S-2 PROBLEMS IN EQUILIBRIUM
Loaded meter stick (lead in one end). Support frame made from lab hardware. Two ball-bearing pulleys. String. Two weight hangers. Slotted weight set. Beam balance.
Does the accuracy of the above method depend on the choice of the two suspension points? If so, what points are best. You must do the error analysis to answer this.
(2) PROBLEM: Consider the stick suspended from strings at 30 and 90 cm. If 500 grams were placed on each of these hangers, the stick would, of course, be unbalanced. What single additional force applied to the stick would bring it into balance, and where must that force be applied?
(1) Does a spring balance correctly measure forces in all positions; hanging down; upside down; and suspended horizontally?
No, as was shown by experment. The spring balance is calibrated to read correctly when hanging down in the usual manner. In this position the spring supports part of its own weight. In the horizontal position it doesn't support its own weight, and its reading will be too low. Scale readings will be much too high when the banance is hanging upside down. This is easy to check in lab, and the student should show evidence (data) that this was done.
(2) A student doing a lab problem with a balanced meter stick as in parts A and B notes that all of the weight hangers have equal mass. The student concludes that the hanger masses may safely be ignored, since their effects would balance out. Is this correct? Explain.
No, except in the very special case when the additional weights on the hangers are all equal. The ratios of forces determine equilibrium. Adding equal amounts to each hanger will not preserve the ratios except in the trivial case where all forces initially had the same size.
(3) Prove that the torque of a force is equal to the sum of the torques of the components of that force.
Place the Cartesian origin on the torque axis. The position vector of the force, that the force vector itself determin a a plane. So we can do the analysis using only two dimensions. From the diagram we see that:
τ = FRcosδ = FRcos(β-α)
τx = Fcosα , Rx = Rcosβ
τy = Fsinα , Ry = Fsinβ
τ = FRsin(β-α) = FR[sinβcosα- cosβsinα]
= Fsinα(Rcosβ) - [Fcosα(Rsinβ)] = τx + τy
The last step uses a familiar trig identity. This is the sum of two torques, one being a negative torque in this diagram.
The student familiar with vector algebra may prefer:
τ = R×F = (Rx+Ry)×(Fx+Fy) = Rx×Fx + Rx×Fy + Ry×Fx + Ry×Fy
= Rx×Fy + Rx×Fy = τx + τy
Here we are speaking of vector components, rather than the usual cartesian components (scalars). The vector components of a vector are any set of non-parallel vectors which have a sum equal to that vector. In this case we chose orthogonal components.
(4) If only two forces, forming a couple, act on a body, show that that body cannot be brought into equilibrium by adding just one more force.
A couple has zero net force. Therefore adding a non-zero force to a couple will give a non-zero net force, which cannot be in equilibrium.
(5) Prove that the size of a couple is independent of the choice of the center of torques. You may limit the discussion to a two-dimensional situation.
Draw a line perpendicular to the forces F. Locate a center of torques C anywhere on this line. The net torque about this center is
τ = Fx2 - Fx1 = F(x2 - x1)
This depends only on the separation of the two forces and their size, not on the location of the center of torques. Displacement of the line up or down obviously has no effect on this argument.
(6) Prove that if a body is in equilibrium under the action of several forces, among which is a couple, that the body will still be in equilibrium if the couple is moved to any other location, provided the couple's size is kept unchanged. You may limit the discussion to a two-dimensi onal situation.
This question merely checks to see whether the ideas of questions 4 and 5 were understood. The net force of a couple is zero, so its location has no effect on the net force. We have seen that the torque of a couple is determined only by the separation of thetwo forc es which make up the couple, and not on the couple's location relative to the center of torques, so moving it cannot change its torque or the net torque. In all of these questions we have assumed everything to be in two dimensions. If the couple is moved so that its plane does not coincide with the plane of ther est of the forces, this will change the equilibrium.
(7) [Bernard and Epp] If, in any part of the procedure, the meter stick were balanced resting at an angle, rather than in a horizontal position, would the meter stick be in equilibrium?
Yes. If the stick is at rest, it certainly is in equilibrium.
(8) [Bernard and Epp] Suppose the stick made a 10° angle with the horizontal in procedure A, but this fact was not taken into account in the calculations. When the oversight is discovered the situation is recalculated by explicitly including the 10° angle and correctly calculating the true lever arms. How do the "right" and "wrong" results compare?
One might suppose that since all of the lever arms would be modified by a constant cosine factor, no correction would be necessary. But that's not quite true.
Consider why the stick could balance this way. Conversely, why does it usually balance "naturally" in the horizontal position? The reason is because the fulcrum is cleverly located just a bit above the stick's center of mass. If it were below the center of mass, the stick would topple off. This displacement of the fulcrum above the center of mass determines the sensitivity of the balance. Precision laboratory balances use the same principle.
So the tilted stick would have it's center of mass slightly displaced horizontally from its pivot point. This introduces a small systematic error into all measurements of the hanger positions.
© 2004 by Donald E. Simanek.