# Calculation of the forces and the work of these forces

• JrK
In summary, the conversation discusses a scenario in which a circle moves in horizontal translation without rotation, while being controlled by a cylinder and a motor. The circle is in contact with a rotating red wall, with no friction present. The mass is low and the bodies involved are rigid, except for the elastic being used to apply a constant force (F). The energy required to push the circle and rotate the wall is calculated, along with the energy stored in the elastic. There is also a needle involved, with a very small diameter, being pulled by the elastic but prevented from moving closer to the contact point by lateral walls. Enlargements and diagrams are provided to illustrate the positions and forces involved in the scenario.
JrK
Like AT asked I post a new thread because it is a different example, here there is no friction and I use a needle and an elastic to have the forces.

The circle moves in horizontal translation, the circle doesn't rotate around itself, the red wall rotates around A0, and A0 is fixed to the ground. The circle is always at the contact with the red wall. The mass is very low and there is no friction. There is no acceleration/deceleration. The bodies are rigid except the elastic. The force of the elastic is constant : F. I drew the start and the final position, the circle and the red wall are controlled in position with a cylinder and a motor for example, I count the energy I need to give to push the circle and to rotate the red wall and I count the energy stored inside the elastic, I also compute the energy from the force F3 and F4 on the lateral walls (I think my mistake comes from these forces). There are 4 objects: the circle and the wall, the elastic and the needle. The needle has a diameter very small: some atoms, to be near the dot of contact, I cannot be exactly at the dot of contact but from start to end with a very small diameter for the needle, the distance from the dot of contact will not change to reach the distance d1 in the calculation of the energy stored inside the elastic. The sum of forces on the needle is 0: the elastic pulls it but the lateral walls prevent it to move closer to the dot B. Please, look the enlargements to watch the needle more farther.

The example:
The example from start to end:

Enlargement of the previous image:

Enlargement of the area where there is the elastic, I drew the start and final position, the elastic is taken by the dots A and B, the dot B is fixed on the red wall, the dot A is fixed on the needle, the needle is near the dot of contact between the circle and the red wall:
To watch the forces of the elastic F1 and F2, I drew the start and final position, at start the elastic has a length at 0, and at final its length is d2:

To watch the needle (the needle is perpendicularly to the screen and its diameter is very small, one atom if I can), I drew only the final position, the needle is pulled by the elastic but the lateral walls prevent it to move closer the dot B:

Enlargement, I drew the final position, to watch the forces from the needle to the lateral walls (circle and red wall):

The program :
Code:
 #include <stdio.h>
#include <math.h>

int main()
{
long double pi=M_PI;

long int N=100000000L;
long double R=.1;
long double D=1.;
long double xinf=1.;
long double xsup=xinf/tanl(44.99/180.0*pi);

long double px1=0,py1=0,a1=0,px2=0,py2=0,a2=0,dl=0,wd=0,wf=0,ix1=0;
long double iy1=0,ix2=0,yi2=0,qx1=0,qy1=0,s=0,xl1=0,yl1=0,xl2=0,yl2=0;
long double sx=0,sy=0,qx2=0,qy2=0,xi1sav=0,yi1sav=0,xsav=0,suma=0,l=0,asav=0;
long double xf=0,yf=0,wp=0,af=0,sumdx=0,dlrc=0,dlpc=0;

long int i;
qy1=D;

for(i=0;i<N;i++)
{
//a1=(long double)(asup/180.0*pi-ainf/180.0*pi)/N*i;
qx1=xinf+(xsup-xinf)/N*i;
a1=atanl(qy1/qx1); if(i==0) a2=a1;
s=R/tanl(a1/2.);
px1=qx1-s;
py1=qy1-R;
if(i==0 || i==N-1) printf("\npx1=%Lf",px1);
ix1=px1+R*sinl(a1);
iy1=py1-R*cosl(a1);
if(i<1) {xi1sav=ix1;yi1sav=iy1;xsav=px1;asav=a1;}
//printf("\npx1=%Lf , py1=%Lf , a=%Lf , ix1=%Lf , iy1=%Lf , a1=%Lf",px1,py1,180/pi*atanl((fabsl(iy1-yi2)-fabsl(py1-py2))/(fabsl(ix1-ix2)-fabsl(px1-px2))),ix1,iy1,180/pi*a1);
suma+=a1;

if(i>0)
{
sumdx+=(fabsl(px1-px2)-fabsl(ix1-ix2))*cosl(a1);

