### Redbelly98

Staff Emeritus
Where does the 30 degrees come from? We can't see the figure, if there is one, or on which side of the pivot masses A and C are located.

3. Nov 16, 2008

The seesaw is at an angle of 30 degrees from the horizontal. Mass A is on the left side and Mass C is on the right side

4. Nov 16, 2008

### Redbelly98

Staff Emeritus
Okay.

There is a problem here. Yes, the angle of the seesaw to the horizontal is 30 degrees. But what is the angle θ in the equation:
torque = r F sinθ ?

5. Nov 16, 2008

I got the equationn from the book since for the force what is need is the component on the horizontal

6. Nov 16, 2008

### Redbelly98

Staff Emeritus
Check the book formula and discussion again, what exactly does it say that angle is?

7. Nov 16, 2008

### Redbelly98

Staff Emeritus
Hint: if the seesaw were exactly horizontal then the forces are vertical and still exert a torque, even the the horizontal force component is zero.

8. Nov 16, 2008

so in that case i dont need the angle because even those there is an angle in the seesaw the force is not being applied at an angle

9. Nov 16, 2008

### Redbelly98

Staff Emeritus
Well, basicly there are two things going on:

1. You don't really need the angle because every term in the torque equation has the same angle factor, so these can be cancelled out of the equation.

2. You were using the wrong angle dependence when you calculated the torques, but that doesn't really matter (see 1).

So let's get back to the problem ...

Mass A is on the left, so we could say that it is located at -1.0m from the pivot (note the - sign)
Mass C is to the right, so it's located at +3.0 m.

Can you set up the torque equation you wrote before:

τABC= 0

using what you know about the masses and positions of A and C? You may leave out the angle sin and/or cos stuff.

10. Nov 16, 2008

oh ok it makes sense
so now that everything is clear
I set up the equation

-1(60g)+3(10g)+ τc=0

-588+294 + τc= 0

r F = 294
r (30g) = 294
r = 1

(g = 9.8)

this means that mass B must be place 1 m away from the pivot ( same side as mass C) to have no net torque in the system

11. Nov 16, 2008

### Redbelly98

Staff Emeritus
Yes. You got it

12. Nov 16, 2008

thank you! :)