How is normal force distributed along the surface of contact?

Question

Help me settle this argument.

A mass $m$ is placed on a thin diving board. The base of the diving board has mass $M>>m$. Does the board tip over?

I drew the following FBD and concluded there is no net torque.

My friend thinks the normal force $N$ will be applied at a different $x$-coordinate, perhaps $x=D_1/2$. This would result in a net torque. How can I prove him wrong using the laws of classical mechanics?

Of course, the normal force is not applied at a single point. It is distributed across the entire surface of contact. Is it possible to compute the force distribution $N(x)$? We could measure this experimentally by placing many small scales under the base.

For a complete answer (a step or two beyond where you are in the progression) you need to account for the distortion of all the relevant parts. If you are on an engineering track you'll probably take statics and they will address the problem explicitly. If you are on a physics track they probably won't ever show you the full version; in that case borrow a statics book from the nearest friendly engineer. — dmckee --- ex-moderator kitten, Oct 04 '15 at 20:39

score 3 · Accepted Answer · answered Oct 04 '15 at 20:28

I think that you're making this problem more complicated than it has to be in order to simply determine if the assembly will tip over or not. You don't really need the spatial distribution of the forces being exerted by the table or ground on the assembly. All you need to note is that if the pivot point is at x=D1 then the ground will exert whatever counter-torque is needed in order to prevent the assembly from rotating counter-clockwise about the pivot point.

So all you need to do is to calculate the torques contributed by mass m and mass M about the pivot point. If the sum of those two torques acts in the counter-clockwise direction, then the ground will exert a counter-torque of the same magnitude but in the opposite direction to prevent the whole assembly from turning counter-clockwise. On the other hand, if the sum of the two torques from m and M acts in the clockwise direction, the ground plays no role in providing a counter-torque and the whole assembly tips over in the clockwise direction around the pivot point.

Calculating the total torque due to m and M should not be difficult. For the purposes of calculating the torque due to M, you can assume just a single force of magnitude Mg (where g is gravitational acceleration) acting downward at its center-of-mass.

This is the simplest answer and is a good approximation when neither material will be notably distorted by the forces at work in the event that it does tip. — dmckee --- ex-moderator kitten, Oct 04 '15 at 20:40
+1 from me. This is much simpler than whatever the heck I had written up. It makes intuitive sense, too. — HDE 226868, Oct 04 '15 at 20:48

score 2 · Answer 2 · answered Oct 04 '15 at 20:26

If there is no torque, then $$\sum \tau = \mathbf{r_1}\times\mathbf{F_M}+\mathbf{r_2}\times\mathbf{F_m}=0$$ Therefore, $$\mathbf{r_1}\times M\mathbf{g}+\mathbf{r_2}\times m\mathbf{g}=0\tag{1}$$ where $\mathbf{r_i}$ denotes the position of the center of mass of the combined system relative to the force applied. If we give the box dimensions $h$ and $l$, the ball a radius $R$, and say that the ramp extends a distance $d$ from $x=0$, then $$\mathbf{x_{cm}}=\frac{\frac{1}{2}\mathbf{l}M+\left(\frac{1}{2}\mathbf{l}+\mathbf{d}-\frac{1}{2}\mathbf{R}\right)m}{M+m}$$ and $$\mathbf{y_{cm}}=\frac{\frac{1}{2}\mathbf{h}m+\left(\mathbf{h+\frac{1}{2}R}\right)M}{M+m}$$ ($x_{cm}$,$y_{cm}$) are the coordinates of the center of mass. You can then compute $\mathbf{r_1}$ and $\mathbf{r_2}$. If $(1)$ holds, then you are correct; if not, your friend is correct.

Does this account for torque exerted by the normal force N? – tba Oct 05 '15 at 17:26 — tba, Oct 05 '15 at 17:26

score 1 · Answer 3 · answered Oct 05 '15 at 04:52

1

Torque? Why do you think you need to think about torque?

Is the center of mass over the base of support? It is if $0 < M D_1/2 + m D_2 < D_1$. I.e., if $\frac{-M D_1}{2 D_2}< m < \frac{D_1}{D_2} (1-M/2)$.

answered Oct 05 '15 at 04:52

Eric Towers

1,767

(+1) This is a great concise answer, but I'm looking for a solution which can be derived directly from the laws of classical mechanics. – tba Oct 06 '15 at 22:18
This is disingenuous. The position of the centre of mass is derived by considering moments of forces. And the condition for stability - that the CM must be "over" the base of the object - is also derived by considering moments. This answer applies a general condition (CM "over" the base) without explaining or acknowledging where it comes from. – sammy gerbil Jun 22 '18 at 14:47
@sammygerbil : I reject your contention that all answers must be derived from first principles. We are not required to forget all derived results just because we are posting anwers on SE. – Eric Towers Jun 22 '18 at 17:48

How is normal force distributed along the surface of contact?

3 Answers3

Linked