l=(sqrtl(ix1*ix1+iy1*iy1)+sqrtl(ix2*ix2+yi2*yi2))/2.;
xf=ix2+l*sinl(a1)*fabsl(a2-a1);
yf=yi2-l*cosl(a1)*fabsl(a2-a1);
af=fabsl(atanl((iy1-yf)/(ix1-xf)));
wf+=fabsl(px2-px1)*cosl(af);
wd+=sqrtl((ix1-xf)*(ix1-xf)+(iy1-yf)*(iy1-yf))*cosl(a1-af);
//printf("\nl=%Lf , xf=%Lf , yf=%Lf , af=%Lf , diffa=%Lf",l,xf,yf,180/pi*af,180/pi*(a1-af));
wp+=fabsl(sinl(a1-af))*l*fabsl(a2-a1);
dlpc+=(px2-px1)*sinl(a1);
dlrc+=l*fabsl(a2-a1);

}
px2=px1; py2=py1; qx2=qx1; qy2=qy1; ix2=ix1; yi2=iy1; xl2=xl1; yl2=yl1; a2=a1;

}
printf("\nN=%ld , R=%Lf , D=%Lf , xinf=%Lf , xsup=%Lf",N,R,D,xinf,xsup);
printf("\ndlp=%Lf , dli=%Lf, dlpc=%.18Lf , dlrc=%.18Lf, diff=%.18Lf",px1-xsav,ix1-xi1sav,dlpc,dlrc,(dlpc-dlrc));
printf("\nsuma=%Lf , cos=%Lf , wf=%Lf , wd=%Lf , wp=%Lf , diff=%Lf , eff=%Lf , ddx=%Lf , sumdx=%Lf\n",(asav-a1)*180/pi,cosl(suma/N),wf,wd,wp,wf-wd-wp,(wd+wp)/wf, fabsl(px1-xsav)-fabsl(ix1-xi1sav),sumdx);
return 0;
}

The results:

The energy to move the circle: lg*cos(44.5°)*F = d1*F (in the computation I calculate the integral)
The difference of energy stored inside the elastic: d2*F
The energy needed from F3 is equal to the energy recovered from F4

d1 is greater than d2.

JrK said:
The energy to move the circle: lg*cos(44.5°)*F = d1*F
What is that 44.5°? Please use symbols that are defined in your diagrams, not some numbers.

Make a free body diagram of the contact force vectors between the 3 bodies. With the needle diameter at least 200 pixels, and the circle small enough, so the wall-circle-contact and the circle center are also in the diagram. Denote the angles in that diagram.

Last edited:
The 44.5° is the mean of the angle 'a' (the angle of the red wall relatively to the horizontal) I drew from the start position at 45° to the end position at 44° it is a rapid calculation but I take the numerical integral inside the program. I noticed the angles, from start: 'a2', to the end 'a1' :

I drew what you asked (the needle I would like to use is very small in reality but here I think it is to watch the angles):

F1 is the force the elastic applies on the red wall
F2 is the force the elastic applies on the needle
F3 is the force the needle applies to the red wall
F4 is the force the needle applies to the circle
I notices the angles relatively to the red wall, a3, a4 and a5.

JrK said:
The 44.5° is the mean of the angle 'a' (the angle of the red wall relatively to the horizontal) I drew from the start position at 45° to the end position at 44° it is a rapid calculation but I take the numerical integral inside the program. I noticed the angles, from start: 'a2', to the end 'a1' :

View attachment 262553

I drew what you asked (the needle I would like to use is very small in reality but here I think it is to watch the angles):

View attachment 262556

F1 is the force the elastic applies on the red wall
F2 is the force the elastic applies on the needle
F3 is the force the needle applies to the red wall
F4 is the force the needle applies to the circle
I notices the angles relatively to the red wall, a3, a4 and a5.
Can we have the left end of the elastic fixed to a point with distance r_needle from the wall, such that the elastic is parallel to the wall? Or is non-zero angle a3 important to you?

A.T. said:
Can we have the left end of the elastic fixed to a point with distance r_needle from the wall, such that the elastic is parallel to the wall? Or is non-zero angle a3 important to you?
Yes, it is important to have at start the dot A and B at the dot of contact and the angle a3 exists. And the length of the red arm cannot be infinite. The size of the elements is the size I drew at start in the first message.

Edit: but the needle has not the size I drew in the last message (I drew it bigger because you asked), the diameter of the needle is very small (like in the first message), the radius of the circle is 0.1 m and the diameter of the needle is one atom (I cannot take less) like 0.3 nm. So the angle a3 is near the angle of the red wall.

