Calculating Tire/Tyre Forces in Weight Distribution of 1400kg Car

AI Thread Summary
To calculate the forces on each tire of a 1400kg car, the initial values given are T1 = 2820N and T2 = 4050N, which do not sum to the total weight of the car (mg = 13734N). The user proposes using the ratio d1/d2 = T2/T1 and the equation T1 + T2 = mg to find corrected values, resulting in T1 = 2418N and T2 = 11315N, which are still incorrect. Acknowledging the car has four wheels, the discussion shifts to a new problem involving a 100kg weight added to the car, requiring recalculation of the normal forces at the three wheels. The user is seeking guidance on how to approach this more complex problem, particularly in using torque and force distribution principles effectively.
danielI
Messages
16
Reaction score
0
This is a lill awkward question but I need to know. A car has the mass 1400kg. What force will be distributed on each tire/tyre?
Code:
           ________________
         /                 \
       /                    \
 ____/                       \___
 |       |----d1--|G|-d2-|      |
 |-------O---------------O---------
         T1             T2

G is the mass center and d1 = 1386mm, d2 = 964mm. First of all, it (the paper) says T1 = 2820N and T2 = 4050N. But 4050+2820 = 6870 != mg = 13734. How could this be and how would I calculate T1 and T2? My logic (and I guess my best, of many, shots)told me that maybe d1/d2 = T2/T1. We also know that T1+T2=mg. So I used this and got T1 = 2418N and T2 = 11315N. Not quite correct.

/daniel
 
Last edited:
Physics news on Phys.org
it appears that if t1+t2 were doubled it would account for the extra force missing and for the other two wheels on the other side of the car.
 
A car has 4 wheels?:biggrin:

Cheers mate, cheers!:-p
 
Now to a little harder problem.

A 100kg weight is added on the red spot. Calculate the changes in the normal force reactions at the three wheels due to the weight of the box.

http://img110.imageshack.us/img110/7644/car7pu.png

In the last one I could use the equations d1/d2 = T2/T1 and T1+T2=mg. But for this one I only can use the last condition (A+B+C=mg). The only tactics I could come up with was to split the box up in (three?) pieces and move them (hopefully under each wheel) and add torques if necessary. But this seems so comprehensive, and I'm not even sure it will work.

Any help please?
 
Last edited by a moderator:
Please? I'm not finding any other relationships.
 
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