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F = ma will be how much impact (i.e. force) the mass will produce.
It is the increased kenetic energy a heavier mass will carry as compared to a lighter one given the same velocity that would have to be dissipated by the braking system, so:
K = 1/2 * (MV^2)
So given same velocity (V), the higher mass (M) will produce higher kenetic energy to be dissipated by the braking system.
Man all this talk brings me back to college physics, which I'm kinda rusty on now. Correct me if I'm wrong in any of this. And if you wanna get really technical, you could find out what the kinetic friction is for your tires on dry asphalt (0.5 to 0.8 generally) and brake pads on rotors and factor that in. Oh, and don't forget wind resistance. Man, you could get really technical.
I'll try my best to explain things here. I'm going to assume that whoever reads this post has little to no knowledge of classical mechanics. For those of you who took a course or two in physics, no offense. Just hang tight!
Aviography is pretty much correct in everything he says, but in this scenario we really don't have to factor in energy, whether it be kinetic or potential (in regards to classical mechanics). For those of you who don't know, energy is a measurement of work equal to force times distance (W=F*x), measured in Joules (J). We could measure the amount of work done by the decelerating vehicle, but it is really not necessary.
What we're really concerned with here is finding out the distance the vehicle will travel with the added weight compared to the vehicle without the added weight. In my scenario we will find out the distance, and as an added bonus, we will find out the time it takes to decelerate at a constant rate from a given velocity to rest with and without the added weight.
If you know the formula for force, acceleration, and velocity, and you can manipulate the formulas to find out unknowns, you're pretty much in luck. Anyway, let's begin.
First off, let's start off by using the SI system (which is mostly in metric units). Let's assume you have a 1200 kg vehicle and plan on adding 100 kg in the trunk later to test your theory.
As you know, velocity (V) is equal to distance divided by time: V=x/t
In the SI system, distance (x) is measured in meters, and time is measured in seconds (s)
So let's assume you accelerate at a constant rate from 0 km/h to 100 km/h (62 mph) in 8 seconds. 100 kilometers per hour in meters per second is approximately 28.77 m/s. Therefore, after 8 seconds, your final velocity is 28.77 m/s.
Acceleration (a) is equal to velocity divided by time: a=V/t
Acceleration can also be calculated by change in velocity divided by time: (final velocity - initial velocity)/t
The SI unit for acceleration is meters per second per second, or commonly phrased as meters per second squared (m/s^2). Our initial velocity was 0 km/h, or 0 m/s, and our final velocity is 28.77 m/s, so if we plug our numbers into the acceleration formula, we get roughly 3.6 m/s^2 (28.77 m/s - 0 m/s)/8 s.
Force (F) is equal to mass multiplied by acceleration: F=m*a
The SI unit for mass is measured in kilograms, and the SI unit for force is measured in Newtons (N), named after Sir Issac Newton.
If you plug in 1200 kg for your mass and 3.6 m/s^2 for your acceleration, it would take 4320 N of force to accelerate your 1200 kg vehicle from 0 km/h to 100 km/h in 8 seconds.
Following me so far? Simple physics, and we're not factoring in friction or wind resistance.
Now here's a question: how much force would it take to decelerate your 1200 kg car from 100 km/h to 0 km/h in exactly 8 seconds?
The answer is 4320 Newtons.
Deceleration is merely negative acceleration. Technically, the answer could also be -4320 N depending on which directions you consider positive and negative. Basically you have to apply a certain force to accelerate a mass, and you would have to apply an equal amount of force in the opposite direction to bring that mass back to rest (4320 N - 4320 N = 0 N).
To put this in terms of vehicle acceleration and deceleration, consider stepping on the gas pedal positive acceleration and hitting the brakes deceleration, or negative acceleration.
Now let's add 100 kg to your 1200 kg car to bring a total mass of 1300 kg.
Before you added the extra weight to your car, you accelerated at a rate of 3.6 m/s^2, and you decelerated at the rate of 3.6 m/s^2. Interpreting this, you hit the brakes with a certain amount of force that caused you to decelerate at 3.6 m/s^2. In this next scenario, you're going to decelerate at the same rate of 3.6 m/s^2.
How much force would be generated if a 1300 kg mass were to be decelerated at a rate of 3.6 m/s^2? Assume acceleration to be positive in this scenario.
Applying the equation F=m*a, we get 4680 Newtons of force.
Interpreting this, before adding the extra weight, it only required 4320 N to stop your car, but after adding an extra 100 kg to your vehicle, it required 4680 N of force to stop the car.
Now think about what you would have to do to stop your car with more force. Figured it out yet? You'd have to press the brake pedal harder. If you haven't already figured it out yet, if your acceleration remains constant, your vehicle is traveling the same distance regardless the mass of your vehicle. It just requires more force to reach that distance in a certain amount of time when the vehicle is heavier.
If you're wondering "What distance? We didn't define any distance!" then let's backtrack.
We're assuming you accelerated at a constant rate, but don't worry about that. You traveled from 0 km/h to 100 km/h. 100 km/h is 28.77 m/s, right? You traveled for 8 seconds, right? So 28.77 meters times 8 seconds is 230.16 meters (or approximately 755.12 feet).