The elastic can have the length not at 0 to have the angle a3 like the red wall, but remember the diameter of the needle is 1 atom:

Last edited:
JrK said:
...the angle a3 exists. ...
OK, what about the contact forces between circle and wall? Your wall is not in equilibrium right now. And neither is your circle, so you can add the driving force on it while you are at it.

A.T. said:
OK, what about the contact forces between circle and wall?...
The circle and the wall are controlled in position by two external devices (cylinder+motor), I counted the energy in/out. I have the contact but I suppose there is no force directly from the circle to the wall.

JrK said:
The circle and the wall are controlled in position by two external devices (cylinder+motor), I counted the energy in/out. I have the contact but I suppose there is no force directly from the circle to the wall.
So now you have two devices pushing against each other, via three bodies in mutual contact. You made this way more complicated than you original friction scenario. You should try to simplify things, if you want to find your errors.

A.T. said:
So now you have two devices pushing against each other,
No, it is because the example is unstable, I need to control the position of the circle and the rotation of the wall. The needle pushes the circle and the wall, I need something to control the position. It is not an additional devices, I explained that in my first message.

JrK said:
No, it is because the example is unstable, I need to control the position of the circle and the rotation of the wall. The needle pushes the circle and the wall, I need something to control the position. It is not an additional devices, I explained that in my first message.
Well, if you want to track all the energy, you need to include all the forces in your diagrams. And when computing work, pay attention to the sign.

I drew the forces:

I drew the force F1 bigger than it is in reality ! Note I place the dot B at A0 like that I can consider the angle of the elastic like the red wall if the diameter of the needle is 0.3 nm for example.
The sum of forces on the needle is 0 at each time so I didn't draw the forces on the needle. The needle receives the force F2 (from the elastic) and the forces -F3 and -F4 (in vector).
The forces F3 and F4 are the forces from the needle on the lateral walls near the dot of contact because the needle is very small.
Even there is the contact between the circle and the red wall there is no force directly from the circle to the wall.
The force F6, given by an external device, compensates the force F4, so the sum of forces on the red wall is 0 but the external device recovers the energy from F6.
The force F5, given by another external device, compensates the horizontal part of the force F3, so like that the sum of forces on the circle is 0, but the external device needs to gives an energy to move the circle.
I count the energy stored inside the elastic.

I calculate the energy from F5 and F6, and the difference of energy stored inside the elastic. I need only to calculate the sum of energy from 44° to 45° for example (or less). Note the force F3=F2-F4 (in vector) and a part of F2 is in the horizontal part of F3, it is visible with the big needle you asked.

JrK said:

I calculate the energy from F5 and F6, and the difference of energy stored inside the elastic.
You should put all the forces and angles in one diagram, not split them across several, that have inconsistent angles. This will just confuse you.

Based on the definitions above, what are the expressions for the work by F5 and F6, and for the energy stored in the elastic?

A.T. said:
Based on the definitions above, what are the expressions for the work by F5 and F6, and for the energy stored in the elastic?

From 45° to 44° (a2 to a1) for the red wall, I noted the mean of the angle am=(a1+a2)/2:

I measure the distance dx (the movement of the red wall) and I reported its length it to dx' to compare the movement of the circle:

Like that I know the distance moved by the circle is dx'/cos(am) note am is the mean of the angles from start to end (a2 to a1), note that dx' is near d1. The energy recovered from F6 is dx*F6. I can decomposed the force F5 in two parts, the part that comes from F2 and the part that comes from the force F4 (the value of F6). The part that comes from F2 needs the energy dx'/cos(am)*cos(am)=dx' because at mean the force F2 is at an angle am relatively to the horizontal. The part that comes from the force F4 has the same orientation than F4, so the energy needed is dx'/cos(am)*sin(am) near equal to dx/cos(am)*sin(am). The difference of energy stored inside the elastic is d2*F.

cos(am)/sin(am) = 0.983
d2/d1 = 0.39/0.45 = 0.866

Note, for a smaller difference of angle of rotation, from 45° to 44.99°, for example, the ratio cos(am)/sin(am) is 1.0003 but the ratio of d2/d1 is always the same. At final, I have the energy needed to move the circle at d1*F and the difference of the energy stored inside the elastic at d2*F.

