# Table on inclined plane - normal force

So, I'm struggling to understand how the normal force works in case of the inclined plane. It's obvious for me what happens if the body is, for instance, a block or a cylinder, however, I can't got any answer how it behaves in case of some more "sophisticated" case. $$\\$$ What if there is a table of mass $$M$$ moving down the plane (the mass in evenly spaced on the surface, and the table is assumed to be square)? Well, I know that if a table like this one in question stands on a flat surface, then the normal force on each of the legs is $$\frac{Mg}{4}$$, and the center of mass is in the middle of the surface, somewhere above the ground (the table legs have mass too). However, I doubt whether this is still the case when we put it on the inclined plane. Intuitively, I would say that the normal force in such situation is bigger due to the first pair of legs - it's more likely for them to break. Is it true? And if so, how can I prove it? Finally, what are the normal forces due to each leg? I also attached a rather badly made picture of the situation in question. Any help appreciated! Thanks in advance! • imagine a line through the table's COM to the Earth's COM, assuming constant density of table materials, this line will be closer to the lower legs, so yes, they will have more weight on them Aug 7 '19 at 21:08
• To handle this problem with the table at rest you need to know about rotational equilibrium as well as translational equilibrium. Have you gotten to the chapter on torques yet? Mind you working out what is going on if you put a table on a frictionless incline is a fairly interesting exercise as well. Aug 7 '19 at 23:29

EDIT: As pointed out by dmckee this is only true for sliding without friction. If we have friction (e.g. table at rest) then the frictional force $$F_r$$ exerts a torque $$M=l\cdot F_r$$, where $$l$$ is the length of a leg, relative to the center of mass. The torque pushes the front legs into the plane. Since the table should not flip over this torque has to be counteracted by the plane. Let $$\frac{L}{2}$$ be the distance from the center of mass to the front leg. Then, the rotational equilibrium gives $$\frac{L}{2}F=lF_r.$$ Rearranged, we have $$F=\frac{2l}{L}F_r$$ for the additional normal force on the front legs.

The normal force is perpendicular to the surface per definition. The downhill force is perpendicular to the normal force and hence does not influence the normal force. In this case it is given by $$F_\mathrm{n}=Mg\cos\varphi$$ where $$\varphi$$ is the inclination of the plane. Since the table dose not move into the plane the normal force is countered by the plane. As the center of mass is in the middle the normal force is distributed between the legs equally if there is no friction. If the center of mass would not be in the middle a torque equation will be needed to solve.

• What is the "downhill force"?
– user196418
Aug 7 '19 at 21:33
• @ggcg It is the component of the weight parallel to the plane. Aug 7 '19 at 21:39
• If the table is at rest then "As the center of mass is in the middle the normal force is distributed between the legs equally." is simply incorrect. You have to tackle this problem with the full infrastructure of statics---pointedly including rotational equilibrium and the effects of frictional forces in your consideration. The downhill legs will experience a larger normal force. Aug 7 '19 at 23:28
• @dmckee Yes, you are right. I neglected friction. If there is friction there will be a torque exerting more force on the front legs. I’ll edit the answer. Thank you! Aug 7 '19 at 23:36

You need to balance forces and balance torques to find out what happens. Welcome to the wonderful world of statics. The free-body diagram above describes all the forces acting on the table. Each leg has its own contact force $$N_A$$ and $$N_B$$ that are different in magnitude. I have also concentrated all the friction on the front leg which simplifies the problem.

The balance of forces and torques is three equations with three unknowns, the two normal forces, and friction required to remain static.

\begin{aligned} N_A + N_B - W \cos\theta & = 0 \\ -F_A + W \sin \theta & = 0 \\ -b (N_B) + \tfrac{b}{2} (W \cos\theta) - h (W \sin \theta) & = 0 \end{aligned}

Where $$b$$ is the separation of the legs, and $$h$$ the center of mass height from the ground. The solution looks like this

\begin{aligned} F_A &= W \sin \theta \\ N_A &= \tfrac{W}{2} \left( \cos \theta + \tfrac{h}{b} \sin \theta \right) \\ N_B &= \tfrac{W}{2} \left( \cos \theta - \tfrac{h}{b} \sin \theta \right) \\ \end{aligned}