How much torque is required to move an object weighing 150lb?

[uncoordinated]

How much torque is required to move an object weighing 150lb?Torque is required to move an object weighing 150lbs. Its not a homework question, I'm trying to build an electric scooter but I have not yet reached high-school or taken any advanced mechanical engineering classes or any at all for that matter. Someone said use the equation F=umg or force is equal to coefficent of static friction times the mass times the gravitational force? I found something that said .9 is a good coeffiecent of static friction but I don't know the acceleration. This is all very confusing please help.


MUCH appreciated,
Nicholas

 

[firavia]

The motor you want to install on your scooter should have a graph torque over angular speed , first thing to do is decide how much speed you want , thn you have to calculate the torque that is acting against your wheels and in this case it is due to friction and the torque is equal to Fx Radius of your wheels and as I saw your equation of F it is correct.
but you have to be carefull and add your body mass to the mass of the scooter frame.
After calculating the torque select a motor with the required torque to overcome the torque caused by friction on your wheels with the appropriate speed you want.
the f = umg means , u=0.9 , m= total mass , g = gravitational accelaration = 9.80665 m/s2 (approx. 32.174 ft/s2).

P.S:I rather to be confirmed by another user for what I said before you do anything.

 

[Unrest]

F=umg or force is equal to coefficent of static friction times the mass times the gravitational force?

What you have is the force required to start sliding something without wheels.

Instead, make u the coefficient of rolling resistance. 0.9 sounds kind of high but I guess it depends on the surface and wheels. u=1 means it takes as much force to push it along as it does to lift it straight up. u<1 means rolling is easier than lifting. http://en.wikipedia.org/wiki/Rolling_resistance" has a table showing coefficients for bicycle wheels, etc.

After you have F=umg, you can get the torque from T=Fr where r is the radius of the wheel.

That only gives you enough torque to just keep moving on a horizontal surface. You'll need to add more force to accelerate at a reasonable rate and climb hills: F=ma and F=m g sin(incline angle)

You can see if your motor can provide the required torque at the desired speed according to what firavia said.

 

[uncoordinated]

So that would mean if I input my information F=u(.005)*m(160lb I am 120 plus 40lb frame)*(32.14 ft/s2)

and I get 25.712 units(whats the unit?)

your saying if the radius of my wheel is 4inch. I multiply that by 25.712, so i get 102.848 torque needed to break the static of friction and maybe more for hill.?

 

[mpaige1]

Units:
F = lb*ft/s^2 = Newton
T = F*r = Newton/inch

 

[mpaige1]

your saying if the radius of my wheel is 4inch. I multiply that by 25.712, so i get 102.848 torque needed to break the static of friction and maybe more for hill.?

If you were starting on the hill, yes the torque needed would be greater. If you have started before the hill, you would need to calculate the force to keep moving up the hill, which would obviously be smaller than starting from a stop on the hill. hope that helps

 

[jack action]

It is not about how much torque you need, but how much power you need.

And how much you need is determined by http://hpwizard.com/performance.html" and at what speed.

Check http://hpwizard.com/car-performance.html" for more info.

 

[mender]

The force needed to keep 160 lbs of person and vehicle is quite small; multiply the weight by a reasonable coefficient of rolling resistance ( 0.005 is pretty low but might be applicable if you're using large bicycle wheels) and you'll get the force - in this case, the force is 0.8 lbs force. Even at 0.020 (more likely for little scooter wheels), that's only 3.2 lbs.

The force needed to accelerate is F=MA; a moderate acceleration rate is about 0.1g, which means that you'll need 16 lbs of force to achieve that. As you can see, the force needed to accelerate is the number you really need to consider as it is much larger than that required for rolling resistance.

 

[Unrest]

So that would mean if I input my information F=u(.005)*m(160lb I am 120 plus 40lb frame)*(32.14 ft/s2)

and I get 25.712 units(whats the unit?)

These units are a bit dangerous, so you have to be rigorous. Treat the units like numbers and multiply them along with everything else:

F = 0.005 * 160lb * 32ft/s²
F = 25.6 lb ft/s²
lb ft/s² is a unit of force. The weight of a 1lb mass is 32 lb ft/s² = 1 lbf
so
F = 0.8 lbf
It's a whole lot easier in SI ;)

your saying if the radius of my wheel is 4inch. I multiply that by 25.712, so i get 102.848 torque needed to break the static of friction and maybe more for hill.?

Again, carry the units through to be sure what they should be:
25 lb ft/s² * 0.33ft = 8.3 lb ft/s² * ft
= 0.26 ft-lbf

- or -

0.8lbf * 0.33ft = 0.26 ft-lbf



And change "break static friction" to "overcome rolling resistance". They're different types of friction.

 

[mender]

These units are a bit dangerous, so you have to be rigorous. Treat the units like numbers and multiply them along with everything else:

F = 0.005 * 160lb * 32ft/s²
F = 25.6 lb ft/s²
lb ft/s² is a unit of force. The weight of a 1lb mass is 32 lb ft/s² = 1 lbf
so
F = 0.8 lbf
It's a whole lot easier in SI ;)

Actually, it's much simpler in Imperial; F = 0.005* 160 lb = 0.8 lb.

One step; it's only in SI that you need to convert back and forth a few times before you get the answer in the correct units.

 

[Unrest]

Actually, it's much simpler in Imperial; F = 0.005* 160 lb = 0.8 lb.

One step; it's only in SI that you need to convert back and forth a few times before you get the answer in the correct units.

Hehe, true, and torque just as easy. As long as you don't forget that "lb" is now a force instead of a mass. Not growing up with imperials I have to do it the long way to be safe, it also causes me to put an "f" on the end for force because I can never tell the difference from context.