Last edited:
A.T. said:
Can we have the left end of the elastic fixed to a point with distance r_needle from the wall, such that the elastic is parallel to the wall?
I thought you wanted the length [A0;A] infinite but not in fact you want only to have the dot B far enough to consider the elastic at the same angle than the red wall, and yes it is possible to think like that. In that case, the difference of forces F3 and F4 is near 0 by limit ?

JrK said:
... to consider the elastic at the same angle than the red wall, and yes it is possible to think like that. In that case, the difference of forces F3 and F4 is near 0 by limit ?
Making the elastic parallel to the wall will certainly simplify things a lot. But it doesn't make F3 and F4 equal. You should use this type of diagram, but with the elastic parallel to the wall:

Yes, it is better for the details, so I decomposed the force F3 in F3a and F3b (F3a is perpendicularly to the red wall and F3b is parallel to the red wall):

If the dot B is very far (and the needle very small), the angle of the elastic has the same orientation of the red wall, in that case the value of F3a is equal to the value F4 and the value of F3b is equal to the value of F1 ?

Last edited:
But if F3a=F4 and F3b=F1 (in value) the work needed to move the circle is d1F and the difference of energy stored inside the elastic is d2F (with F=F1 in value) so is it really possible to say the angle of the elastic is like the red wall ?

Because when I calculated (numerical integral with a loop) the work needed for the force F3a I do: -abs(px2-px1)*sin((a1+a2)/2)) with px2 and px1 the abcisses between two steps of the dot P (P: center of the circle) and to calculate the work recovered by the force F4 I do: l*abs(a2-a1) with 'l' the mean length of the red wall from A0 to the dot A. At each step I have: l*abs(a2-a1) = abs(px2-px1)*sin((a1+a2)/2))

JrK said:
...is it really possible to say the angle of the elastic is like the red wall ?
You can attach the elastic to a point that is 1 needle radius away from the wall, but fixed to the wall. Then the elastic is exactly parallel to the wall.

JrK
Ok, and in that case I have a little torque on the red wall around A0, but it is not enough to reach the difference of energy between d1F and d2F or the work from F3a is not equal to the work from F4 ?

The sum of forces at each time on the needle is zero, so I don't think I need an external device to control its position, but there is no friction otherwise it will be difficult to move the needle. So, there is:

1/ the energy to move the circle: represented graphically by d1*F
2/ the energy recovered by the difference of the length of the elastic: represented graphically by d2*F
3/ the energy from the torque on the red wall around A0 from F1 (I place B at a radius of the needle), it is 0 in theory if the size of the needle is small and the dot B far from A0 (and far from A)
4/ the energy needed to move F3a: -abs(px2-px1)*sin((a1+a2)/2))
5/ the energy recovered from F4: l*abs(a2-a1), the absolute value is equal to 4/

I cleaned the calculations:

Code:
#include <stdio.h>
#include <math.h>

int main()
{
long double pi=M_PI;
long int N=10000000L;
long double R=.1;
long double D=1.;
long double xinf=1;
long double xsup=1.1;//xinf/tanl(44.99/180.0*pi);

long double px1=0,py1=0,a1=0,px2=0,py2=0,a2=0,dl=0,wredwall=0,wcircle=0,ix1=0;
long double iy1=0,ix2=0,iy2=0,qx1=0,qy1=0,s=0,xl1=0,yl1=0,xl2=0,yl2=0;
long double sx=0,sy=0,qx2=0,qy2=0,suma=0,l=0,asav=0;
long double xf=0,yf=0,wp=0,af=0,wF4=0,wF3=0;
long int i;
qy1=D;
py1=qy1-R;

for(i=0;i<N;i++)
{
qx1=xinf+(xsup-xinf)/N*i;
a1=atanl(qy1/qx1); if(i==0) a2=a1;
s=R/tanl(a1/2.);
px1=qx1-s;
ix1=px1+R*sinl(a1);
iy1=py1-R*cosl(a1);

if(i>0)
{
l=(sqrtl(ix1*ix1+iy1*iy1)+sqrtl(ix2*ix2+iy2*iy2))/2.;
xf=ix2+l*sinl(a1)*fabsl(a2-a1);
yf=iy2-l*cosl(a1)*fabsl(a2-a1);
af=fabsl(atanl((iy1-yf)/(ix1-xf)));
wcircle+=fabsl(px2-px1)*cosl((a1+a2)/2);
wredwall+=sqrtl((ix1-xf)*(ix1-xf)+(iy1-yf)*(iy1-yf))*cos(a1-af);
wF3+=fabsl((px2-px1)*sinl((a1+a2)/2));
wF4+=l*fabsl(a2-a1);
printf("cercle=%.18Lf , mur=%.18Lf",wF3,wF4);
char c = getchar_unlocked();

