Balance proves gravity isn't a force

[georger]

Consider every mobile that Calder ever made. They are all at rest and among the various mobiles you will see all angles between 0 and 90 degrees depending on the weights at each end and the lengths of the arms. (The arms are free to move so friction is not involved) Not sure its the same question as the balance but the problem is the same: what's the formula for the angle the arm makes (with the horizontal) as a function of the masses and the lengths of the arms. I tried the sum of the torques = 0 but something else is going on.

 

[russ_watters]

Welcome to PF.

Summing the torques is the proper approach. My guess is you're making the mistake of using the standard equation for torque without any transformations. But the equation requires the applied force to be perpendicular to the lever arm and in this case it is not. So you need to use trigonometry to either transform the force or the lever arm so they are perpendicular to each other. My preference would be to transform the lever arm length by calculating the horizontal component of the distance between the applied force's location and the fulcrum.

 

[georger]

Welcome to PF.

Summing the torques is the proper approach. My guess is you're making the mistake of using the standard equation for torque without any transformations. But the equation requires the applied force to be perpendicular to the lever arm and in this case it is not. So you need to use trigonometry to either transform the force or the lever arm so they are perpendicular to each other. My preference would be to transform the lever arm length by calculating the horizontal component of the distance between the applied force's location and the fulcrum.

Excellent! Thats what I was missing. Let me think about this for a bit. Whats not clear is: what is the vector perpendicular to the arm: Is it the vector whose vertical component is Mg or is it the vector in the corrdinate system of the lever arm rotated from the vertical?? Sorry... I'm not a physics student. Thanks again!

 

[russ_watters]

If you want to use the arm as-is, you need to calculate the component of mg perpendicular to it. Mg is the resultant vector, not a component vector. The two legs of the component vector are drawn parallel to and perpendicular to the arm.

It would certainly help you out if you drew yourself a picture. Bear-in-mind, though, that the arms on these devices are not straight lines and/or don't connect directly to the fulcrum, so you'll also need to draw-in and calculate the length and angle of the "effective" arm from the fulcrum to the weight.

 

[georger]

If you want to use the arm as-is, you need to calculate the component of mg perpendicular to it. Mg is the resultant vector, not a component vector. The two legs of the component vector are drawn parallel to and perpendicular to the arm.

It would certainly help you out if you drew yourself a picture. Bear-in-mind, though, that the arms on these devices are not straight lines and/or don't connect directly to the fulcrum, so you'll also need to draw-in and calculate the length and angle of the "effective" arm from the fulcrum to the weight.

I'll give that approach a try...and indeed I have drawn a picture. By the way its easy to make a simple "mobile" which shows the effect: take a stiff wire (galvanized 16 gage works great) put a loop roughly in the middle (wrap the wire around say a philips screwdriver so its easy to slip off) and attach two peices of cardboard to each end. Hang it buy a thread thru the loop and watch it come to rest at some angle. I just want to solve the case of a straight wire although there is a more complicated case that takes into account the weight of the wire and the shape of the wire. There is also a dynamics problem here : push one end down and watch it rock. Whats the equation of that motion (now ignoring friction)? I guess its just two pendulums stuck together

 

[Dadface]

The balance is designed so that its centre of gravity is vertically below the pivot point when the balance is unloaded.For zero load the weight of the balance itself(beam,scale pans etc) exerts zero moment.If a weight is added this will set up a moment which will cause the beam to turn but the centre of gravity of the balance now shifts horizontally away from the pivot point and sets up a moment in the opposite direction.The additional moment caused by the added weight decreases as the beam turns but the moment caused by the weight of the balance increases.

Whoops originally I wrote vertically above when it should have been vertically below.Now corrected.