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Easy question about weight transfer of cars

  1. Feb 22, 2009 #1
    Why does the weight of a car transfer to the back when accelerating and the front when braking?

    I'm assuming it has to do with torque, but where is the force acting and how?

    I'm probably missing something really obvious
  2. jcsd
  3. Feb 23, 2009 #2


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    The weight isn't really being transferred, except for a small amount of play in the suspension allowing movement of the center of mass with respect to the contact patches of the tires.

    The downforce on the tires is related to the weight of the car supported by each tire, and any force related to torque due to acceleration of car. The torque is related to the vertical component of distance from ground to the center of mass of the car, times the horizontal force from the ground.

    When taking off, the ground applies a forward force at the contact patch, which accelerates the car and creates an pitching up related torque, reducing the net downforce on the front tires and increasing the net downforce on the rear tires. The opposite happens under braking.
  4. Feb 23, 2009 #3
    Ok, lets look at an airplane. There are several forces action the airplane which are drag, lift, thrust, and gravity; a car only has drag, thrust, and gravity. Newton's Third Law of motion states: "To every action there is an equal and opposite reaction." When an object moves, there is going to have a force opposing it like drag on an airplane that opposes thrust. When you are braking, you suddenly transfer your weight to the front of the car, like you said.

    The reason is because your body's inertia is the same as the velocity of your car until you come to a stop. When you brake your car, the car's inertia will return to 0 as well as yours which is the force that you feel that the seat belt applies to return your inertia back to rest and to keep you from hitting the windshield while you come to a stop.

    Lets say that you are have accelerating at 19.6 m/s2 (43.845 mph2). With this acceleration, you will be experiencing 2G or 2 times the force of gravity. Lets say that you have a mass of 45.3597 kg (100 lbs) and a weight of 444.525 N. If you were accelerating at 19.6 m/s2, your weight will be 889.05 N or 90.72 kg which is your drag before you brake your car neglecting the drag and weight of the car. In Newton's Third Law of motion states: "To every action there is an equal and opposite reaction," and Newtons First Law of motion states: "An object that is not subject to any outside forces moves at a constant velocity, covering equal distances in equal times along a straight-line path," or in other words its where inertia is stated.

    So, inertia is the tendency of an object to resist a change of motion. When you brake your car, your inertia like Newton's First Law of motion states that if there was no friction an object will move at a constant velocity which your body is doing while you are in a constant velocity. When you brake your car, your seat belt is returning that inertia back to zero or your weight of 889.05 N back to 444.525 N. The force that moves your car is Fthrust and the force that stops your car is Ffriction which your tires are doing to stop your car. So if the street was frictionless then you couldn't stop your car because you will be moving at a constant velocity or acceleration like Newton's First Law states. I think this is the reason why.
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