Statics, bike on inclined plane

[jonjacson]

Homework Statement

I´ll need an image to show the problem:

A bike with length L is on an inclined plane, the distance between the center of gravity and center wheel centers A,B is L/2.

Brakes are acting on the wheels so we have a static position.

In the picture is possible to see normal forces perpendicular to the plane Na,Nb, friction forces Fa,Fb and the gravity force acting at the CG point.

I have to calculate the value of the forces and moments on the system in order to get the equilibrium.

I´ll use the coordinate system X', Y' with origin A. And angle h, the lenghts L,f are known.

Homework Equations

Sum Fx= 0

Sum Fy=0

Sum Ma=0

The Attempt at a Solution

In order to calculate the moments I´ll need the moment arms. Since the reference point is A forces Na, Fa, Fb won´t be calculated since their moment arms are 0.

The gravity has a moment arm that I showed with a strong red line.

I´m going to calculate it:

Distance AZ= L/2 ; BZ=f/ tg (h) ; So the distance AB= L/2 - f/tg(h)

In the triangle AVB I see:

cos(h)=AV/AB----> AV=moment arm=cos(h)*AB

And the moment arm of the force Nb is simply L, finally:

Sum Ma= -mg*(cos(h)*(L/2- f/tg(h))) + Nb*L = 0

Sum Fx= Fa + Fb -mg sin(h)=0

Sum Fy= Na + Nb -mg cos(h)=0

And my problem is that I have three equations and four unkown quantities: Na, Nb, Fa, Fb.

Is this one of those cases of static indeterminacy?

Have I done any mistakes?

Maybe I should use a condition like Fa=Fb or Na=Nb but I can´t see any physical reason behind those conditions so I won´t use them. Experience tells me that Na will be higher than Nb on this inclined plane.

So I´m surprised, or I´m doing something wrong or it´s not possible to calculate this problem without any extra assumption, I´ll wait for your answers to check what´s happening here.

Thanks!

 

[PhanthomJay]

Homework Statement

I´ll need an image to show the problem:

A bike with length L is on an inclined plane, the distance between the center of gravity and center wheel centers A,B is L/2.

Brakes are acting on the wheels so we have a static position.

In the picture is possible to see normal forces perpendicular to the plane Na,Nb, friction forces Fa,Fb and the gravity force acting at the CG point.

I have to calculate the value of the forces and moments on the system in order to get the equilibrium.

I´ll use the coordinate system X', Y' with origin A. And angle h, the lenghts L,f are known.

Homework Equations

Sum Fx= 0

Sum Fy=0

Sum Ma=0

The Attempt at a Solution

In order to calculate the moments I´ll need the moment arms. Since the reference point is A forces Na, Fa, Fb won´t be calculated since their moment arms are 0.

The gravity has a moment arm that I showed with a strong red line.

I´m going to calculate it:

Distance AZ= L/2 ;

looks good

BZ=f/ tg (h) ;

error here...BZ = f*(tan h)

So the distance AB= L/2 - f/tg(h)

In the triangle AVB I see:

cos(h)=AV/AB----> AV=moment arm=cos(h)*AB

And the moment arm of the force Nb is simply L, finally:

Sum Ma= -mg*(cos(h)*(L/2- f/tg(h))) + Nb*L = 0

Sum Fx= Fa + Fb -mg sin(h)=0

Sum Fy= Na + Nb -mg cos(h)=0

And my problem is that I have three equations and four unkown quantities: Na, Nb, Fa, Fb.

Is this one of those cases of static indeterminacy?

Have I done any mistakes?

Maybe I should use a condition like Fa=Fb or Na=Nb but I can´t see any physical reason behind those conditions so I won´t use them. Experience tells me that Na will be higher than Nb on this inclined plane.

So I´m surprised, or I´m doing something wrong or it´s not possible to calculate this problem without any extra assumption, I´ll wait for your answers to check what´s happening here.

Thanks!

You are correct that Na and Nb cannot be equal here, from your sum of moments equation. There is yes indeterminancy in the friction force calculations for Fa and Fb, but the usual assumption is that they are equal, on the basis that the brakes are applied on each wheel similtaneously. I suppose that if you had brakes applied on one wheel only, say A, then Fb would be zero, or if you applied the brakes on A first, the bike stops, then you apply the brakes on B, Fb would still be zero, but there is no way to determine this, so assume Fa=Fb. Your work looks very good other than the trig error.

 

[haruspex]

There is yes indeterminacy in the friction force calculations for Fa and Fb, but the usual assumption is that they are equal, on the basis that the brakes are applied on each wheel simultaneously.

I'm not so sure. If that is the case then the crossbar would be under tension. I think an argument can be made that typically the frictional forces would be in proportion to the normal forces. Certainly that will be the case when the limit of static friction is reached.