Now your goal was to decrease your stopping distance by adding extra weight. Hey, it was a good theory, but let me explain some more. You were probably assuming you could use the same amount of brake force (or less) to stop your moving vehicle with the added weight, right?
Remember 4320 Newtons? That's the amount of force required to stop your 1200 kg moving vehicle from 100 km/h to 0 km/h in 8 seconds. Now let's say you apply 4320 Newtons of force to stop your vehicle with an extra 100 kg of added weight, or 1300 kg vehicle. What would your deceleration rate be?
Rearranging F=m*a to solve for a, you get a=F/m, or force divided by mass.
4320 N/1300 kg equals approximately 3.32 m/s^2.
So, now that we have figured out your deceleration rate, let's figure out how long it would take to stop your vehicle with 4320 N of force.
As you know, acceleration is equal to velocity divided by time (a=V/t). We know the acceleration (3.32 m/s^2) and velocity (28.77 m/s), but the unknown is time. Still following?
If you rearrange the formula for acceleration to find time, you get t=V/a. let's plug and chug.
28.77 m/s / 3.32 m/s^2 = 8.66 seconds approximately.
It's not a huge difference, but as you can see, it would take longer to stop a heavier mass with the same amount of force.
How much farther would you travel in 8.66 seconds?
Remember velocity is equal to distance divided by time (V=x/t), and if you rearrange the equation to solve for distance, you get x=V*t.
28.77 m/s * 8.66 s = 249.15 meters approximately (or 817.42 feet).
Without the added weight, you traveled a distance of 230.16 meters (or approximately 755.12 feet). The difference in distance is 18.99 meters, or 62.3 feet.
Average length of a vehicle (according to a study done in Portland, Oregon measuring 390 vehicles) is roughly 13.52 feet. 62.3 feet is roughly 4.61 car lengths. If you used the same amount of force applied to the brakes on your vehicle with the 100 kg added weight as you did without the added weight, you would have traveled 4.61 extra car lengths to stop, assuming you traveled from 100 km/h to 0 km/h with an average deceleration rate of 3.32 m/s^2.
I hope this little physics lesson and scenario helps you in your decision on whether or not to add extra weight to your vehicle to help braking time and distance. There are many other factors that would also affect your braking distance and time, including the coefficient of kinetic friction between your tires and the road (The coefficient of friction will vary due to material of tires and the road and whether the road is wet, dry, or has sand, gravel, or snow on it.), the coefficient of static and kinetic friction between your brake pads and rotors, the force of air resistance, and gravitational resistance due to an incline or decline (slope) of the road you are traveling on. Just remember that this scenario did not factor friction or external forces, and acceleration (and deceleration) were assumed to be a constant rate.
As far as adding a few sandbags or bricks in your trunk, you will add more force against your rear axle (due to the force of gravity), and indirectly, your rear wheels. Since your car is front wheel drive, the back axle is merely pulled whereas the front axle is pushed by the engine. More force will be required from the engine to pull the extra weight, meaning that your vehicle will accelerate at a lower rate than normal in a certain gear at a certain RPM. If you want to accelerate at the same rate as you normally do, you could compensate by increasing your RPMs and applying more force, or press down harder on your gas pedal, but you could potentially lose traction on your front wheels and spin out until you gain traction again. Think of this as towing and applying too much gas causes you to spin out.
Next, suppose you take a hard left turn (without braking) at a given velocity without the added weight in the trunk. In this scenario you do not lose traction in your rear wheels and spin out. The inertia of the vehicle wants to move forward, but your front axle is trying to force the vehicle to move left. The vehicle eventually curves to the left safely due to rotational inertia.
Now suppose you take the same turn at the same velocity with the added weight. You could potentially sling your rear end around and fishtail. Why? As you try to curve your vehicle to the left, more force (I say force because in this scenario you are fighting wind resistance and friction from the tires and the surface.) and inertia is trying to keep the vehicle moving straight. Again, as you are turning, rotational inertia comes into play. Imagine a tennis ball tied to a string, and you are swinging the ball in circles from the string with your hand. Where the string is being held in your hand is the axis of the rotation. Your hand is a fixed axis. If you release the string, the ball will travel straight instead of curving. Now imagine swinging a bowling ball tied to a string. You swing the ball around and let the string go again. Notice how the bowling ball requires more force to rotate it? Now let's apply rotational inertia to your car with the extra weight added.
Suppose your front axle is axis of rotation, the car itself is the string, and the extra weight in the trunk is the ball. More force is going to be required to rotate the vehicle around the left turn with the added weight instead of making the turn without the weight. Why? Inertia wants to keep the car moving forward.
A certain amount of frictional force is applied between your tires and the road as you make a turn. Net force is the sum of all forces, and frictional force is considered negative since it is fighting your positive force. In this scenario, F(net) = F(applied) - F(frictional). While your vehicle with the added weight is making the turn, if your net force is greater than 0 Newtons, chances are you will lose traction and spin out in the direction your vehicle was traveling when frictional force was overpowered. This can be related to you letting go of the string attached to the ball. Remember on snowy or rainy roads, less frictional force is being applied to your vehicle. Therefore, adding extra weight to the rear of your vehicle would be a valid cause to lose traction during a curve or turn at a given velocity on less frictional roads. Adding snow tires would help increase friction between the tires and the road, though.
Anyway, like I said, I hope this helps! You guys chime in if you want!