Minimum spring force required to keep a car level

[at570]

Homework Statement

This is an attempt to solve a problem I asked about here https://www.physicsforums.com/threads/can-a-car-ever-sit-like-this.929453/

Homework Equations

F = kx spring force

The Attempt at a Solution

using the largest angle down it can get with the front spring compressed all the way.
distance from the center of gravity to the front wheels / distance between the wheels = weight on the front wheels
convert from kg to n and divide by 2 wheels = minimum spring force to keep this from happening
but if the spring force is strong enough to push it up at the farthest angle down, will it be stronger than the weight the entire way back? As the spring gets longer the force will decrease and as the body rotates back the weight on the front wheels will also decrease.

 

[haruspex]

Homework Statement

This is an attempt to solve a problem I asked about here https://www.physicsforums.com/threads/can-a-car-ever-sit-like-this.929453/

Homework Equations

F = kx spring force

The Attempt at a Solution

using the largest angle down it can get with the front spring compressed all the way.
distance from the center of gravity to the front wheels / distance between the wheels = weight on the front wheels
convert from kg to n and divide by 2 wheels = minimum spring force to keep this from happening
but if the spring force is strong enough to push it up at the farthest angle down, will it be stronger than the weight the entire way back? As the spring gets longer the force will decrease and as the body rotates back the weight on the front wheels will also decrease.

Start from the other end. Consider the car body level, then subjected to a small perturbation (rotation).
You will likely need variables for car mass, spring constants, distance between springs and height of mass above rotation centre.

 

[at570]

Im using
weight of sprung mass 160 kg
spring constant I am using 6000 n/m = 6 n/mm all 4 springs are the same
distance between springs 1m
height of mass above rotation center - I am a little fuzzy on this one. I think for a car with straight vertical suspension its on the ground under the center of gravity when its upright. Height = 1m

Im trying someone pushes down on the front so the center of gravity moves to 51% on the front wheels and let's go.
weight on front wheels = 160 kg * .51 = 81.6 kg
81.6 kg *10 = 816 n
816 n / 2 wheels = 408 n spring force
408 / 6000 = .068 m spring compression
Im not sure where to go from here

 

[haruspex]

height of mass above rotation center

A small perturbation from the equilibrium position would depress one set of springs by the same extent that the other set would expand. So the centre of rotation is halfway between the top ends of the springs.
If the perturbation is through an angle Δθ, what is the restoring torque from the springs? The centre of mass will now not be above the centre of rotation. What torque (tending to increase the perturbation) comes from gravity?

 

[at570]

Im getting 0 restoring force
Im doing
51% on front wheels
weight on front wheels = 160 kg * .51 = 81.6 kg
81.6 kg *10 = 816 n
816 n * .5 m = 408 nm
weight on back wheels = 160 kg * .49 = 78.4 kg
78.4 kg * 10 = 784 n
784 n * .5 m = 392 nm
408 nm - 392 nm = 16 nm
weight of car = 160 kg * 10 = 1600 n
1600 n * .01 m offset from roll center = 16 nm
16 nm - 16 nm = 0

52% on front wheels
weight on front wheels = 160 kg * .52 = 83.2 kg
83.2 kg *10 = 832 n
832 n * .5 m = 416 nm
weight on back wheels = 160 kg * .48 = 76.8 kg
76.8 kg * 10 = 768 n
768 n * .5 m = 384 nm
416 nm - 384 nm = 32 nm
weight of car = 160 kg * 10 = 1600 n
1600 n * .02 m offset from roll center = 32 nm
32 nm - 32 nm = 0

 

[haruspex]

Im getting 0 restoring force
Im doing
51% on front wheels
weight on front wheels = 160 kg * .51 = 81.6 kg
81.6 kg *10 = 816 n
816 n * .5 m = 408 nm
weight on back wheels = 160 kg * .49 = 78.4 kg
78.4 kg * 10 = 784 n
784 n * .5 m = 392 nm
408 nm - 392 nm = 16 nm
weight of car = 160 kg * 10 = 1600 n
1600 n * .01 m offset from roll center = 16 nm
16 nm - 16 nm = 0

52% on front wheels
weight on front wheels = 160 kg * .52 = 83.2 kg
83.2 kg *10 = 832 n
832 n * .5 m = 416 nm
weight on back wheels = 160 kg * .48 = 76.8 kg
76.8 kg * 10 = 768 n
768 n * .5 m = 384 nm
416 nm - 384 nm = 32 nm
weight of car = 160 kg * 10 = 1600 n
1600 n * .02 m offset from roll center = 32 nm
32 nm - 32 nm = 0

The trouble with plugging in numbers straight away is that you cannot see what is going on. Do it all algebraically.

