There is a building under construction next to the physics building. I can’t help but pause each day to observe the progress. Currently, crews are using a massive crane to hoist steel pieces in place. It’s like a giant Lego set.

But what about the crane as a physics example? In this case, we can use it to examine the forces on an object in equilibrium.

Static Equilibrium

We can define static equilibrium with the following two equations:

Don’t freak out, but yes, that is a vector above the zero. It’s the zero-vector. Since force is a vector, it can only be equal to another vector. If you didn’t want to write this as a zero vector, you could also write the force equation as:

La te xi t 1

Now we have three scalar equations. These essentially say that for an object in equilibrium, the forces in all three directions must each add up to zero Newtons. Technically, it could still be moving at a constant speed, but we call that equilibrium.

Torque also is a vector (technically). However, 3-D torque is a bit complicated so introductory physics textbooks usually stick to cases that are two-dimensional with the torque about some fixed axis (labeled “o”). In this case, torque can be calculated as (the scalar torque):

La te xi t 1

In this definition, r is the distance from the point “o” to the location that a force is applied to some rigid object. F is the magnitude of this force and θ is the angle between the force and r. If a torque would make the object rotate counter clockwise, we call that a positive torque. Clockwise torques would then be negative.

So, in two dimensions we really just have the three following equations:

La te xi t 1

It’s important to notice that for an object in equilibrium, it doesn’t matter where the point “o” is for the net torques. You can pick whatever point you like—I suggest picking a point that makes the calculations easier. But in the end we have these three equations and can use them to solve for things (usually forces) that we don’t know.

I guess there is one other thing to talk about: gravity and the center of mass. If there is a rigid object, we have to calculate both the force due to gravity as well as the torque. The gravitational force actually pulls on all parts of a rigid object—but we can simplify this force by assume that the gravitational force only pulls at one location. We call this location the center of gravity. In a constant gravitational field (like near the surface of the Earth), the center of gravity is the same as the center of mass. Here is an older post that connects the torque and the center of gravity.

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Finding the Scale and Other Assumptions

Before I look at the forces on this crane, I need to know the size (or at least the approximate size). I tried a quick Google search to find the size of this particular crane, but I gave up. Let me instead use things I know—in particular the angular field of view for my camera and the distance to the image.

First, let me start with a different picture (to get the size of the vertical steel beams).

Spring 2016 Sketches key

Now I can approximate the distance from the camera location to the location of the beam with Google Maps.

Map Distance

Using an approximate distance of 141 meters and an angular size of 0.12 radians for a beam length of 16.9 meters. With the length of a beam, I can get a scale measurement of the crane.

There are three more things I need. First, I must know the location of the center of mass for the crane arm and the mass of the crane arm. I’m going to assume the arm has uniform density (probably not true) and it’s made of steel that is 1 cm thick.

Now what about the load that is lifted? Again, I will assume it lifts a steel beam with a mass of 2,000 kg. Don’t worry, I am going to solve this problem and only put in values at the end so that you can use your own estimations if you like.

What Is the Force on the Piston?

If I want to use the equilibrium equations, I have to first pick an object that is in equilibrium. Yes, I really want the force that piston exerts so I need to look at an object that is both in equilibrium and has that force. The only choice is the crane arm itself.

What forces are acting on this arm? Here is sketch of the crane arm with all the forces on it.

Spring 2016 Sketches key

Yes, that diagram looks a bit messy but it has just about everything we need. Here are some notes:

  • I am assuming the location of the center of mass is at L/2.
  • The distance from the point where the piston pushes to the pivot point (at the bottom) is labeled “s”.
  • The pivot point can push both vertically and horizontally.
  • α is the angle between the piston and the arm and θ is the angle between the vertical forces and the arm. θ is also the angle that the arm is tilted.
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Hopefully everything else is clear. Now I can write down my two force equilibrium equations. I’m going to call the horizontal direction “x” and the vertical “y”.

La te xi t 1

You might notice that I already made a mistake. I chose the direction of force F2 to be to the right. But in this case, it would be impossible for the total forces in the x-direction to add up to zero. Don’t worry, in the end I will get a negative value for F2 and everything will be fine.

For the net torque equation, I need to first pick a point about which to calculate the torque. Any point should work, but the easiest thing is to chose the pivot point at the bottom of the crane. With this as my torque point, both F1 and F2 contribute zero torque since the distance from the force to the point is zero. This gives the following torque equation.

La te xi t 1

Here you can see why the piston is so important. There are three forces that exert a torque about point o, but only one of them is in the counter clockwise direction—that from the piston. So this force has to create a torque to balance both the torque due to the weight of the arm and due to the load.

I can solve this torque equation for the force from the piston (maybe piston isn’t the correct technical term, but now it’s too late).

La te xi t 1

If I just want the force from the piston, I only need the torque equation (and not the net forces equation). Now for some values. From the image (and using Tracker Video to measure distances and angles), I get the following values:

  • L = 48 meters.
  • s = 12.5 m.
  • θ = 0.642 rad.
  • α = 0.354 rad.

Estimating the thickness of the arm, I get a value of right around 1 meter. Assuming it’s steel with a shell thickness of 1 cm, this gives it a mass of 15,000 kg. That seems a bit much. I’m just going to reduce that by a factor of three and go with 5,000 kg.

Using a load mass of 2,000 kg, I get a piston force of 2.9 x 105 Newtons. Although I’m still not quite sure about my values, I’m absolutely certain that the forces on that piston are fairly high. What if you didn’t even have a piston? In this case, the only force to keep the crane arm in equilibrium would be at the location of the hinge. Since the distance from the torque point would be very small (less than half a meter), the required force would be ginormous.

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