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Consider a moving car and brakes are applied on the road.It feels the car's front wheel are more pressurized than the back ones. Why?

I tried to convince myself using difference in normals on both tires due to road. Call the normal on back tyre to be $N_b$ and that on front to be $N_f$. Assuming center of mass of car lies midway between back and front tires.

Balancing torque about lowest point of front tire gives, $$N_b=\frac{Mg}{2}$$ As fricition and $N_f$ passes through that point. And same about lowest point of back tire gives,

$$N_f=\frac{Mg}{2}$$

I considered that the car stopped suddenly after application of brake. So mo acceleration of center of mass took in account. This gives me $N_f=N_b$

But am sure that car dips front. Does normal decides why will it dips front?

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  • $\begingroup$ This may help - Toppling of a cylinder on a block $\endgroup$
    – mmesser314
    Commented Jul 21, 2023 at 14:51
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    $\begingroup$ Possible duplicate of What causes the back of a bike to lift when the front brake is applied? $\endgroup$
    – Ruslan
    Commented Jul 22, 2023 at 7:35
  • $\begingroup$ The reason for front wheels to get more pressurized is that in some cars the brakes are directly affecting the front wheels to have them stop. In some other cars it's the rear wheels that are directly affected. $\endgroup$ Commented Jul 23, 2023 at 4:29
  • $\begingroup$ @SnackExchange all reasonably modern cars use all four wheels for braking, and though it's true that they apply more force on the front this is because any braking shifts the loading to the front, and therefore the rear wheels would easily lock into a skid that could cause the car to spin out. $\endgroup$ Commented Jul 24, 2023 at 0:23

3 Answers 3

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Take the centre of mass as the pivot point. Weight passes through the centre of gravity, which is practically the same position as the centre of mass, and so contribute nothing to the angular acceleration.

If things are balanced, all 4 wheels would feel equal amounts of upward force, that balances the weight.

However, the car is decelerating. That means that the friction between the ground and the wheels are directed backwards. This creates a torque that rotates the car so that the front part is pushed into the ground and the back part is lifted off the ground. Thus, the front wheels will now see a stronger upward force than the back wheels.

If you take the pivot points as the point of contact between wheels and ground, then the acceleration of the car itself will need to be considered, and that is annoying, but you will get the same answers.

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    $\begingroup$ Why would it be annoying? Can you give me a hint how could it be solved about POC. Will the pseudo force of the acceleration pass through center of mass? $\endgroup$
    – Chesx
    Commented Jul 21, 2023 at 12:57
  • $\begingroup$ Yes, it will have to pass through the centre of mass for the scheme to make sense. $\endgroup$ Commented Jul 21, 2023 at 13:06
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    $\begingroup$ Addition: The increase of vertical force at the front wheels and the decrease at the back work together to provide the necessary torque that compensates the torque induced by the horizontal forces. $\endgroup$ Commented Jul 21, 2023 at 22:25
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The front-end dipping of the car upon braking is called dive. The torque described by naturallyinconsistent as causing dive can be managed with a front end suspension designed so the front end of the car does not dip down- or actually causes the front end to rise instead of diving upon braking- but then the steering system would need serious redesign. So, almost all cars dive upon braking.

Motorcycles intended for use with sidecars have front fork suspensions which are designed to prevent dive because with a sidecar arrangement, dive causes the vehicle to steer unintentionally and crash. The most common of these dive-proof front suspension arrangements is called the Earles fork and is common on older BMW motorcycles which were often operated with sidecars.

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When the car brakes, the normal reaction force on the front wheels will be greater than the normal reaction force on the rear wheel. This is necessary to oppose the counter-clockwise moment created about the center of mass (COM) by the static friction braking forces acting backward on the wheels (without locking the brakes). See the Figures below (which neglects the effects of air resistance).

For simplicity, it is assumed the contact points with the road between the front and rear wheels are symmetric about the COM. FIG 1 shows the vehicle without braking or accelerating. Given the symmetry about the COM the normal force acting on each wheel is one force the weight of the vehicle so that the sum of the moments about the COM are zero.

FIG 2 shows the vehicle braking. The situation on the other two wheels (not shown) is identical by symmetry. Each wheel is subject to a rearward static friction force applied by the road. For equilibrium we have

$$\sum M_{COM}=N_{R}L+2fh-N_{F}L=0\tag{1}$$

$$N_{F}=N_{R}+\frac{2fh}{L}\tag{2}$$

$$\sum F_{V}=-\frac{Mg}{2}+N_{F}+N_{R}=0\tag{3}$$

$$N_{F}+N_{R}=\frac{Mg}{2}\tag{4}$$

Combining equations (2) and (4)

$$N_{F}=\frac{Mg}{4}+\frac{fh}{L}\tag{5}$$

$$N_{R}=\frac{Mg}{4}-\frac{fh}{L}\tag{6}$$

We see that braking reduces the normal force on the rear wheels (and thus the pressure on the rear wheels) and increases the normal force on the front wheels (and thus the pressure on the front wheels) from what they were before braking (or accelerating). For this example due to symmetry the static friction force $f$ on each wheel is one forth the total static friction force $f_{tot}$ acting on all wheels, or

$$f=\frac{f_{tot}}{4}=M\frac{a}{4}$$

Where $a$ is the deceleration of the car during braking (assumed constant).

Hope this helps

enter image description here

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  • $\begingroup$ Nice answer! I am not fully convinced about the fact that $f$ is a static friction force... If such forse were acting on the wheel, doesn't the wheel accelerate its rotation instead of slowing it down? $\endgroup$
    – Matteo
    Commented Jul 22, 2023 at 9:57
  • $\begingroup$ So if the wheels were locked (no rolling motion) and the car was slipping on a surface with a dynamic friction, then I am pretty sure this is the correct description. On the other hand I am not fully convinced that this is 100% accurate for a car braking under normal circumstances (with rolling wheels); but maybe I'm missing something $\endgroup$
    – Matteo
    Commented Jul 22, 2023 at 10:02
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    $\begingroup$ @Matteo Regarding your first comment, the static friction force the road applies to the wheel is the road’s Newton 3rd law equal and opposite reaction force to the force the wheel applies to the road. During acceleration, the torque applied to the wheel is clockwise and the wheel exerts a force back on the road. The static friction force thus acts forwards, enabling the vehicle to accelerate without slipping (skidding). $\endgroup$
    – Bob D
    Commented Jul 22, 2023 at 13:19
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    $\begingroup$ During breaking the torque is opposite (counter clockwise) and the wheel exerts a force forward on the road. The static friction force thus acts backwards (as shown in the figures) enabling the car to decelerate without skidding $\endgroup$
    – Bob D
    Commented Jul 22, 2023 at 13:19
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    $\begingroup$ @Matteo The wheels do not accelerate under braking, because you put the brakes on! The rotary force on the wheel from the brakes appears at the point of contact with the road, and the (equal and opposite) force on the brake calipers is transferred to the horizontal elements of the suspension, and from there to the chassis. (Slight correction -- the brakes also have to dissipate the considerable angular momentum of the wheels.) $\endgroup$ Commented Jul 23, 2023 at 8:55

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