What is the force required to upright a tipped drum?

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In summary, a 518 pound drum tipped over will require approximately 1/2 of its weight (weight of the drum) to lift it upright.
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MacMurchyEHS
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55 gallon drum weighing 518 pounds is tipped over. How much force is needed to upright the tipped drum?
Summary: 55 gallon drum weighing 518 pounds is tipped over. How much force is needed to upright the tipped drum?

I am an environmental, health and safety professional in the food manufacturing industry. We recently had two employees that came upon a 55 gallon drum that was tipped over. The weight of the drum and contents was 518 pounds. They attempted to lit the drum by hand to the upright position. One of the employee injured their lower back attempting to lift the drum. I have been tasked with determining the required amount of force to lift and upright the tipped drum. Please let me know if you need additional information to assist with my task. Thank you.
 

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MacMurchyEHS said:
Summary: 55 gallon drum weighing 518 pounds is tipped over. How much force is needed to upright the tipped drum?

Summary: 55 gallon drum weighing 518 pounds is tipped over. How much force is needed to upright the tipped drum?

I am an environmental, health and safety professional in the food manufacturing industry. We recently had two employees that came upon a 55 gallon drum that was tipped over. The weight of the drum and contents was 518 pounds. They attempted to lit the drum by hand to the upright position. One of the employee injured their lower back attempting to lift the drum. I have been tasked with determining the required amount of force to lift and upright the tipped drum. Please let me know if you need additional information to assist with my task. Thank you.
Welcome to the PF. :smile:

The force as a function of tilt angle is pretty easy to calculate, and we can help you with that. However, keep in mind that we do not support accident investigations or reconstructions. That is for your local expert to support.

What math have you had so far? Have you had trigonometry? Do you know how to calculate the sum of torques on a rotating object?

Also, we prefer that attachments not be in Word or Excel type formats -- that can lead to security issues. It's better if you convert them to PDF format (using a free utility like PrimoPDF) before attaching them to posts. Thanks.
 
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We need dimensions for thee drum and need to know if it is full. If not full, what portion is filled?
 
  • #4
Dr.D said:
We need dimensions for thee drum and need to know if it is full. If not full, what portion is filled?
I think the dimensions are pretty standard. I'd also assumed it was mostly uniformly filled, but it's still a good question for the OP.

1565374311950.png
 
  • #5
Ullage is important. How full is the drum?
Does the drum contain a liquid, a viscous syrup, or a granular material?
If granules, what is the rest angle of the surface?

To stand a heavy drum by hand it is easiest to roll one end up a ramp. But that cannot be done easily with a drum that is partly filled with a granular material. A drum handling trolley capable of standing a fallen drum should be provided.

The easiest drum (by weight) to stand is one partly filled with a liquid. The centre of mass of the contents initially moves very rapidly towards the base of the drum as the drum begins to tilt. That experience can give a worker a false impression of the effort that will be needed later to stand a full drum.
The position of the centre of mass, of a truncated cylindrical wedge is an interesting (solved) problem.
 
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Worst case, full drum, laying on a horizontal surface:
  • the Center-Of-Gravity (COG) is at the physical center of the drum
  • therefore, the Torque needed will be = {weight of drum} x 1/2 {length of drum}
  • lifting from one end gives you a lever arm twice the length of the lever from the COG
  • therefore the force needed to lift one end of the drum will be 1/2 {weight of drum}
Note that {weight of drum} is the gross weight, not net weight.

Ouch!

Cheers,
Tom
 
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  • #7
MacMurchyEHS said:
Summary: 55 gallon drum weighing 518 pounds is tipped over. How much force is needed to upright the tipped drum?
I am an environmental, health and safety professional...
I have been tasked with determining the required amount of force to lift and upright the tipped drum
That's it?
Calculating , knowing the Cg of the drum, the vertical lifting force is fairly simple, but drums are complicated creatures. They are cylindrical with a round surface acting as a pivot point, and inadequate lifting grip points.

1. the drum will have a tendency to roll on its rounded pivot point touching the floor. In addition to the lifting force, the "employee" will in addition have to provide lateral forces to ensure a sideways stability, which he/she may not be able to adequately provide due to insufficient availability of strength of the employee if all muscle power is being used for lifting,
2. the drum will have a tendency to slip forward on the floor at some angle of lift, if the static friction is inadequate, or the drum is not blocked from moving forward, or if the employee does not compensate for this effect ( very difficult to compensate ), or a combination of all of the above. A polished concrete floor is an example where "all of the above" work against the drum being able to be righted, and at some point of lift the drum will just attempt to slide out from under you.

One way out is to lay a board near the pivot point so that when lifting the drum bottom surface has two points of contact to provide lateral stability; and if the friction between board and floor is enough, reduce or eliminate tendency of the drum to slide forward. The downside is that one is now lifting up and over a curb - in essence the whole mass of the drum has to lifted upwards the height of the curb.

Reason I mention this is that for future considerations, if you build a machine to right a drum, or do a write-up on how to right a drum, these items have to be taken into account for the robustness and safety of operation of the machine.
 

1. What is the definition of force?

Force is a physical quantity that describes the strength or intensity of a push or pull. It is measured in units of newtons (N) in the metric system.

2. How is the force required to upright a tipped drum calculated?

The force required to upright a tipped drum can be calculated using the formula F = m x a, where F is force, m is mass, and a is acceleration. The mass of the drum and the acceleration due to gravity must be known in order to calculate the force.

3. What factors affect the force required to upright a tipped drum?

The main factors that affect the force required to upright a tipped drum are the mass of the drum, the angle at which it is tipped, and the coefficient of friction between the drum and the surface it is resting on. Other factors such as the shape and size of the drum may also play a role.

4. Can the force required to upright a tipped drum be decreased?

Yes, the force required to upright a tipped drum can be decreased by reducing the mass of the drum, decreasing the angle at which it is tipped, or increasing the coefficient of friction between the drum and the surface it is resting on. Additionally, using tools or equipment such as levers or pulleys can also help decrease the required force.

5. How does the force required to upright a tipped drum relate to the principle of conservation of energy?

The force required to upright a tipped drum is related to the principle of conservation of energy in that the work done to upright the drum is equal to the change in potential energy. This means that the force required to upright the drum is dependent on the height at which it is tipped and the mass of the drum, as both of these factors contribute to the potential energy of the drum.

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