 

[at570]

l = length between front and back spring
l_g = moment arm of gravity
w = weight of car
d = distance center of gravity has moved closer to front wheels
x_f = compression distance of front spring
x_b = compression distance of back spring
l_sf = moment arm of front spring
l_sb = moment arm of back spring
k = spring constant
m = mass of car
t_g = torque from gravity
t_sf = torque from front spring
t_sb = torque from back spring
t_r = restoring torque
t = overall torque

f_g = mg
t_g = f_g x l
t_g = mg x l

w = mg
w_sf = wl / 2 + d
w_sb = wl / 2 - d

f_sf = w_sf
f_sb = w_sb

f_sf = wl / 2 + d
f_sb = wl / 2 - d

t_sf = (l / 2) x f_sf
t_sb = (l / 2) x f_sb

t_r = t_sf - t_sb
t = t_r - t_gI was wondering since the suspension is attached solidly to the body the suspension would rotate too when the body rotates and the front wheel would move closer to the center of gravity so the center of gravity wouldn't have to move as far to get the same weight on the front wheels. I am not sure if its needed though. I would do it like this

f_sfo = spring force on the front springs on one spring
f_sbo = spring force on the back springs on one spring
x_l = compression of springs at level
x_sfd = compression difference from level to angled for a spring
a = angle from a line from the roll center to the top of the front spring to a line from the roll center to the distance down the spring its compressed from rolling
h = height of roll center off the ground
d_w = distance the front wheels have moved back
d_a = actual distance the center of gravity moved relative to the roll center

spring force
f = kx
x = f / k

f_sfo = f_sf / 2
f_sbo = f_sb / 2

x_l = .067 m

x_sf = f_sf / k
x_sb = f_sb / k

x_sfd = x_sf- x_l

tan(a) = x_sfd / ( l / 2)
a = tan^-1 (x_sfd / ( l / 2) )

d_w = tan(a) / h

d_a = d - d_w

 

[at570]

I tried a new approach and I think it works better. If the body rotates angle θ, first get the original compression of the springs at level by doing m is the sprung mass and do f_g = mg and divide by 2 to get the force on the front springs and / 2 to get the force of one front spring. Then use the spring equation f = kx to get x_o the original compression at level. Then if it rotates by θ make a triangle from the roll center to the front spring. The angle at the roll center is θ and the opposite is x_θ the amount the front spring compresses from the rotation. Get x_f the compression of the front spring by doing x_f = x_o + x_θ and get the force for the front spring by doing f = kx_f. Get the torque from the 2 front springs by doing τ_f = ( f_f * 2 ) × l₁ and l₁ is the distance from the roll center to the spring. Then make another triangle for the back spring. Do x_b = x_o - x_θ since the back spring rotates by the same amount but up. Use f_b = kx_b for the force of one back spring. Do τ_b =
( f_b * 2 ) × l₂ and l₂ is the distance from the roll center to the back spring.
Then make another triangle from the roll center to the center of gravity. The angle at the roll center is θ and the hypotnuse is h, the height above the roll center of the center of gravity when the cars level. Get the opposite side and that's l₃, the moment arm for the force of gravity. Do f_g = mg and then τ_g = f_g × l₃ to get torque from gravity.
Then do τ_s = τ_f - τ_b to get the torque from the springs. Then τ = τ_s - τ_g to get the overall torque.
I tried it and if everythings good there, gravity is stronger than the springs under the right conditions

 

[at570]

Can someone confirm this is the way to do it?

 

[haruspex]

I tried a new approach and I think it works better. If the body rotates angle θ, first get the original compression of the springs at level by doing m is the sprung mass and do f_g = mg and divide by 2 to get the force on the front springs and / 2 to get the force of one front spring. Then use the spring equation f = kx to get x_o the original compression at level. Then if it rotates by θ make a triangle from the roll center to the front spring. The angle at the roll center is θ and the opposite is x_θ the amount the front spring compresses from the rotation. Get x_f the compression of the front spring by doing x_f = x_o + x_θ and get the force for the front spring by doing f = kx_f. Get the torque from the 2 front springs by doing τ_f = ( f_f * 2 ) × l₁ and l₁ is the distance from the roll center to the spring. Then make another triangle for the back spring. Do x_b = x_o - x_θ since the back spring rotates by the same amount but up. Use f_b = kx_b for the force of one back spring. Do τ_b =
( f_b * 2 ) × l₂ and l₂ is the distance from the roll center to the back spring.
Then make another triangle from the roll center to the center of gravity. The angle at the roll center is θ and the hypotnuse is h, the height above the roll center of the center of gravity when the cars level. Get the opposite side and that's l₃, the moment arm for the force of gravity. Do f_g = mg and then τ_g = f_g × l₃ to get torque from gravity.
Then do τ_s = τ_f - τ_b to get the torque from the springs. Then τ = τ_s - τ_g to get the overall torque.
I tried it and if everythings good there, gravity is stronger than the springs under the right conditions

That's the process, yes.
You did not describe the conditions under which the gravitational torque beats the restoring torque, but I would guess it's when the height to the mass centre is large in proportion to the distance between the springs.

