# How Much Torque is Needed to Lift a 1000 lb Load with a Power Screw?

• orangeincup
In summary: The 1mm per revolution seems to be correct. Now you need to decide if the efficiency is 35% (ie 35% of the power goes into lifting the load) or if that's the torque loss (ie only 65% of the torque is available to lift the load).Once you have made that decision you can calculate the power required to lift the load (ie 4450N at 1mm per second) and then calc the torque required to provide that power (and account for efficiency/losses).In summary, the conversation discusses a power screw with a pitch of 1 mm that is directly coupled to a stepper motor rotating at 60 rpm. The efficiency of the power screw is 35
orangeincup

## Homework Statement

A power screw with a pitch of 1 mm is directly coupled to a stepper motor rotating at 60 rpm. If
the efficiency of the power screw is 35%, what is the stepper motor torque required to raise a
4450 N (1000 lb) load?

P=τ*ω
τ=F*r

## The Attempt at a Solution

τ=4450*(.0005)=2.22 N-m (I assume this is wrong)

The answer is suppose to be 2.023 N-m

You need to use all the given data; the pitch to get mechanical advantage and the efficiency to account for torque loss.

I think torque efficiency is just dividing my torque by .37 once I have solved the torque required? I really have no idea how to get torque though...

Yes, that method will work for the torque loss.

orangeincup said:

τ=4450*(.0005)=2.22 N-m (I assume this is wrong)

Where did 0.0005 come from?

You don't need T=Fr to solve this.

Do you know how to calc. the velocity of the load? via the velocity ratio of a screw thread?

Once you know how fast the load is being lifted you calc. the power required (via P=Fv) then finally the torque.

billy_joule said:
Yes, that method will work for the torque loss.
Where did 0.0005 come from?

You don't need T=Fr to solve this.

Do you know how to calc. the velocity of the load? via the velocity ratio of a screw thread?

Once you know how fast the load is being lifted you calc. the power required (via P=Fv) then finally the torque.

I thought the pitch was diameter, so I divided by 2 to get radius. And no I don't think I know how to do that, I just started power screws and I mostly only know about how gears work.

No, pitch is not diameter.

You need to study power screws before attempting exercises on them.
They are just like an inclined plane which you should have covered in physics.
Good luck!

4450=(2pi*0.35*T)/.001
T=2.028 N-m

Is this correct?

No.
It looks like you are plugging numbers into random equations and hoping for the right answer, that calc. doesn't include the motor speed so how could it be right?

Do you know what pitch is yet? Can you calculate the speed of a power screw from an angular velocity and a pitch?

Do it step by step:

Calc velocity of the load.
calc power required to lift the load at that velocity.
calc torque required (and account for losses) for the calculated power and given angular velocity.

You need to include all units.

billy_joule said:
Do it step by step:

Calc velocity of the load.
calc power required to lift the load at that velocity.
calc torque required (and account for losses) for the calculated power and given angular velocity.

I agree but would suggest breaking as follows...

Calc. velocity of the load.
Calc. power required to lift the load at that velocity.
Calc. the power required after accounting for losses (aka efficiency)
Calc. torque required for the calculated power and given angular velocity.

How do I calculate velocity based off pitch and angular velocity? I can't find any information on it.

Would it just be v=(pitch*angular velocity)/2pi?

v=(.001*6.28)/2*pi
=.001
(.001*4450/.35)/6.28= 2.023

Last edited:
How do I calculate velocity based off pitch and angular velocity? I can't find any information on it.

Would it just be v=(pitch*angular velocity)/2pi?

v=(.001*6.28)/2*pi
=.001
(.001*4450/.35)/6.28= 2.023

No... or at least it might be but I don't get 2.023 as the velocity.

I refer you to Billy's post. Specifically...

It looks like you are plugging numbers into random equations and hoping for the right answer..
and
You need to include all units.

The values in the problem (eg 1mm pitch and 60rpm) appear to have been chosen to make things easy.

For example 60rpm = 1rps. If the pitch is 1mm how far does the load move per revolution or per second?

That is my calculation for Torque. My calculation for velocity is below. I'm not sure if it's right. I'll write it more clearly.

Pitch=.001
60*2pi/60=6.28 angular velocity

Solving for velocity...

V=(.001*6.28)/2*pi
V=.001

orangeincup said:
I'll write it more clearly.

We still don't know what any of those numbers represent...You need to include units.

You have the right number but any professor of mine would give zero marks.
is it 0.001 Ω? furlongs/fortnight? cubits/afternoon? what about all the other figures?

What Billy said. You must state the units.

If the screw is rotating at one revolution per second and the pitch is 1mm then seems rather obvious (eg no need for equations) that the load will move at 1mm per second.

## 1. How does a power screw lift a load?

A power screw works by converting rotational motion into linear motion. When a motor or other power source turns the screw, the threads of the screw engage with the threads of a nut or other object, causing the screw to move in a linear direction. This linear motion can then be used to lift a load.

## 2. What are the advantages of using a power screw to lift a load?

Power screws have several advantages over other types of lifting mechanisms, such as pulleys or hydraulic lifts. They are relatively simple and easy to manufacture, they can lift heavy loads with less force, and they have a high mechanical advantage, meaning they can lift the load with less input force.

## 3. What factors affect the lifting capacity of a power screw?

The lifting capacity of a power screw is affected by several factors, including the pitch of the threads, the diameter of the screw, the friction between the threads, and the efficiency of the power source. In general, a screw with a larger diameter and finer threads will have a higher lifting capacity.

## 4. Can a power screw be used to lower a load?

Yes, a power screw can be used to lower a load by reversing the direction of the motor or other power source. When the screw rotates in the opposite direction, the threads will disengage and the load will be lowered. This is a useful feature for precise and controlled lowering of heavy objects.

## 5. What are some common applications of power screws in lifting systems?

Power screws are commonly used in various lifting systems, such as scissor lifts, car jacks, and elevators. They are also used in industrial machinery for moving heavy objects, and in robotics and automation for precise positioning of components. Power screws are versatile and can be found in various industries, from construction to manufacturing to healthcare.

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