Recent content by Jasonn

Yes, thank you for your help. :)

I see where I made a mistake in the equations. Rather, the two torques should be equal, correct? So I need to find a distance from the axis, which I can place the mass and still remain in equilibrium?

K, I understand that much. You obviously need the mass to the left of the axis (finger) if the center of the stick is to the right. However, I don't understand WHERE to place the mass to the left exactly. I do see how the torque changes depending on where you place it from the axis.

If the mass is hanging from a point .25m from the end of one stick, then I can see why you can't do that (the stick wouldn't stay on your finger for one :-p ). Does that mean the weight should be placed in the center?

Alright. I have a .1kg mass and a 1m uniform stick. The length of the stick is given. The axis (my finger) will be placed .25m from one of the ends. Measuring the center of the stick to this will yield a .25m distance. The mass will be placed .25m from the OTHER end. Measuring this place on the...

We will not be able to measure the mass of the meter stick. Only the mass of the weight (approx. .1kg). We are trying to FIND the mass of the meter stick. Alright, first, you will draw a diagram of everything involved/labeled. The meterstick, the forces (force of weight of the stick and mass...

I've already explained this though. I've explained that I would place an axis somewhere on the stick, calculate the forces perpendicular to the axis for both the Fw of the stick and the mass. You've told me that these equations/answers are either way off or getting there. Aside from what I've...

No numbers...hmm We know that F = F(D) (perpendicular) Torquenet = F*d - F*d ... = 0 F*d...force has to be perpendicular to the distance involved (which in this case it is (Force of weight of the meterstick is perpendicular to the distance to the axis). Manipulations? I don't think I...

Alright, then just place the finger slighly off center (say .25m from the center), and solve for distances accordingly? T = -Fw1 * .25m - (-Fw2 * D) = 0 (where D is the distance from the force of weight of the mass to the axis)..let's say .5m T = -Fw1 * .25m +.981N * .5m = 0 T = -Fw1 *...

No no you are right...wasn't thinking straight at all. If the axis is where my finger is (.5m) from both ends, how would the equation fit T= -Fw1*Distance -Fw2*distance = 0 should the distance for Fw1 be 0 because it is at the axis or .5m?

Alright, the stick will be balanced on my finger, .5m from both ends. The mass will be suspended from a point (the axis?) on the stick. Using T=F*d - F*d ... = 0, Fw of the stick should be solveable Distances would be how far each force is from the axis, which can change depending on...

I made a mistake... [T = F*distance (perpendicular)] [T = (negative force of weight * distance) - (negative force of weight * distance) = 0] Torque = (-Fw1)*.5m - .981N*.25 = 0 [Solve.] = -Fw1*.5 + .736 = 0 = -Fw1*.5 = - .736 Fw = 1.4715N m = .15kg ...should have been what I posted.

K, so the axis will have to be placed above the mass... Fw1 is the force of weight of the meter stick Fw2 is the force of weight of the .1kg mass (.981N) L = .25m (placement of mass) Torque = (-Fw1)*.5m - .981N*.25 = 0 = -Fw1*.5 - .736 = 0 = -Fw1*.5 = .736 Fw = 1.4715N m = .15kg...

Should the axis be placed at the very end of the stick? Fw1 = ? at .5m Fw2 = ~.981N at .25m (example) L = 1m Sum of y: -Fw1 - Fw2 = 0 -Fw1 = Fw2 ...that can't be right. Shouldn't I be solving for tourqe though? Maybe this is incredibly too easy and I'm not on the right page. :-p

I was thinking sort of the same thing...maybe hanging it off a table at a fixed point. And yes, the meter stick is uniform.