How to calculate friction of wheels of car?

AI Thread Summary
To calculate the friction of a car's wheels, the discussion presents a formula that incorporates mass, acceleration, distance, and coefficients for wheel and air friction. The suggested values include a mass of 1000 kg, an acceleration of 10 m/s², and coefficients of friction k=0.01 and c=0.0001. The total energy required to accelerate to 50 m/s is calculated to be 62,625 J. Additionally, it is noted that air resistance complicates calculations as it varies with speed, while kinetic friction remains constant. The discussion emphasizes the importance of accurately measuring energy loss to determine friction effectively.
vissarion.eu
Messages
49
Reaction score
0
If car of mass m=1000 kg I want to accelerate to speed v=50 m/s then to me need to calculate friction of car axis on which puted wheels and friction of wheels into ground and air friction. So I suggesting such formula
m*a*t+k*m*a*S+c*m*a*S=m*a*t+k*0.5*m*v^2+0.5*c*m*v^2, where a is acceleration of car say a=10 m/s/s and S is distance which car made until got speed v=50 m/s and k is friction koeficient of car wheels inside car and to the ground and c is koeficient for air friction. I would give k=0.01 and c=0.0001 and t is 5 seconds. So total energy of car:
m*a*t+k*0.5*m*v^2+0.5*c*m*v^2= 1000*10*5+0.01*0.5*1000*50*50+0.0001*0.5*1000*50*50= 50000+12500+125= 62625 J.
If you can calculate better just calculate and show and don't blame my formula.
 
Physics news on Phys.org
vissarion.eu said:
If car of mass m=1000 kg I want to accelerate to speed v=50 m/s then to me need to calculate friction of car axis on which puted wheels and friction of wheels into ground and air friction. So I suggesting such formula
m*a*t+k*m*a*S+c*m*a*S=m*a*t+k*0.5*m*v^2+0.5*c*m*v^2, where a is acceleration of car say a=10 m/s/s and S is distance which car made until got speed v=50 m/s and k is friction koeficient of car wheels inside car and to the ground and c is koeficient for air friction. I would give k=0.01 and c=0.0001 and t is 5 seconds. So total energy of car:
m*a*t+k*0.5*m*v^2+0.5*c*m*v^2= 1000*10*5+0.01*0.5*1000*50*50+0.0001*0.5*1000*50*50= 50000+12500+125= 62625 J.
If you can calculate better just calculate and show and don't blame my formula.
To calculate the friction loss due to wheels you can give the car a known quantity of energy and then see how far it will coast on the flat. mv^2/2d = F = \mu_kN (You have to be careful not to give it too much speed to begin as the air resistance could be significant).

The problem with factoring in the air resistance is that it is proportional to speed or speed^2. Kinetic friction force on the wheels is a constant force.

AM

AM
 
TL;DR Summary: I came across this question from a Sri Lankan A-level textbook. Question - An ice cube with a length of 10 cm is immersed in water at 0 °C. An observer observes the ice cube from the water, and it seems to be 7.75 cm long. If the refractive index of water is 4/3, find the height of the ice cube immersed in the water. I could not understand how the apparent height of the ice cube in the water depends on the height of the ice cube immersed in the water. Does anyone have an...
Thread 'Variable mass system : water sprayed into a moving container'
Starting with the mass considerations #m(t)# is mass of water #M_{c}# mass of container and #M(t)# mass of total system $$M(t) = M_{C} + m(t)$$ $$\Rightarrow \frac{dM(t)}{dt} = \frac{dm(t)}{dt}$$ $$P_i = Mv + u \, dm$$ $$P_f = (M + dm)(v + dv)$$ $$\Delta P = M \, dv + (v - u) \, dm$$ $$F = \frac{dP}{dt} = M \frac{dv}{dt} + (v - u) \frac{dm}{dt}$$ $$F = u \frac{dm}{dt} = \rho A u^2$$ from conservation of momentum , the cannon recoils with the same force which it applies. $$\quad \frac{dm}{dt}...
Back
Top