 

[jonjacson]

looks good error here...BZ = f*(tan h) You are correct that Na and Nb cannot be equal here, from your sum of moments equation. There is yes indeterminancy in the friction force calculations for Fa and Fb, but the usual assumption is that they are equal, on the basis that the brakes are applied on each wheel similtaneously. I suppose that if you had brakes applied on one wheel only, say A, then Fb would be zero, or if you applied the brakes on A first, the bike stops, then you apply the brakes on B, Fb would still be zero, but there is no way to determine this, so assume Fa=Fb. Your work looks very good other than the trig error.

I'm not so sure. If that is the case then the crossbar would be under tension. I think an argument can be made that typically the frictional forces would be in proportion to the normal forces. Certainly that will be the case when the limit of static friction is reached.

looks good error here...BZ = f*(tan h) You are correct that Na and Nb cannot be equal here, from your sum of moments equation. There is yes indeterminancy in the friction force calculations for Fa and Fb, but the usual assumption is that they are equal, on the basis that the brakes are applied on each wheel similtaneously. I suppose that if you had brakes applied on one wheel only, say A, then Fb would be zero, or if you applied the brakes on A first, the bike stops, then you apply the brakes on B, Fb would still be zero, but there is no way to determine this, so assume Fa=Fb. Your work looks very good other than the trig error.

Ups yes I made that mistake. I have recalculated everything using these four equations:

1. fa=fb

2. Sum Ma= -mg*(cos(h)*(L/2- f*tg(h))) + Nb*L = 0

3. Sum Fx= Fa + Fb -mg sin(h)=0

4. Sum Fy= Na + Nb -mg cos(h)=0

From the third equation is easy to obtaing fa and fb:

fa=fb=mg sin(h)/2

Using the fourth equation I get:

Nb= mgcos(h) - Na

And I use that on the second equation so:

-mg*cos(h)*(L/2- f*tg(h))) + (mgcos(h) - Na )*L = 0

Na= mg cos(h) (1- (L/2 -f * tng(h))/L )

Using this value now I can calculate Nb:

Nb=-mg cos(h)/L * (L/2 -f*tg(h))

Using f=1 meter ; Mass= 115 kg; h=45º ; L=1meter I get these results:

Na=1195,5 N (121 kg)

Nb= 398,5 N (40 kg)

These results look like quite surprising to me because only at 45º you have three times more weight at the rear than at the front.

I'm not so sure. If that is the case then the crossbar would be under tension. I think an argument can be made that typically the frictional forces would be in proportion to the normal forces. Certainly that will be the case when the limit of static friction is reached.

I tried to use your assumptions, so I had these four equations:

1. Nb*Fa=Fb*Na

2. Sum Fx= Fa + Fb -mg sin(h)=0

3. Sum Fy= Na + Nb -mg cos(h)=0

4. Sum Ma= -mg*(cos(h)*(L/2- f*tg(h))) + Nb*L = 0 = A+ Nb*L

I´ll try to show how I calculated the unknown quantities, using the second equation I have fa as function of fb:

fa= mg sin(h) – fb , relation 1

From the first equation:

Nb=Na * fb/fa

From the fourth equation, where I defined A to write less:

A + Na * fb * L / fa = 0 ----> Na= -A * fa/fb * L

Substituting on the third equation:

-A*fa/fb*L - A*fa*fb/fb*fa*L= mgcos (h), I cancel fa fb on the second term and I get fa as function of fb:

fa= (-1-Lmg cos(h)/A)*fb , relation 2

Using relations 1 and 2:

mg sin(h) – fb = ( -1 -Lmg cos(h)/A)*fb

fb( 1-( 1+ L mg cos(h))/A)= mg sin(h)

Now I use the value of A and this equations leads to:

fb= tg(h)/L * ( -mg cos(h) * (L/2 – f*tg(h))) , but tg(h)= sin (h) / cos (h) so:

fb=-mg sin(h) (½ – f*tg(h)/L)

To calculate fa I use relation 1 but knowing the value of fb so :

fa= mg sin(h) + (mg sin(h)/L *( L/2 -f tg(h))= mg sin(h) (3/2 – f tg(h) / L)

Using the fact that Na= -A fa/ fb L , I obtain:

Na=(mg cos(h) (L/2 – f tg(h))/L)* (-1 * (3/2 – f tg(h)/L)/(1/2 -f tg (h)/L))

Nb= mgcos(h) +Na

With the same values as before f=1m, h=45º, L=1m, I get these answers for Na and Nb:

Na= (115* 9,81*cos(45) (0,5 -1*1))/1 *(-1 *(3/2-1)/(1/2 -1))= -398,86 N

Nb= 115*9,81*cos(45)-398,86= 398,86 N

I think these results are a complete nonsense because the minus sign at Na it´s absurd, and the modulus should not be the same Na should be much bigger.

