# Center of Mass of a Box

If I have the following area-split box:

Depending on the mass distribution of the box, the center of mass (CM) could be inside the light gray area labeled A, the red area labeled B, the blue area labeled C, the pink area labeled D, the green area labeled E, the yellow area labeled F, or the dark gray area labeled G. These areas come close to, but do not touch, the dashed force lines that bound them.

During quasistatic pushing, the contact forces acting on the box, from the pushing finger and the table, must balance the downward gravity force. If the contact mode is SrSr (the box slides to the right without tipping) as the finger moves to the right, in which areas might the center of mass be?

This is for a course I'm taking now, and I have tried every possible answer and it didn't work.

• I don't understand the question. Where is the finger pushing, and in which direction? How is the box oriented? Is it laying on the table, such that all regions are touching the table, or upright, so only the bottom of G is touching the table? Does friction not count? I'm not understanding how a pushing finger and contact forces have to equal gravity; if this is the case then the box doesn't move at all, right? If the sum of all forces is equal to m*g then there's no "leftover" force to accelerate the box in a horizontal direction. Mar 26 at 14:16
• The finger pushing is from the left "next to B region" to the right, and friction counts. The box is set on a table in the configuration shown in the image. Mar 26 at 14:29

Let's consider the rectangle to be in a vertical plane, all one single rigid part. The bottom edge is the one resting on the table. Since you specified quasistatic, we only care about the first infinitesimal motion. If the box starts sliding, we don't care about tipping. If it starts tipping, we don't care about sliding.

Nothing in the problem statement quantifies the coefficient of friction, so we'll have to carry it around symbolically. We might have to do the same with the finger height.

We will compute the forces and torques that resist tipping and sliding. If tipping occurs, the rectangle will rotate about the bottom-right corner, so we'll put the origin there.

Definitions

• point $$O$$: the bottom-right corner. This is the origin of the coordinate system. Coordinate $$y$$ increases going up, $$x$$ increases to the left.
• point $$C$$: the center of mass
• $$X$$, $$Y$$: coordinates of point $$C$$
• $$h$$: the height of the finger above table
• $$F$$: the horizontal force of the finger pushing
• $$W$$: the weight (mass * g)
• $$\mu$$: coefficient of (static) friction at the table.

Tipping

First, consider the tipping torques at point $$O$$. The finger is trying to rotate the box clockwise. The rectangle's weight is trying to rotate it counterclockwise.

The force $$F$$ pushing from the left (between area B and area D) is equivalent to a force pulling at the right (between area C and area F). Operating at height $$h$$, that applies a torque of $$F h$$.

Consider the weight $$W$$ to be concentrated at the center of mass $$C$$ (that is, at $$(X,Y)$$). It's trying to rotate the rectangle counter-clockwise about $$O$$. The downward weight vector operates about $$O$$ with moment arm $$X$$. So the torque due to weight is $$W X$$.

The box will tip (if it doesn't slide) if the torque at $$O$$ due to the finger exceeds the torque due to gravity $$F h > W X$$. Call the limiting tipping force $$F_{tip}$$:

$$F_{tip} = W X/h$$

Sliding

Now examine the friction at the table. The max friction force is $$W \mu$$, pushing horizontally to the left (countering the finger). It cannot exert any torque at $$O$$. Thus the box will slide (if it doesn't tip) when the finger force exceeds the friction. Call the limiting force $$F_{slide}$$:

$$F_{slide} = W mu$$

Comparison

Imagine the finger gradually pushing harder and harder. At first, nothing happens. Eventually, the box will start to move. It will either tip or slide, depending on which of $$F_{tip}$$ or $$F_{slide}$$ is reached first -- that is, which one is less. In comparing the two forces, both have a factor of $$W$$ which we can divide out. So comparing $$X/h$$ with $$\mu$$:

slide if $$X > h*\mu$$

We can sanity-check this by looking at limiting cases. High friction (high $$\mu$$) means $$X$$ can't be large enough so it will tip. With no friction (zero $$\mu$$) any $$X$$ will let it slide. If $$X=0$$ ($$C$$ anywhere on the right edge) it will always tip.

Conclusion

Restated, the box will slide if $$C$$ is to the left of $$h \mu$$. The boundary between the tip region and the slide region will be vertical. (It is curious, perhaps suspicious, that the results do not depend on $$Y$$, the height of point $$C$$.)

Recall that the problem statement doesn't specify anything about $$\mu$$. Neither does it say those angles that look like 45 degrees really are, so we can't relate $$h$$ to the width. Without that critical information, I don't see any way to connect analysis to geometry in order to answer the question about the regions.

• Ok, thanks a lot Bryan for your time and effort Apr 6 at 16:11