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A Crane, A Tank, and the Center of Gravity

Hey, let’s just move this really heavy object over here. It should be easy, right? Well, I guess things didn’t go as well as they would have liked.

Why did this crane tip over as it turned but not at the beginning? One way to look at this problem is by looking at the center of gravity (on Earth this is essentially the same as the center of mass).

What is the center of gravity?

Suppose I have some rigid object made of 4 very tiny masses that are connected by zero mass sticks. Yes, this sounds crazy but this is what we do to derive things. We start with things that seem impossible but in the end it will work for many other cases (see spherical cow).

Here are my four tiny masses. They have different masses, just for fun.

Fall 13 Sketches.key 4

Now suppose I push up on this rigid object at the location of mass 4. In this case, there is a gravitational force on each of the 4 masses along with the force pushing up.

Fall 13 Sketches.key 5

When dealing with rigid objects, there are really two things that we are interested in – what is the net force and and what is the net torque? First, for the forces. Clearly, there are only forces in the vertical direction (I will call this the y direction.)

La te xi t 1

In case you can’t tell, I am using M to represent the sum of the 4 masses. What we get from this equation is an expression for the acceleration. Maybe the rigid object will fall or maybe it won’t – but there will be some value for the vertical acceleration.

Now, what about torque? Clearly, this rigid object could change its rotational motion. I can calculate the torque about an axis that runs through mass 4. For, this case, I can use the simplified calculation of torque as:

La te xi t 1

Torque should actually be described as a vector, but this definition will work here. In this case, the r is the distance from the point (or axis) that the torque is being calculated and θ the angle between r and the force. If I am calculating the torque about an axis through mass number 4, then both the force, F and the gravitational force on mass 4 would have zero torque. Both of these forces have a zero value for r since they both push AT this calculation point. But what about the rest of the torque? All of the forces are perpendicular to the x-axis so that sin(θ) = 1. This means that the net torque would be:

La te xi t 1

Net torque and net force are really all that we need to describe what happens to this thing.

One Force to Rule Them All

No one wants to deal with four gravitational forces. For real objects, there are more than 4 point masses. If I had just 12 grams of carbon, that would be more than 1023 atoms that have gravitational forces on them. No one has time to calculate that many forces (it’s too hard even for a computer).

What if I could use just ONE gravitational force instead of many? The key is that this force would have to give the same net force and the same net torque as the 4 individual gravitational forces. Let’s look back at the net force. Clearly a net gravitational force of F = (m1 + m2 + m3 + m4)g would give the same total gravitational force.

When dealing with torques, it’s not just the magnitude of the force that matters but also the location. Where could I put this one gravitational force so that it gives the same torque about some point (point o in this case)?

La te xi t 1

Notice that I put mass 4 back into the calculation. It dropped out before because x4 was zero, but it’s more generic to have it in there. And here is the definition of the center of gravity in the x-direction. Of course, I could make a similar calculation for the y and z-direction.

Tipping Crane

Now we can look at the tipping crane. Instead of looking at the gravitational force and torque on all the individual masses in the crane plus tank system, I can just look at the center of gravity. Here is a diagram showing the forces on the system before it tips over.

Fall 13 Sketches.key

Clearly the ground pushes up at the location of the tires and this total ground force must have the same magnitude as the total gravitational force. But what about torque? In order for the crane to NOT tip over, the net torque about any point must be zero. Let’s look at that wheel on the left as the point about which we calculate the torque. The gravitational force at the center of mass would cause a clockwise torque on the system. This would be balanced by the counterclockwise torque from the ground pushing on the right wheel. It doesn’t tip over. Oh, you don’t like that? You could do the same thing with the torque about any point.

When the crane rotates, the tank is on the SIDE of the crane. The side is not as wide as the length of the crane. Here is a front view of the crane with the tank on the side.

Fall 13 Sketches.key

In this case it doesn’t matter what point you calculate the torque about – but as an example, let’s look at the left tire. The left tire exerts no torque about this point since the torque arm is zero. Then both the gravitational force at the center of gravity and the ground pushing on the right tire create counterclockwise torques. There is no way these two torques can cancel. The crane tips over.

This is why you need to have the center of gravity in between the two support points (or directly over one of them) so that the object will not tip over.

Here are some other examples of tipping or not tipping over with the center of gravity (for your reading pleasure).

Center of Gravity or Center of Mass?

I tried to always say “center of gravity” but in real life I often say “center of mass”. Mostly, it doesn’t matter. But what is the difference? The center of mass is just the mass weight sum of all the point masses that make up an object. It’s the same formula I have above.

The center of gravity is point in an object that you could use to model the motion due to gravitational forces. In this post, these two locations are the same. But when would they be different? Here is an extreme example. Suppose I have two identical masses connected by a zero mass rod and I have this object near a massive object. For all objects, the gravitational force is not uniform. Instead, the gravitational force decreases with distance from the object. Here is a diagram of the forces on this object.

Fall 13 Sketches.key

The center of mass is in the middle of the two masses since they are identical. For gravity though, it’s different. Since the gravitational force on mass 1 is greater, the center of gravity will be closer to mass 1. Yes, this is an extreme case, but the center of mass isn’t always the same as the center of gravity.