# Car Dynamics – Sliding and Overturning

In this post we will take a look at car performance in curves. Of central importance for our considerations is the centrifugal force. Whenever a body is moving in a curved path, this force comes into play. You probably felt it many times in your car. It is the force that tries to push you out of a curve as you go through it.

The centrifugal force C (in N) depends on three factors: the velocity v (in m/s) of the car, its mass m (in kg) and the radius r (in m) of the curve. Given these quantities, we can easily compute the centrifugal force using this formula:

C = m · v² / r

Note the quadratic dependence on speed. If you double the car’s speed, the centrifugal force quadruples. With this force acting, there must be a counter-force to cancel it for the car not to slide. This force is provided by the sideways friction of the tires. The frictional force F (in N) can be calculated from the so called coefficient of friction μ (dimensionless), the car mass m and the gravitational acceleration g (in m/s²).

F = μ · m · g

The coefficient of friction depends mainly on the road type and condition. On dry asphalt we can set μ ≈ 0.8, on wet asphalt μ ≈ 0.6, on snow μ ≈ 0.2 and on ice μ ≈ 0.1. At low speeds the frictional force exceeds the centrifugal force and the car will be able to go through the curve without any problems. However, as we increase the velocity, so does the centrifugal force and at a certain critical velocity the forces cancel each other out. Any increase in speed from this point on will result in the car sliding.

We can compute the critical speed s (in m/s) by equating the expressions for the forces:

m · s² / r = μ · m · g

s = sqrt (μ · r · g)

This is the speed at which the car begins to slide. Note that there’s no dependence on mass anymore. Since both the centrifugal as well as the frictional force grow proportionally to the car’s mass, it doesn’t play a role in determining the critical speed for sliding. All that’s left in terms of variables is the coefficient of friction (lower friction, lower critical speed) and the radius of the curve (smaller radius, more narrow curve, smaller critical speed).

However, sliding is not the only problem that can occur in curves. Under certain circumstances a car can also overturn. Again the centrifugal force is the culprit. Assuming the center of gravity (in short: CG) of the car is at a height of h (in m), the centrifugal force will produce a torque T acting to overturn the car:

T = h · C = m · v² · h / r

On the other hand, there’s the weight of the car giving rise to an opposing torque T’ that grows with the width w (in m) and mass m of the car:

T’ = 0.5 · m · g · w

At low speeds, the torque caused by the centrifugal force will be lower than the one caused by the gravitational pull. But at a certain critical speed o (in m/s), the torques will cancel each other and any further increase in speed will result in the car overturning. Equating the above expressions, we get:

m · o² · h / r = 0.5 · m · g · w

o = sqrt (0.5 · r · g · w / h)

Aside from the coefficient of friction, the determining factor here is the ratio of width to height. The larger it is, the harder it will be for the centrifugal force to overturn the car. This is why lowering a car when intending to go fast makes sense. If you lower the CG while keeping the width the same, the ratio w / h, and thus the critical speed for overturning, will increase.

Let’s look at some examples before drawing a final conclusion from these truly great formulas.

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According to caranddriver.com the center of gravity of a 2014 BMW 435i is h = 0.5 m above the ground. The width of the car is about w = 1.8 m. Calculate the critical speed for sliding and overturning in a curve of radius r = 300 m on a dry asphalt road (μ ≈ 0.8).

Nothing to do but to apply the formulas:

s = sqrt (0.8 · 300 m · 9.81 m/s²)

s ≈ 49 m/s ≈ 175 km/h ≈ 108 mph

So with normal driving behavior you certainly won’t get anywhere near sliding. But note that sudden steering in a curve can cause the radius of the your car’s path to be considerably lower than the actual curve radius.

Onto the critical overturning speed:

o = sqrt (0.5 · 300 m · 9.81 m/s² · 3.6)

o ≈ 73 m/s ≈ 262 km/h ≈ 162 mph

Not even Michael Schumacher could bring this car to overturn.

