Where should mass B be placed for rotational equilibrium?

[bmandrade]

Homework Statement

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

Homework Equations

torque = rFsin \Theta

The Attempt at a Solution

I found the torques but i don't know if they are right

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

mass C
\tau=(3)(10g)sin30 = 147 N\cdotm


then I tried to find where to put mass B

\tau_{mass A}+\tau_massB+\tau_{mass C}= 0

i don't know if i should use this equaiton of soemthing else to find where to place B

 

[Redbelly98]

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.

 

[bmandrade]

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

 

[Redbelly98]

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 ?

 

[bmandrade]

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

 

[Redbelly98]

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

 

[Redbelly98]

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.

 

[bmandrade]

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

 

[Redbelly98]

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 canceled 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:

τ_A+τ_B+τ_C= 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.

 

[bmandrade]

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

-1(60g)+3(10g)+ τ_c=0

-588+294 + τ_c= 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

 

[Redbelly98]

Yes. You got it

 

[bmandrade]

thank you! :)