}
px2=px1;  py2=py1;  qx2=qx1;  qy2=qy1;  ix2=ix1;  iy2=iy1;  xl2=xl1;  yl2=yl1;  a2=a1;

}
printf("\nN=%ld , R=%Lf , D=%Lf , xinf=%Lf , xsup=%Lf",N,R,D,xinf,xsup);
printf("\nwF3=%.18Lf , wF4=%.18Lf, diff=%.18Lf",wF3,wF4,wF3-wF4);
printf("\nwcircle=%Lf , wredwall=%Lf , diff=%Lf  , eff=%Lf\n",wcircle,wredwall,wcircle-wredwall,wredwall/wcircle);
return 0;
}

Like d1 is different of d2 (graphics and in the calculations), what I forgot ?

Last edited:
JrK said:
...represented graphically by d1*F ...represented graphically by d2*F
There is neither F nor d1 nor d2 in your graphic.

A.T. said:
There is neither F nor d1 nor d2 in your graphic.
In the third image in the first message, I noted d1 and d2. F is the value of the force of the elastic I noted the message #17. In the calculation I calculated by steps the energies 1/ and 2/ to verify what I found in the drawings.

JrK said:
F is the value of the force of the elastic
Why don't you use the labels in the graphic?

A.T. said:
Why don't you use the labels in the graphic?
Because I would like to dissociate the two forces of the elastic: F1 and F2. F is just the value of F1 (or F2).

So, with:

##F##: the value of the force of the elastic equal to ##F_1## or ##F_2##
##p##: a dot, center of the circle
##i##: a dot, it is the dot of contact between the circle and the red wall
##f##: a dot, it is the new position of the last dot ##i## that I fixed on the red wall
##a##: the angle of the red wall relatively to the horizontal
##l##: the length of the red wall from ##A_0## to the dot of contact ##i##

The circle at start: blue color, the wall at start: red color
The circle at the end: grey color, the wall at the end: yellow color

I decomposed each dot with ##x## and ##y## value (easting/northing) and I used ##2## or ##1## for distinguish the last position ##2## and the new position ##1##

For a small angle of rotation of the red wall, (one step of the numerical integral, ##a_2-a_1## is very small) I have:

1/ the energy to move the circle: ## F (p_{x2}-p_{x1}) \cos{\frac{a_1+a_2} {2}}## , represented graphically by ##-d_1*F##
2/ the energy recovered by the difference of the length of the elastic: ##F \sqrt{(i_{x1}-f_x)*(i_{x1}-f_x)+(i_{y1}-f_y)*(i_{y1}-f_y)}##, with: ##fx=i_{x2}+l \sin{\frac{a_1+a_2} {2}} (a_2-a_1)## and ##f_y=i_{y_2}-l \cos{\frac{a_1+a_2}{2}} (a_2-a_1)## , and with ##l= \frac{ \sqrt{i_{x1}*i_{x1}+i_{y1}*i_{y1}}+\sqrt{i_{x2}*i_{x2}+i_{y2}*i_{y2}}} {2}## , that energy is represented graphically by ##d_2 F##
3/ the energy from the torque on the red wall around ##A_0## from ##F_1## (I place ##B## at a radius of the needle), it is ##0## in theory if the size of the needle is small and the dot ##B## far from ##A_0## (and far from ##A##)
4/ the energy needed to move ##F_{3a}##: ##F_4 (p_{x2}-p_{x1}) \sin{\frac{a_1+a_2} {2}}## because the value of ##F_4## is equal to the value of ##F_{3a}## if the orientation of the elastic is the same than the red wall
5/ the energy recovered from ##F_4##: ##l (a_2-a_1) F_4##, that absolute value is equal to 4/

The value of 3/ is ##0## in theory
The sum of 4/ add 5/ is ##0##, not sure about that calculation !
I have at final only the 1/ and 2/

And maybe the hypothesis : I take the dot ##A## (the needle) at the position of the dot of contact falses all the results. In theory, I could reduce the diameter of the needle as I want but even in reality, with a diameter of 1 atom, the distance from A to the dot of contact is 25µm and from the last position ##2## to the new ##1## that length is constant I think. If I compute from a=45° to a=30° the difference of length ##d_2-d_1## is very big in comparison of the 25µm.

