or (solving for hcrit, the critical height where it barely tips over even as it starts to slip):

Now, does this make sense? If µs → 0 (a frictionless surface) we will never tip it before it starts to slide, although we might well push hard enough to tip it over in spite of it sliding. We note that in this limit, hcrit → ∞, which makes sense. On the other hand for finite µs if we let W become very small then hcrit similarly becomes very small, because the block is now very thin and is indeed rather precariously balanced.

The last bit of “sense” we need to worry about is hcrit compared to H. If hcrit is larger than H, this basically means that we can’t tip the block over before it slips, for any reasonable µs. This limit will always be realized for W H. Suppose, for example, µs = 1 (the upper limit of “normal” values of the coe cient of static friction that doesn’t describe actual adhesion). Then hcrit = W/2, and if H < W/2 there is no way to push in the block to make it tip before it slips. If µs is more reasonable, say µs = 0.5, then only pushing at the very top of a block that is W × W in dimension marginally causes the block to tip. We can thus easily determine blocks “can” be tipped by a horizontal force and which ones cannot, just by knowing µs and looking at the blocks!

7.4: Force Couples

Figure 99: A Force Couple is a pair of equal and opposite forces that may or may not act along the line between the points where they are applied to a rigid object. Force couples exert a torque that is indenpedent of the pivot on an object and (of course) do not accelerate the center of mass of the object.

Two equal forces that act in opposite directions but not necessarily along the same line are called a force couple. Force couples are important both in torque and rotation problems and in static equilibrium problems. One doesn’t have to be able to name them, of course – we know everything we need to be able to handle the physics of such a pair already without a name.

One important property of force couples does stand out as being worth deriving and learning on its own, though – hence this section. Consider the total torque exerted by a force couple in the coordinate frame portrayed in figure ??:

This torque no longer depends on the coordinate frame! It depends only on the di erence between ~r1 and ~r2, which is independent of coordinate system.

Note that we already used this property of couples when proving the law of conservation of angular momentum – it implies that internal Newton’s Third Law forces can exert no torque on a system independent of intertial reference frame. Here it has a slightly di erent implication – it means that if the net torque produced by a force couple is zero in one frame, it is zero in all frames! The idea of static equilibrium itself is independent of frame!

It also means that equilibrium implies that the vector sum of all forces form force couples in each coordinate direction that are equal and opposite and that ultimately pass through the center of mass of the system. This is a conceptually useful way to think about some tipping or slipping or static equilibrium problems.

Example 7.4.1: Rolling the Cylinder Over a Step

Figure 100:

One classic example of static equilibrium and force couples is that of a ball or cylinder being rolled up over a step. The way the problem is typically phrased is:

a)Find the minimum force F that must be applied (as shown in figure 100) to cause the cylinder to barely lift up o of the bottom step and rotate up around the corner of the next one, assuming that the cylinder does not slip on the corner of the next step.

b)Find the force exerted by that corner at this marginal condition.

The simplest way to solve this is to recognize the point of the term “barely”. When the force F is zero, gravity exerts a torque around the pivot out of the page, but the normal force of the tread of the lower stair exerts a countertorque precisely su cient to keep the cylinder from rolling down into the stair itself. It also supports the weight of the cylinder. As F is increased, it exerts a torque around the pivot that is into the page, also opposing the gravitational torque, and the normal force decreases as less is needed to prevent rotation down into the step. At the same time, the pivot exerts a force that has to both oppose F (so the cylinder doesn’t translate to the right) and support more and more of the weight of the cylinder as the normal force supports less.

At some particular point, the force exerted by the step up will precisely equal the weight of the step down. The force exerted by the step to the left will exactly equal the force F to the right. These forces will (vector) sum to zero and will incidentally exert no net torque either, as a pair of opposing couples.

That is enough that we could almost guess the answer (at least, if we drew some very nice pictures). However, we should work the problem algebraically to make sure that we all understand it. Let us assume that F = Fm, the desired minimum force where N → 0. Then (with out of the page positive):

Xp

where I have used the r F form of the torque in both cases, and used the pythagorean theorem and/or inspection of the figure to determine r for each of the two forces. No torque due to N is present, so Fm in this case is indeed the minimum force F at the marginal point where rotation just starts to happen:

−

Next summing the forces in the x and y direction and solving for Fx and Fy exerted by the pivot corner itself we get:

Obviously, these forces form a perfect couple such that the torques vanish.

That’s all there is to it! There are probably other questions one could ask, or other ways to ask the main question, but the idea is simple – look for the marginal static condition where rotation, or tipping, occur. Set it up algebraically, and then solve!