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I want to find out how much torque will be needed from my motor if I want to pull a mass. So imagine a car towing a trailer.How would I calculate the torque need to move the trailer?

I found the friction acting on the wheels and then did sum of the forces in the x direction to find the tension on the hitch.

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  • $\begingroup$ Torque where? At the engine or at the wheels? There’s a gearbox in between... $\endgroup$ – Jonathan R Swift Jun 23 '18 at 10:10
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Torque of the engine does not matter. What matters when pulling is the force exterted by the car at the base of the wheel pushing the car forward. This force depends on many things but the two most important things are:

  • The current speed of the vehicle
  • The power of the engine

In fact ultimately, assuming a perfect transmission and no losses, you can write an equation for the force which is:

$$ f_p = {P \over v}$$

Where:

$f_p$ - is the pulling force of your vehicle,

$P$ - is the power of the engine (at the current rpm/throttle) and

$v$ - is the current speed of the car

$$$$

Once you have the current pulling force you can compare it to the force required to pull your trailer. Note that the pulling force tends to infinity as speed approaches 0, this is correct. In fact even a very low power tractor can still pull tonnes on a trailer because of a very advantageous transmission ratio and consequently very low speeds.

So to answer your question: to pull any trailer you need a torque greater than zero at an angular speed greater than zero. This give us a power which is greater than 0 which will be enough to pull any trailer at a certain speed.

The speed ultimately depends on many other forces which have to be balanced with the initial pulling forces but the most important are air drag, road slope losses, and tires losses.

Air drag depends on the size and shape of the vehicle and increases roughly with the square of the speed.

Slope losses depends on the mass of the vehicle and the slope of the road.

Tires losses depends on the weight of the vehicle and increase roughly linearly with the speed.

A more 'correct' question should be:

At which speed $v$ can I pull a trailer of mass $m_t$ with a car of mass $m_c$ and power $P$ on a road with a slope of $s$ if the air drag coefficient is $k_a$ and the friction coefficient is $k_f$?

Then a very rough answer would be obtained by solving for v the equation:

$${P \over v} - {{k_a}v^2 } - {((m_t + m_c)sin(s))} - ((m_t + m_c){k_f}v) = 0$$

Where:

${P \over v}$ is the pulling force of the car at speed v

${{k_a}v^2 }$ is the air drag force acting on the whole vehicle at speed v

$((m_t + m_c)sin(s))$ is the gravity force acting on the vehicle

$(m_t + m_c){k_f}v$ are the friction losses at speed v

Note that this is a very "simple" model and does not take into consideration many things, for example even friction changes as the slope of the terrain changes.

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Most of car makers talk about torque at engine flywheel. And most have for the first gear approximately 1 to 3.6 gear ratio and then 2 , 1.3, 1, etc. You find these in car specs.

Lets assume you have 16 inch diameter tires, meaning 8 inch radius.

So you have 12/8 = 1.5 mechanical advantage at wheel.

You divide F towing demand by 1.5 to gat torque at wheels, then you divide that by gear ratio, say 3.6 to get engine torque required to tow your trailer.

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