# What is the torque required to rotate a 450 lb., 35' long cylindrical load 90 degrees?

I need to create a simple machine to assist in rotating a cylindrical load, akin to a roll of carpet, from a horizontal position to vertical. The load weighs 450 pounds, which is distributed evenly along a 35 foot length. The machine would use a geared motor to rotate the load from the bottom of the cylinder.

I need two questions answered: 1 how much torque would the motor need to rotate the load, and 2 how heavy would the machine need to be to keep it from tipping over?

The machine would be assisted by human power, but I'd like to begin with the values needed for the machine to do all the work, then work my way backwards from there. Thanks!

• Is the whole system securely fixed in its position? I mean dose it move freely along the horizon ? Because i see wheels under the construction. And how long is the distance between the two wheels? – Sam Farjamirad May 22 '19 at 21:04
• It would be on wheels (I know it's impossible to stabilize), as it needs to be portable. The drawing is basically to scale, but wheels were just added to show that it's not a fixed setup. The motorized rotating mechanism would be attached to something like a Genie material lift: genielift.com/en/material-handling-products/material-lifts – Craig May 22 '19 at 23:28
• You posted a "thanks" comment on the answer from kamran. On SE sites, posting a "thanks" comment is frowned upon. When you really do like any answer though, you can thank the person by upvoting it using the up/down triangles near the top of the answer. And then, after a day or two, if there is one answer that you think is outstandingly good, you, as the Original Poster, can mark it as your selected best answer using the check mark. When you have time, look at the Engineering.SE tour. – Ray Butterworth May 23 '19 at 1:20
• @RayButterworth Who cares! – Sam Farjamirad May 23 '19 at 11:41

## 1 Answer

The CG of the load is at 17.5 feet. $$Torque= 450*17.5=7875 \ lbs.ft$$

And the connection of the base plate to concrete should be able to support this torque (moment) with a say 50% safty factor.

For example if your baseplate connecting the lift to concrete is

$$18\ inches *18\ inches.$$

$$M_{safety.factor }=7875*1.5= 11812.5$$

So if you have 4 anchor bolts attaching the base plate to the concrete slab at 1.25 ft OC,

$$F_{AB }= 11812.5/1.25= 9450\ lbs$$ The plate needs to be engineered bot as a first estimate you can start at a 18 inch square plate with 3/4 thickness.

Or you could have the bottom of you lift attached to a horizontal steel beam leg of at least longer than 18 ft.

**Edit **

After OP' comment.

If the lifter needs to be portable, then you need a base wide and heavy to resist overturning moment.

Say you have a bas of 6ft SQ with the crane mounted near the left side.

The weight needed to keep the balance under the crane is approximately,

12000/6 = 2000 lbs. But you need to have say 500 lbs right above each wheel as well, just for dynamic loading.

I would add grab bars to the crane for operators to hold on to.

• I appreciate the answer, thank you! And although it complicates things greatly, the base of the machine can't be anchored to the floor, it needs to be portable. We will have to add some weight to it to help keep it stable, but I'm sure that it wouldn't be feasible to add enough weight to keep it from tipping over completely. Human power would be needed to help stabilize the load while it's being lifted. – Craig May 22 '19 at 22:54
• You've been a great help, thanks! – Craig May 23 '19 at 0:35