I´ll have a look at the mess of the calculations and if I see the mistake I´ll post here.

Thank you very much for your answers!

 

[PhanthomJay]

Ups yes I made that mistake. I have recalculated everything using these four equations:

1. fa=fb

This is an assumption, without more info, but it doesn't affect your results for the normal force calcs.

2. Sum Ma= -mg*(cos(h)*(L/2- f*tg(h))) + Nb*L = 0

3. Sum Fx= Fa + Fb -mg sin(h)=0

4. Sum Fy= Na + Nb -mg cos(h)=0

From the third equation is easy to obtaing fa and fb:

fa=fb=mg sin(h)/2

Using the fourth equation I get:

Nb= mgcos(h) - Na

And I use that on the second equation so:

-mg*cos(h)*(L/2- f*tg(h))) + (mgcos(h) - Na )*L = 0

Na= mg cos(h) (1- (L/2 -f * tng(h))/L )

Using this value now I can calculate Nb:

Nb=-mg cos(h)/L * (L/2 -f*tg(h))

Using f=1 meter ; Mass= 115 kg; h=45º ; L=1meter I get these results:

Na=1195,5 N (121 kg)

Nb= 398,5 N (40 kg)

These results look like quite surprising to me because only at 45º you have three times more weight at the rear than at the front.

the algebra gets a bit messy the way you have done it, your signs might be messed, but anyway, for the values you have chosen, the bike will tip over, because the cg acts to the left of A (Na points up and Nb points down, but because the normal force at B cannot act downward unless the wheel at B was anchored to the ground, the wheel B will lift and the bike will tip about A.

 

[jonjacson]

This is an assumption, without more info, but it doesn't affect your results for the normal force calcs.
the algebra gets a bit messy the way you have done it, your signs might be messed, but anyway, for the values you have chosen, the bike will tip over, because the cg acts to the left of A (Na points up and Nb points down, but because the normal force at B cannot act downward unless the wheel at B was anchored to the ground, the wheel B will lift and the bike will tip about A.

I didn´t realize that those values could be so wrong but well I measured the distance between the wheels and it was 1 meter, and I don´t know where the CG is in my bike+me but when I´m sitted on the saddle I suppose that it´s more or less at one meter above the ground, of course I´m not sure about that but, What vale would you use for the CG when I´m on the bike?

Of course it wouldn´t be exactly at the center between the wheels L/2, it would be much close to the rear wheel when I´m on the bike I´m aware about that.

About the weight I´m 95kg and the bike it´s 20kg, so this cannot be the mistake value.

edit:

As you can imagine I didn´t see this problem in any book, I was just curious about how much weight is on the rear wheel when I´m riding my bike uphill... I didn´t expect that this apparently "simple" problem could create so many headaches hahaha:rofl:

 

[haruspex]

I tried to use your assumptions,
With the same values as before f=1m, h=45º, L=1m, I get these answers for Na and Nb:

Na= (115* 9,81*cos(45) (0,5 -1*1))/1 *(-1 *(3/2-1)/(1/2 -1))= -398,86 N

Nb= 115*9,81*cos(45)-398,86= 398,86 N

I think these results are a complete nonsense because the minus sign at Na it´s absurd, and the modulus should not be the same Na should be much bigger.

I haven't tried to wade through your working, but there must be a mistake somewhere.
Na+Nb = mg sin(θ)
Moments about the point of contact of lower wheel:
Nb L = mg((L/2) cos(θ) - h sin(θ))
Nb = mg(cos(θ)/2 - (h/L) sin(θ))
whence
Na = mg(sin(θ) - cos(θ)/2 + (h/L) sin(θ))
Fa+Fb = mg cos(θ)

The above is all independent of assumptions re relative strengths of Fa, Fb. Applying Fa/Na = (Fa+Fb)/(Na+Nb) gives
Fa = Na cot(θ) = mg(1 - cot(θ)/2 + (h/L) )cos(θ)
When Nb < 0, cos(θ)/2 < (h/L) sin(θ), tan(θ) > L/2h. At this point the mass centre is beyond the point of contact of the lower wheel and the bicycle will overbalance.

 

[PhanthomJay]

The cg will be a bit lower, but you happen to have chosen a very steep incline that will overturn the bike. You might want to choose a less steep incline like 20 degrees to avoid this. Also, you can solve directly for Nb from your 2nd equation and avoid the more tedious algebra.

