I am looking to calculate the force that would be required to pull a single axle 12000 lb box trailer with a 72 sq ft frontal area up a 10% grade at 80 mph.

It has been quite a long time since I've done any engineering and I want to make sure I'm not way off. I am only looking to find a comfortable upper limit for potential force required to pull any trailer. My numbers estimate a heavy trailer going up a very steep grade faster than one probably should.

For this example I'm estimating 1200 LB of strain from pulling the weight of the trailer and another 1200 LB from the drag of the wind. Is this a fair rough estimate or am I way off?


There are a few things that you need to consider:

10% road inclination

the force from 10% road inclination would be: $$F_{incl} = m\cdot g \sin(\phi)\approx 5313[N]$$


  • m= 12000[lb]
  • $\phi=tan(10\%)= 5.7[deg]$

Air Drag

The air drag is given by:

$$F_D= \frac{1}{2} C_D \rho_{air} A v^2\approx 3200$$ where:

  • $C_D =0.608 $ coefficient of Drag (value is from trucks)
  • $\rho_air$ density of air
  • $A$ frontal area
  • $v$ velocity (quite fast if you ask me)


you haven;t mentioned this but, you need to take it into account. Assuming an acceleration of let's say $1[m/s^2]$, you will need a force of

$$F_a = mass\cdot accel \approx 5500[N] (\approx 2450 lbf)$$


This can be a lot more important, since the deceleration in emergency braking can be in the order of 4-7.5 [m/s^2] (probably for a trailre would be in the lower range to avoid jackknifing bust still..). Fro the 4m/sec^2 the deceleration force would be:

$$F_{Braking} = 22000[N]$$

combination of loads

The worst combination is probably going downhill in an emergency brake situation, then the load would be in the order of

$$F_{combined} = 27000[N].$$

Other considerations:

The value above is the nominal load. You need to increase that with a safety factor N which in my opinion would need to be in the order of 2 and upwards.

Depending on the style of driving (hard/soft braking/accelerating, tight turning at moderate speed), the dynamic loads will be significantly higher and you need to account for fatigue.

Apologies for supplying most values in the metric (its more convenient for me). Please make sure to make the correct conversions, or even better redo my calculations to make sure that they make sense.

  • $\begingroup$ Thank you! This was extremely helpful. I had not thought about the acceleration loads which helps immensely. $\endgroup$
    – kbel
    Mar 9 at 17:37

The general equation of an object moving uphill is FT >= FR, in which FT is the thrust produced by the object, and FR is the resistance force that impede the movement of the object. FR includes,

  1. Downward pull of gravity, F1 = mgsin(\theta), the angle \theta = slope (10 degrees)

  2. Friction of tires on pavement, F2 = mgcos(\theta)(\mu), \mu is the friction coefficient (it varies, but can be conservatively taking as 1)

  3. Aerodynamic drag, F3 = C(\rho)AV^2/2, in which,

  • C is drag coefficient, it is about 0.71 for truck trailer.

  • \rho is density of the air, it can be assumed as 24x10-4 slugs/ft^3 at sea level.

  • A is the frontal surface of the truck trailer.

so FR = F1 + F2 + F3. Note that if your are only interested in the force required to toll the trailer, you shall set mg in the equation equal to the weight of the trailer (12000 lb), otherwise, mg is the total weight of the truck and trailer.

Depends on the condition of your truck, you need to ensure it has adequate power to drive uphill. Good luck.


Your Answer

By clicking “Post Your Answer”, you agree to our terms of service, privacy policy and cookie policy

Not the answer you're looking for? Browse other questions tagged or ask your own question.