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If I apply a single force to an object ‘floating in free space’, then it will either translate (if the force is in line with the Cof G ) or more generally it‘ll rotate about the C of G due to the couple =Force x perpendicular distance. I’m happy so far…. BUT…

... this of course never actually happens (except in space…) and more particularly what if my arbitrary shaped object is sitting on a table (as most blocks do!) Then experience and experimentation says that any offset force will actually cause a momentary rotation about a point other than the ‘centre’ of the object.

Presumably the position of this real rotation centre is such as to balance out the effect of friction resisting the force, and to balance turning moments in some way?

Any suggestions or hints on how to find this real rotation centre for a given frictional coefficient, shape (rectangle/ circle / arbitrary) and pushing force direction?

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Suppose you have a heavy beam resting on a table of friction-coefficient $\mu$, idealized as a 1d line, of length 2L resting between coordinate x=-L and coordinate x=L, with a uniform mass density, pushed at a distance x from the midpoint with a force F. As you increase F, you start moving at the point where the friction force becomes equal to the coefficient of friction times the mass density, which is some critical value $f_0$.

The pivot point position, -L

To have equilibrium, the forces must balance:

$$ F+ f(y+L) - f(L-y)= 0 $$

which gives f in terms of F by $2yf= -F$ (for x positive, y is typically negative, so that f will be positive)

The torques around the pivot point at y must balance:

$$ f{(y+L)^2\over 2} + f {(L-y)^2\over 2} - F (x-y)= 0$$

Where the first two torques are due to the constant f force from y to the end, and both are in the same direction, while the last torque is due to the pushing, and it is in the opposite direction. Substituting for f in terms of F removes F from the equation and gives a quadratic equation for y, which solves to

$$ y = x \pm \sqrt{x^2 + L^2} $$

This has a geometric interpretation. To find the pivot point for the line-mass pushed at position x, draw the circle C whose diameter is the line-mass, and draw a perpendicular bisector to the line mass, and mark where it intesects the circle C as point P. Draw the circle with center at the pushing point x which passes through P, and where it intersects the line mass, that's y.

This immediately shows geometrically (although it is also obvious from the equation) that the point y is uniquely determined for all nonzero x's. When the force is exactly at the center, the two solutions are both physical, they are

$$ y = \pm L $$

which correspond to pivot points at either end of the rod. This situation gives a constant friction force along the rod.

When x is right at the center, the friction force f is everywhere F/2L (using either center). The critical value for motion is when $F= 2Lf_0$, or $ F = \mu Mg$, where M is the total mass of the rod, and g is the acceleration of gravity. This is the standard situation.

When x is right at the edge, the pivot point is at $y=L(\sqrt{2}-1)$, and the friction force f is bigger, it is

$$f = - {F\over 2y} = {F\over 2L} (1+\sqrt{2})$$

So that the critical F is determined to be

$$ F_c = {2L f_c \over 1+\sqrt{2}} = {\mu Mg \over 1+\sqrt{2}} = \mu Mg (\sqrt{2}-1)$$

so that the critical force is reduced to .414 of its usual value, and the pivot point is about 30% from the end of the rod.

The fact that this is a quadratic equation gives two solutions, but the solution outside the body is spurious. Rotation about this point does not produce force/torque balance, because there is no friction outside the body, and the quadratic torque formula fails in this regime. There is one unique solution for the pivot point, except when you are pushing exactly at the center, and then there are two. This limiting ambiguity is resolved by giving the beam a finite thickness.

Rectangle

The case of the thin-line is atypical, because the zero thickness means that the pivoting point is never outside the body. This is fixed by considering a rectangle with uniform mass density resting on a table.

If the rectangle edges are parallel to the axes, and the force is entirely in the vertical direction, acting along the edge at a displaced by an amount x horizontally from the middle, the pivot point is located somewhere on the line bisecting the vertical side, and horizontal location of the pivot point is y.

The force of friction is given by taking the unit vector from (a,b) and joining it to (y,0) (the rectangle runs from -L to L in the first coordinate, and from -H to H in the second coordinate--- I am using y for something else, so the first coordinate is "a" and the second is "b"). The unit vector is:

$$ u = {(a-y,b)\over \sqrt{(a-y)^2+ b^2}}$$

Rotating this vector by 90 degrees $(x,y)\rightarrow (y,-x)$ gives the direction of the friction force, and multiplying by f, its uniform magnitude, gives the local friction force:

$$ f{(b,y-a)\over \sqrt{(a-y)^2 + b^2}}$$

Then the condition of force balance gives

$$ -F = f \int_{a,b} {y-a\over \sqrt{(y-a)^2 + b^2}} da db$$

which determines f from F (and vice-versa), while the condition of torque-balance around the pivot point determines y:

$$ F(x-y) = f \int \sqrt{(a-y)^2 + b^2} da db$$

These two equations have a unique solution for all situations (there are never two solutions). This can be seen formally by writing the equation determining y,

$$ (x-y) \int {(a-y) \over \sqrt{(a-y)^2 + b^2}} da db= \int \sqrt{(a-y)^2 + b^2}da db$$

and rewriting the left hand side using the numerator manipulation

$$(a-y)(x-y) = (a-y)(a-y) + b^2 + (x-a)(a-y) - b^2$$

completely cancels the right hand side, leaving the following condition for y:

$$ \int {(x-a)(a-y) - b^2\over \sqrt{(a-y)^2 + b^2}} da db = 0$$

Now it is easy to see that for positive x, the only solution is at negative y, from the positivity of $b^2$. This is physically obvious, from the direction in which the friction force from pivoting pulls and torques.

The pivot point for x positive is at negative y. There is a limit which is important, namely the limit where the pivot runs away to negative infinite y. This is when x is very close to zero. In this limit, the friction force is nearly vertical everywhere, with the same magnitude up to order angle-squared corrections, and the force balance determines it

$$ f = F/A $$

where A is the area of the rectangle. Then the torque balance gives

$$(x-y) = -{1\over A} \int y + {b^2\over 2y} da db $$

or

$$ y = {1\over 2x } \int b^2 da db$$

This shows that y and x are related by an inversion, with an inversion radius given by the moment of $b^2$, the mean vertical square displacement. In the limit that the vertical width goes to zero, the inversion radius goes to zero, and the pivot point jumps instantaneously through infinity, reproducing the infinitely thin limit of the previous section.

This example is interesting, because you can experiment yourself by pushing writing paper on a desk, to verify the solution for the pivot point. This provides a nontrivial test of the standard constant-friction model.

Frying pan

There is another limit you can easily solve, which is a circle of uniform density resting on a plane, with a handle going far out--- a pancake griddle--- pushed perpendicularly on the handle. In this case, the rotation point moves very close to the center of the pan in the limit that the handle is long, and the force required to start the motion asymptotes to zero as the handle gets longer.

The torque balance, putting the pivot at the center of the pan, and assuming a massless handle, is

$$f 2\pi\int r^2 dr = F(L+R)$$

or

$$ {2\pi\over 3} R^3 f = F(L+R)$$

The force balance is determined by the degree to which the pivot is off-center, $\epsilon$. The off-center pivot means that the friction force on the left doesn't quite cancel the friction force on the right, leaving a residual

$$ \epsilon R \int d\theta f \sin^2\theta = F$$

Which is given by the extra sliver on the left, times the downward force component, plus the equal missing sliver on the right. These two slivers must balance the external force on the handle, F, so doing the integral

$$\pi \epsilon R f = F $$

solving these two equations gives $\epsilon$ and $F$,

$$ \epsilon = {2R\over 3}{1\over 1+{L\over R}}$$

$$ F = {2\pi \over 3} {R^2\over 1+ {L\over R}} f_0 $$

and noting that $\pi R^2 f_0$ is the total friction force on the pan, it is equal to $\mu Mg$, so that the force required to push the pan is

$$ F_c = { 2\mu Mg \over 3 (1 + {L\over R}) }$$

which goes to zero linearly as the handle length L becomes large.

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Thanks for this, I'm working through your reply.. who would have thought that friction makes thisng so complicated? –  Adrian Nov 7 '11 at 10:53
    
It's not as complicated as it could be--- these are all pretty much exact solutions. Generally any problem you choose which is not in a textbook is at least this complicated, usually much more so. –  Ron Maimon Nov 7 '11 at 17:25
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If the body is rotating about a point, then the sum of the forces must pass through the axis of percussion of that point. The relationship between the force axis (line) and center of rotation (point) is called pole-and-polar, and it is directly related to the geometric pole and polar of circles.

Look at section 8.10 of this course notes.

Look also a section 5.4 of this thesis.

To work out the center of rotation of a 2D rigid body measure the velocity components $(v_x,v_y)$ at any point $A$ and compute $$ r = (v_y/\omega,\;\text{-}v_x/\omega) $$ where $\omega$ is the angular velocity in ${\rm rad/s}$. This $r$ is the location of the center relative to point $A$.

You can not do the same for accelerations because they contain an additional centripetal component.

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It's a little similar, but this case specifically has a force which is distributed over a spatial region, which makes it slightly more difficult than the more elementary discussions of point-contact forces in the links. Thank you for these links, by the way--- it is good to have a good freely available elementary mechanics book. –  Ron Maimon Nov 8 '11 at 5:45
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