 

[jonjacson]

I haven't tried to wade through your working, but there must be a mistake somewhere.
Na+Nb = mg sin(θ)
Moments about the point of contact of lower wheel:
Nb L = mg((L/2) cos(θ) - h sin(θ))
Nb = mg(cos(θ)/2 - (h/L) sin(θ))
whence
Na = mg(sin(θ) - cos(θ)/2 + (h/L) sin(θ))
Fa+Fb = mg cos(θ)
The above is all independent of assumptions re relative strengths of Fa, Fb. Applying Fa/Na = (Fa+Fb)/(Na+Nb) gives
Fa = Na cot(θ) = mg(1 - cot(θ)/2 + (h/L) )cos(θ)
When Nb < 0, cos(θ)/2 < (h/L) sin(θ), tan(θ) > L/2h. At this point the mass centre is beyond the point of contact of the lower wheel and the bicycle will overbalance.

Maybe I´m confused but, Why mg sin (θ) is on the y axis? I thought it was projected to the x axis, and that´s the reason I put that on the x-axis equation with Fa and Fb.

The cg will be a bit lower, but you happen to have chosen a very steep incline that will overturn the bike. You might want to choose a less steep incline like 20 degrees to avoid this. Also, you can solve directly for Nb from your 2nd equation and avoid the more tedious algebra.

THat´s a good idea, I´m not interested in Fa, Fb so I´ll try to calculate this only with the equations for Na and Nb:

Na=mg cos(h) - Nb

Nb= mg( cos(h)/2 -f * tg(h)/L)

I get:

Na= 115*9.81*cos(20) - 119.44 = 940N

Nb= 115*9.81*(cos(20)/2 - 0.36)= 119.44 N

Definetly if the tangent at Nb is for a 45º angle the value of Nb is negative, but for 20º it´s positive, maybe it´s too high but experience tells me that I could survive to that angle on my bike without turning around like a boomerang, maybe my intuition is wrong and it wouldn´t be possible to get equilibrium at that angle.

For 20º the weight at A it´s 7,87 times bigger than it is at B, and this is a very conservative approach since the CG is at the midpoint between A and B which is false.

Maybe with more realistic data it´d be possible to show that it´s≈10 times bigger which is quite surprising to me, I didn´t expect such an enormous difference between the two wheels.

 

[haruspex]

Maybe I´m confused but, Why mg sin (θ) is on the y axis? I thought it was projected to the x axis, and that´s the reason I put that on the x-axis equation with Fa and Fb.

Whoops - sorry. Don't how that happened. Let me try that again:
Na+Nb = mg cos(θ)
Moments about the point of contact of lower wheel:
Nb L = mg((L/2) cos(θ) - h sin(θ))
Nb = mg(cos(θ)/2 - (h/L) sin(θ))
whence
Na = mg(cos(θ)/2 + (h/L) sin(θ))
Fa+Fb = mg sin(θ)

The above is all independent of assumptions re relative strengths of Fa, Fb. Applying Fa/Na = (Fa+Fb)/(Na+Nb) gives
Fa = Na tan(θ) = mg(1/2 + (h/L)tan(θ))sin(θ)
When Nb < 0, cos(θ)/2 < (h/L) sin(θ), tan(θ) > L/(2h). At this point the mass centre is beyond the point of contact of the lower wheel and the bicycle will overbalance.

 

[jonjacson]

Whoops - sorry. Don't how that happened. Let me try that again:
Na+Nb = mg cos(θ)
Moments about the point of contact of lower wheel:
Nb L = mg((L/2) cos(θ) - h sin(θ))
Nb = mg(cos(θ)/2 - (h/L) sin(θ))
whence
Na = mg(cos(θ)/2 + (h/L) sin(θ))
Fa+Fb = mg sin(θ)

The above is all independent of assumptions re relative strengths of Fa, Fb. Applying Fa/Na = (Fa+Fb)/(Na+Nb) gives
Fa = Na tan(θ) = mg(1/2 + (h/L)tan(θ))sin(θ)
When Nb < 0, cos(θ)/2 < (h/L) sin(θ), tan(θ) > L/(2h). At this point the mass centre is beyond the point of contact of the lower wheel and the bicycle will overbalance.

Hi Haruspex

Yes maybe the distances choosen are not correct for this problem and we are getting an strange result for the Nb value. I´ll try to measure more realistic data.

There is one thing that I´d like to talk about with you, you used a relation between friction and normal forces that does not show an inverse relation between Na and Nb.

If you write that Na it´s proportional to Na + Nb that won´t show the fact that on a inclined hill the bigger the angle the bigger Na will be compared to Nb, I mean Why didn´t you prefer to use this relation Na*Fb=Nb*Fa ?

I think that this relation shows that the less weight you put on B, the more is going to A, Isn´t that true?

I see that you think that fa and Na increase at the same rate which is interesting, the truth is that I´d like to measure all this things to really know what´s happening.

 

[haruspex]

Why didn´t you prefer to use this relation Na*Fb=Nb*Fa ?

I used the form Fa/Na = (Fa+Fb)/(Na+Nb) because it was more convenient, but it's the same thing. If you multiply it out you'll see that.

 

[jonjacson]

Ups now I understand what you mean, well thanks to everybody for the help :)