The sum is : ##F (p_{x2}-p_{x1}) \cos{\frac{a_1+a_2} {2}}## add ##F \sqrt{(i_{x1}-f_x)*(i_{x1}-f_x)+(i_{y1}-f_y)*(i_{y1}-f_y)}## and it is graphically represented by ##(-d_1+d_2)*F##

Last edited:
I found my mistake. I took the radius from A0 to the dot of contact for the torque but the needle applies the torque a little farther and even the distance is small like the force is high, the result is not 0. So I have well the sum at 0. I gave the program in case it could help someone.

Code:
#include <stdio.h>
#include <math.h>int main()
{
long double pi=M_PI;
long int N=350000L;
long double R=.1;
long double D=1.;
long double xinf=1;
long double xsup=xinf/tanl(44/180.0*pi);

long double px1=0,py1=0,a1=0,px2=0,py2=0,a2=0,dl=0,wredwall=0,wcircle=0,ix1=0;
long double iy1=0,ix2=0,iy2=0,qx1=0,qy1=0,s=0,xl1=0,yl1=0,xl2=0,yl2=0;
long double sx=0,sy=0,qx2=0,qy2=0,suma=0,l=0,asav=0;
long double fx=0,fy=0,wp=0,af=0,wF4=0,wF3=0,k=0,F=0,Rball=0,wball=0;
long int i;
qy1=D;
py1=qy1-R;
F=1.;
Rball=5e-3;
k=1/tanl(acosl((R-Rball)/(R+Rball)))*F;
for(i=0;i<N;i++)
{
qx1=xinf+(xsup-xinf)/(long double)N*(long double)i;
a1=atanl(qy1/qx1); if(i==0) a2=a1;
s=R/tanl(a1/2.);
px1=qx1-s;
ix1=px1+R*sinl(a1);
iy1=py1-R*cosl(a1);

if(i>0)
{
l=(sqrtl(ix1*ix1+iy1*iy1)+sqrtl(ix2*ix2+iy2*iy2))/2.;
fx=ix2+l*sinl((a1+a2)/2.)*(a2-a1);
fy=iy2-l*cosl((a1+a2)/2.)*(a2-a1);
af=atanl((iy1-fy)/(ix1-fx));
wcircle+=(px2-px1)*cosl((a1+a2)/2);
//wredwall+=sqrtl(ix1*ix1+iy1*iy1)-sqrtl(fx*fx+fy*fy);
wredwall+=sqrtl((ix1-fx)*(ix1-fx)+(iy1-fy)*(iy1-fy));//*cos(a1-af);
wF3+=(px2-px1)*sinl((a1+a2)/2)*k;
wF4+=(l+(R+Rball)*sinl(acosl((R-Rball)/(R+Rball))))*(a2-a1)*k;
wball+=Rball*(a2-a1);

//char c = getchar_unlocked();

}
px2=px1;  py2=py1;  qx2=qx1;  qy2=qy1;  ix2=ix1;  iy2=iy1;  xl2=xl1;  yl2=yl1;  a2=a1;

}
printf("\nN=%ld , R=%Lf , D=%Lf , xinf=%Lf , xsup=%Lf , k=%Lf",N,R,D,xinf,xsup,k);
printf("\nwF3=%.18Lf , wF4=%.18Lf, wball=%.18Lf , sum=%.18Lf",wF3,wF4,wball,(wF3+wF4));
printf("\nwcircle=%Lf , wredwall=%Lf , diff=%.18Lf  , eff=%Lf\n",wcircle,wredwall,wcircle+wredwall+fabsl(wball)+(wF3+wF4),fabsl((wredwall+fabsl(wball)+(wF3+wF4) )/wcircle));
return 0;
}

## 1. What is the definition of force?

Force is a physical quantity that describes the interaction between two objects. It is a vector quantity, meaning it has both magnitude and direction.

## 2. How is force calculated?

Force is calculated by multiplying an object's mass by its acceleration. The formula for force is F=ma, where F is force, m is mass, and a is acceleration.

## 3. What is the unit of force?

The unit of force in the International System of Units (SI) is the Newton (N). Other commonly used units of force include pounds (lb) and kilograms (kg).

## 4. What is work in relation to forces?

Work is a measure of the energy transferred to or from an object by a force. It is calculated by multiplying the force applied by the distance over which the force acts.

## 5. How is work related to the concept of power?

Power is the rate at which work is done. It is calculated by dividing the work done by the time taken to do it. In other words, power is a measure of how quickly work is being done.