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How would the critical speeds change if we drove the 2014 BMW 435i through the same curve on an icy road? In this case the coefficient is considerably lower (μ ≈ 0.1). For the critical sliding speed we get:

s = sqrt (0.1 · 300 m · 9.81 m/s²)

s ≈ 17 m/s ≈ 62 km/h ≈ 38 mph

So even this sweet sport car is in danger of sliding relatively quickly under these conditions. What about the overturning speed? Well, it has nothing to do with the friction of the tires, so it will still be at 73 m/s.

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This was an excerpt from More Great Formulas Explained. Interested in more car dynamics? Take a look at my post on How to Compute Maximum Car Speed. For other interesting physics articles, check out my BEST OF. I hope you enjoyed and drive safe!

# 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?

(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.

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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.

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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.

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This was an excerpt from More Great Formulas Explained.

Check out my BEST OF for more interesting physics articles.

# What is Torque? – A Short and Simple Explanation

Often times when doing physics we simply say “a force is acting on a body” without specifying which point of the body it is acting on. This is basically point-mass physics. We ignore the fact that the object has a complex three-dimensional shape and assume it to be a single point having a certain mass. Sometimes this is sufficient, other times we need to go beyond that. And this is where the concept of torque comes in.

Let’s define what is meant by torque. Assume a force F (in N) is acting on a body at a distance r (in m) from the axis of rotation. This distance is called the lever arm. Take a look at the image below for an example of such a set up.

(Taken from sdsu-physics.org)

Relevant for the rotation of the body is only the force component perpendicular to the lever arm, which we will denote by F’. If given the angle Φ between the force and the lever arm (as shown in the image), we can easily compute the relevant force component by:

F’ = F · sin(Φ)

For example, if the total force is F = 50 N and it acts at an angle of Φ = 45° to the lever arm, only the the component F’ = 50 N · sin(45°) ≈ 35 N will work to rotate the body. So you can see that sometimes it makes sense to break a force down into its components. But this shouldn’t be cause for any worries, with the above formula it can be done quickly and painlessly.

With this out of the way, we can define what torque is in one simple sentence: Torque T (in Nm) is the product of the lever arm r and the force F’ acting perpendicular to it. In form of an equation the definition looks like this:

T = r · F’

In quantitative terms we can interpret torque as a measure of rotational push. If there’s a force acting at a large distance from the axis of rotation, the rotational push will be strong. However, if one and the same force is acting very close to said axis, we will see hardly any rotation. So when it comes to rotation, force is just one part of the picture. We also need to take into consideration where the force is applied.

Let’s compute a few values before going to the extremely useful law of the lever.

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We’ll have a look at the wrench from the image. Suppose the wrench is r = 0.2 m long. What’s the resulting torque when applying a force of F = 80 N at an angle of Φ = 70° relative to the lever arm?

To answer the question, we first need to find the component of the force perpendicular to the lever arm.

F’ = 80 N · sin(70°) ≈ 75.18 N

Now onto the torque:

T = 0.2 m · 75.18 N ≈ 15.04 Nm

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If this amount of torque is not sufficient to turn the nut, how could we increase that? Well, we could increase the force F and at the same time make sure that it is applied at a 90° angle to the wrench. Let’s assume that as a measure of last resort, you apply the force by standing on the wrench. Then the force perpendicular to the lever arm is just your gravitational pull:

F’ = F = m · g

Assuming a mass of m = 75 kg, we get:

F’ = 75 kg · 9.81 m/s² = 735.75 N

With this not very elegant, but certainly effective technique, we are able to increase the torque to:

T = 0.2 m · 735.75 N = 147.15 Nm

That should do the trick. If it doesn’t, there’s still one option left and that is using a longer wrench. With a longer wrench you can apply the force at a greater distance to the axis of rotation. And with r increased, the torque T is increased by the same factor.

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This was an excerpt from my Kindle ebook More Great Formulas Explained.

Check out my BEST OF for more interesting physics articles.