 

[at570]

Yep. I want to make it into something where you can find out if your cars going to do this based on the set up. But I want to work on an extension of it first

 

[at570]

An extension of the question: The springs are solidly attached to the body so when the body rotates the springs also rotate. When the springs rotate angle θ theyre not pointing straight up anymore so the force is less than what's needed to oppose gravity. What happens to the springs and body when this happens?
The springs forces upward have to equal gravity. Make a triangle at the connection of a spring and the body. The angle at the connection is θ. The hypotnuse is the spring force and adjacent is the part of the spring force that opposes gravity. I think what happens is either the body rotates again or it goes straight down until all the spring forces equal gravity. I am not really sure where to go from here

 

[haruspex]

The springs are solidly attached to the body so when the body rotates the springs also rotate.

But how are they attached to the axles? Are they rigidly attached there or pivoted?

 

[Bystander]

Google "what is a toggle mechanism."

See also "trip balance/roberval principle/steelyard (balance)."

 

[at570]

These particular ones are rigidly attached which gave me some trouble figuring out where the roll center is since they don't really have instant centers, or its at infinity.
Also I see about lever arms and stuff and how it has to do with this. Everything has to be in balance both mass and distance for it not to rotate

 

[haruspex]

These particular ones are rigidly attached

To both the chassis and the axles? So the springs bend as well as expand and contract. Not sure how that affects the effective spring constant. But within the normal range of movement, I would think you can ignore this.

 

[at570]

I think its like a solid connection with a spring, so the solid part keeps it pretty much straight and also let's the spring move and compress

 

[haruspex]

I think its like a solid connection with a spring, so the solid part keeps it pretty much straight and also let's the spring move and compress

If the spring stays straight then the ends must be pivoted.

 

[at570]

Why?

 

[haruspex]

Why?

Because the chassis rotates while the axles stay put. How is that possible if the springs are attached to fixed points on each, do not bend or pivot, and the chassis structure does not stretch?

 

[at570]

The wheels can rotate wouldn't that take the strain off of the axles?

 

[haruspex]

The wheels can rotate

That has nothing to do with the geometric relationship between the axles, the springs and the chassis.

 

[at570]

Im not really understanding. The way the suspension looks there's a tube that sort of slides that keeps it straight, with the spring and there's no other connections to the wheel and theyre independent

 

[at570]

Oh also the way I drew it in the other post I guess the springs actually are at an angle when the body is at an angle

 

[haruspex]

Im not really understanding. The way the suspension looks there's a tube that sort of slides that keeps it straight, with the spring and there's no other connections to the wheel and theyre independent

We can abstract it as two fixed points, A and B, at the lower ends of thesprings, with a rigid block attached to the top ends of the springs at C and D.
If we rotate the block, compressing one spring more than the other, the horizontal distance from C to D reduces but the horizontal distance from A to B remains constant. It follows that the springs AC and BD cannot both remain vertical. If a spring does not remain vertical then it is either pivoted at the lower end or bends.

 

[at570]

Oh. So if the brakes are on the friction of the tires would try to keep it from rotating and it couldn't rotate like that because it would get out of shape. But if the brakes are not on wouldn't the wheels at A and B be able to roll to keep it from getting out of shape?

 

[haruspex]

Oh. So if the brakes are on the friction of the tires would try to keep it from rotating and it couldn't rotate like that because it would get out of shape. But if the brakes are not on wouldn't the wheels at A and B be able to roll to keep it from getting out of shape?

I don't see what the rolling of the wheels has to do with it. They do not alter the geometrical relationship between the four points I defined.

 

[at570]

Also the car looks more like this

But if the effect on the springs is small mabey we can ignore it?

 

[at570]

Actually, wouldn't the wheels be like pivots at A and B? So we can treat A and B like pivots I think

 

[haruspex]

Actually, wouldn't the wheels be like pivots at A and B? So we can treat A and B like pivots I think

So the springs are pivoted. Whether the brakes are on is irrelevant.
Don't know why I thought the distance between the wheels was fixed - that was dumb.