1. Questions

physics!

The mass of an automobile is 1160kg, and the distance between its front and rear axles is 2.54m. The center of gravity of the car is between the front and rear tires, and the horizontal distance between the center of gravity and the front axle is 1.02m. Determine the normal force that the ground applies to (a) each of the two front wheels and to (b) each of the two rear wheels.

PLEASE GIVE ME SOME HINTS TO DO IT...THANKS A LOT!

  1. 👍
  2. 👎
  3. 👁
Leah

  1. The total force on the four tires is the weight of the car:
    m g = 1160 * 9.8
    Now do a see saw
    You have m g down at 1.02 meters behind the front axle
    You have 2Fb up at the rear tires,2.54 meters behind the front axle (I called it two Fb so I will not forget to divide by two to get each of the rear tires.)
    take moments (torques) about the front axle
    2Fb * 2.54 = m g * 1.02
    solve that for Fb, the force on the back axle
    then the total force up = total force down
    so
    2Fb + 2Ff = m g
    You know mg and your know 2Fb so now you know 2Ff
    divide 2Ff by two to get the force on each front wheel
    divide Fb by two to get the force on each rear wheel

    1. 👍
    2. 👎
    Damon

  2. The total force on the four tires is the weight of the car:
    m g = 1160 * 9.8
    Now do a see saw
    You have m g down at 1.02 meters behind the front axle
    You have 2Fb up at the rear tires,2.54 meters behind the front axle (I called it two Fb so I will not forget to divide by two to get each of the rear tires.)
    take moments (torques) about the front axle
    2Fb * 2.54 = m g * 1.02
    solve that for 2Fb, the force on the back axle
    then the total force up = total force down
    so
    2Fb + 2Ff = m g
    You know mg and your know 2Fb so now you know 2Ff
    divide 2Ff by two to get the force on each front wheel
    divide 2Fb by two to get the force on each rear wheel

    1. 👍
    2. 👎
    Damon

  4. Ok, so after struggling through the above explanation I decided to repost a (imho) better worded, generalized explanation:
    Firstly let me define my variables: Fb=force of BOTH back wheels on the ground=2F(1backwheel)
    Ff=force of BOTH front wheels on the ground=2F(1frontwheel)
    d1= distance of back wheels from front
    d2= distance of CM from front wheels

    We should first recognize that the force of the ground on both back wheels is equal in magnitude (but opposite indirection) to the force of the wheels on the ground or Fb=F(ground.on.back.wheels)and the force of the ground on the front wheels is equal to the force of the front wheels on the ground or Ff=F(ground.on.front.wheels).

    We know that the force of the ground on all four wheels is equal to the force of weight of the car (mg)because the car isn't falling through the ground.

    So F(ground.on.back.wheels)+F(ground.on.front.wheels)= mg = Fb+Ff

    Define down as positive:
    Now if we act as though the car could rotate around the front axle and recognize that the net torque is 0 then we know that Torque at CM + torque at back caused by ground=0=d2*mg+ d1*(-F(ground.on.back.wheels) so F(g.o.b.w)=mg(d2/d1)= magnitude of Fb
    which is twice the magnitude of the force of each back wheel.
    Use Fb+Ff=mg to find Ff which is twice the magnitude of the force of each front wheel

    1. 👍
    2. 👎
    Arno

Respond to this Question

First Name

Your Response

Similar Questions

  1. Physics

    Workers have loaded a delivery truck in such a way that its center of gravity is only slightly forward of the rear axle. .63m in front of the rear wheels and 2.3m behind the front wheels. The mass of the truck and its contents is

  2. math

    A road perpendicular to a highway leads to a farmhouse located 8 mile away.An automobile travels past the farmhouse at a speed of 80 mph. How fast is the distance between the automobile and the farmhouse increasing when the

  3. dynamics of rigid bodies

    7. A 1800 kg car drives up an inclined of 8% from rest at uniform acceleration to a velocity of 60 kph after traveling 75 m. Its wheels are 3 m apart with its center of gravity midway of the wheels and 500 mm from the ground. If

  4. physics

    The shock absorbers in the suspension system of a car are in such bad shape that they have no effect on the behavior of the springs attached to the axles. Each of the identical springs attached to the front axle supports 294 kg. A

  1. Physics PLEASE HELP

    The two barges shown here are coupled by a cable of negligible mass. The mass of the front barge is 2.22 ✕ 10^3 kg and the mass of the rear barge is 3.22 ✕ 10^3 kg. A tugboat pulls the front barge with a horizontal force of

  2. Physics

    A mechanic jacks up the front of a car to an angle of 6.2° with the horizontal in order to change the front tires. The car is 2.91 m long and has a mass of 1100 kg. Gravitational force acts at the center of mass, which is located

  3. math

    For the opening night of an opera house, a total of 1000 tickets were sold. Front orchestra seats cost $80 each, rear orchestra seats cost $60 each, and front balcony seats cost $50 each. The combined number of tickets sold for

  4. Physics

    An advertisement claims that a particular automobile can "stop on a dime". What net force would actually be necessary to stop an automobile of mass 850kg traveling initially at a speed of 38.0km/h in a distance equal to the

  1. Physics

    A penny farthing is a bicylce that was popular between 1870 and 1890. As the drawing shows, this type of bicycle has a large front wheel (R=1.20m) and a smaller rear wheel (r=.34m). A bicyclist, riding at a linear velocity of 12.5

  2. Physics

    An advertisement claims that a particular automobile can "stop on a dime". What net force would actually be necessary to stop an automobile of mass 850Kg traveling initially at a speed of 43.0 km/h in a distance equal to the

  3. physics

    An automobile has a mass of 2150 kg and a velocity of +18 m/s. It makes a rear-end collision with a stationary car whose mass is 1850 kg. The cars lock bumpers and skid off together with the wheels locked. What is the velocity of

  4. Physics

    A set of fossilized triceratops footprints discovered in Texas show that the front and rear feet were 3.2 m apart. The rear footprints were observed to be twice as deep as the front footprints. Assuming that the rear feet pressed

You can view more similar questions or ask a new question.