# How much torque will I need to pull a 2250Kg load?

I need to use this to find a motor that will be able to pull at most 2250 kg at a 1.5 m/s pace. At peak acceleration, I would like this to accelerate at 0.5 m/s2. The area that the load will be pulled is a concrete sidewalk with, at its most extreme, a 20° incline.

• This is not a do your homework or project site - show what you have done so far. Commented Mar 15, 2018 at 17:48
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– Wasabi
Commented Mar 15, 2018 at 18:44
• sliding, on wheels, or what? Commented Mar 15, 2018 at 18:55

Ignoring friction, because you have not provided any details regarding this in your question, and assuming that the object starts from rest on a $20°$ incline, and accelerates at a peak rate of $0.5m/s^2$ on it's way to a max speed of $1.5m/s$

The force required to stop a static object from rolling down a hill is the same as that required to keep it at a constant speed on that hill, $F_1=mg\sin\theta$.

The additional resultant force required to provide an acceleration of $0.5m/s^2$ can then simply be added using $F=ma$, therefore,

$$F=m(g\sin\theta+a)$$

Which equates to a peak force in your case of:

$$F=2250(9.81\sin20+0.5)=8674=8.7kN$$

$P= mgh/s= 2250(9.8)sin(20)1.5$
• "At Peak acceleration" suggests that the acceleration will drop off after a certain time, i.e. speed does not increase infinitely. Acceleration is critical to calculating the force required - remember $F=ma$! Commented Mar 16, 2018 at 14:24