# 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. Mar 15 '18 at 17:48
• Welcome to Engineering! This looks like a "homework question" (notice the quotation marks). In order for such questions to be answered in this site, we need you to add details describing the precise problem you're having. What have you tried to solve this yourself? Please edit your question to include this information.
– Wasabi
Mar 15 '18 at 18:44
• sliding, on wheels, or what? Mar 15 '18 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 $F=ma$! Mar 16 '18 at 14:24