Gear torque --- help me please

[th888]

I am trying to make a lift for my sister's farm. Please help with this problem....thank you.

20 inch diameter geared cable spool is driven by a 2 inch spur gear. The load is 1500 pounds.

How many ft/lbs torque is require at the 2" gear to lift the load at 30 feet per minute?

Thanks in advance for your time and help.

[Kelly_Bramble]

r₁ x F₁ = r₂ x F₂

or

(r₁ x F₁)/r₂ = F₂

Where:

r₁ = radius large gear
r₂ - radius small gear
F₁ = force applied small gear
F₂ = force applied large gear

torque= r x F

torque large gear = F₂ x r₂

[th888]

F1 value is unknown. What I am trying to do is find how much motor torque is needed (2 inch pinion) to lift 1500 pounds connected to a 20 inch diameter cable drum.

I am I looking at this wrong? Sorry I'm not a ME.

Thank you kindly for you response

r₁ x F₁ = r₂ x F₂

or

(r₁ x F₁)/r₂ = F₂

Where:

r₁ = radius large gear
r₂ - radius small gear
F₁ = force applied small gear
F₂ = force applied large gear

torque= r x F

torque large gear = F₂ x r₂

[Kelly_Bramble]

?

torque small gear = F₁ x r<sub>1

Instructional Video:

Gear to Gear Torque Calculations</sub>

[th888]

Thanks Kelly.

[th888]

?

torque small gear = F₁ x r<sub>1

Instructional Video:

Gear to Gear Torque Calculations</sub>

The driven gear is 20" dia. The driving gear is 2" dia.
The ratio is 10:1
To lift a 1500 lb load on the 20" dia gear, 150 in/lb torque will be required.

Correct??????????