mechanics

Law Of The Lever – Explanation and Examples

Imagine a beam sitting on a fulcrum. We apply one force F'(1) = 20 N on the left side at a distance of r(1) = 0.1 m from the fulcrum and another force F'(2) = 5 N on the right side at a distance of r(2) = 0.2 m. In which direction, clockwise or anti-clockwise, will the beam move?

Great Formulas II_html_m2eb595ec

(Before reading on, please make sure that you understand the concept of torque)

To find that out we can take a look at the corresponding torques. The torque on the left side is:

T(1) = 0.1 m · 20 N = 2 Nm

For the right side we get:

T(2) = 0.2 · 5 N = 1 Nm

So the rotational push caused by force 1 (left side) exceeds that of force 2 (right side). Hence, the beam will turn anti-clockwise. If we don’t want that to happen and instead want to achieve equilibrium, we need to increase force 2 to F'(2) = 10 N. In this case the torques would be equal and the opposite rotational pushes would cancel each other. So in general, this equation needs to be satisfied to achieve a state of equilibrium:

r(1) · F'(1) = r(2) · F'(2)

This is the law of the lever in its simplest form. Let’s see how and where we can apply it.

—————————

A great example for the usefulness of the law of the lever is provided by cranes. On one side, let’s set r(1) = 30 m, it lifts objects. Since we don’t want it to fall over, we stabilize the crane using a 20,000 kg concrete block at a distance of r(2) = 2 m from the axis. What is the maximum mass we can lift with this crane?

First we need to compute the gravitational force of the concrete block.

F'(2) = 20,000 kg · 9.81 m/s² = 196,200 N

Now we can use the law of the lever to find out what maximum force we can apply on the opposite site:

r(1) · F'(1) = r(2) · F'(2)

30 m · F'(1) = 2 m · 196,200 N

30 m · F'(1) = 392,400 Nm

Divide by 30 m:

F'(1) = 13,080 N

As long as we don’t exceed this, the torque caused by the concrete block will exceed that of the lifted object and the crane will not fall over. The maximum mass we can lift is now easy to find. We use the formula for the gravitational force one more time:

13,080 N = m · 9.81 m/s²

Divide by 9.81:

m ≈ 1330 kg

To lift even heavier objects, we need to use either a heavier concrete block or put it at a larger distance from the axis.

—————————

The law of the lever shows why we can interpret a lever as a tool to amplify forces. Suppose you want use a force of F'(1) = 100 N to lift a heavy object with the gravitational pull F'(2) = 2000 N. Not possible you say? With a lever you can do this by applying the smaller force at a larger distance to the axis and the larger force at a shorter distance.

Suppose the heavy object sits at a distance r(2) = 0.1 m to the axis. At what what distance r(1) should we apply the 100 N to be able to lift it? We can use the law of the lever to find the minimum distance required.

r(1) · 100 N = 0.1 m · 2000 N

r(1) · 100 N = 200 Nm

r(1) = 2 m

So as long as we apply the force at a distance of over 2 m, we can lift the object. We effectively amplified the force by a factor of 20. Scientists believe that the principle of force amplification using levers was already used by the Egyptians to build the pyramids. Given a long enough lever, we could lift basically anything even with a moderate force.

—————————

This was an excerpt from More Great Formulas Explained.

Check out my BEST OF for more interesting physics articles.

Physics: Free Fall and Terminal Velocity

After a while of free fall, any object will reach and maintain a terminal velocity. To calculate it, we need a lot of inputs.

The necessary quantities are: the mass of the object (in kg), the gravitational acceleration (in m/s²), the density of air D (in kg/m³), the projected area of the object A (in m²) and the drag coefficient c (dimensionless). The latter two quantities need some explaining.

The projected area is the largest cross-section in the direction of fall. You can think of it as the shadow of the object on the ground when the sun’s rays hit the ground at a ninety degree angle. For example, if the falling object is a sphere, the projected area will be a circle with the same radius.

The drag coefficient is a dimensionless number that depends in a very complex way on the geometry of the object. There’s no simple way to compute it, usually it is determined in a wind tunnel. However, you can find the drag coefficients for common shapes in the picture below.

Now that we know all the inputs, let’s look at the formula for the terminal velocity v (in m/s). It will be valid for objects dropped from such a great heights that they manage to reach this limiting value, which is basically a result of the air resistance canceling out gravity.

v = sq root (2 * m * g / (c * D * A) )

Let’s do an example.

Skydivers are in free fall after leaving the plane, but soon reach the terminal velocity. We will set the mass to m = 75 kg, g = 9.81 (as usual) and D = 1.2 kg/m³. In a head-first position the skydiver has a drag coefficient of c = 0.8 and a projected area A = 0.3 m². What is the terminal velocity of the skydiver?

v = sq root (2 * 75 * 9.81 / (0.8 * 1.2 * 0.3) )

v ≈ 70 m/s ≈ 260 km/h ≈ 160 mph

Let’s take a look how changing the inputs varies the terminal velocity. Two bullet points will be sufficient here:

  • If you quadruple the mass (or the gravitational acceleration), the terminal velocity doubles. So a very heavy skydiver or a regular skydiver on a massive planet would fall much faster.
  • If you quadruple the drag coefficient (or the density or the projected area), the terminal velocity halves. This is why parachutes work. They have a higher drag coefficient and larger area, thus effectively reducing the terminal velocity.

This was an excerpt from the Kindle ebook: Great Formulas Explained – Physics. Mathematics, Economics. Check out my BEST OF for more interesting physics articles.