- #1

SevenToFive

- 56

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I am lost on how to calculate the torque needed to pull this cart up their little incline.

Thanks to those who can guide me in the right direction.

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- Thread starter SevenToFive
- Start date

- #1

SevenToFive

- 56

- 3

I am lost on how to calculate the torque needed to pull this cart up their little incline.

Thanks to those who can guide me in the right direction.

- #2

billy_joule

Science Advisor

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- #3

benny_91

- 47

- 5

It would be much easier to help you if you can post a picture or diagram of your setup with necessary dimensions like the diameter of the pulleys etc.

I am lost on how to calculate the torque needed to pull this cart up their little incline.

Thanks to those who can guide me in the right direction.

- #4

- 2,704

- 5,621

The pulling force ##F## needed will the part due to gravity (##mg \sin \theta##) and the rolling friction (##f_r mg \cos \theta##), or:

[tex]F = mg (\sin\theta+f_r\cos\theta)[/tex]

Where:

##mg## = cart weight

##\theta## = slope angle

##f_r## = rolling friction coefficient (0.002 for steel wheels on steel rails)

So putting your numbers in, you need to pull 7132 lb ##= 80 000\times (\sin (5) + 0.002 \cos (5))##.

The power needed is the force times the speed. With the force ##F## in [lb] and the velocity ##v## in [ft/s], you get the power ##P## in [hp]:

[tex]P = \frac{Fv}{550}[/tex]

For example, if you need to pull the cart 30 feet in 60 s, then your average speed is 0.5 ft/s (= 30 / 60), thus you

For torque & rpm, it is only a matter of ratio. For example, if your motor produces 6.5 hp at 1750 rpm, then it produces 19.5 lb.ft at that rpm (= 5252 X 6.5 / 1750; 5252 is a to keep consistency between units like 550 in the previous equation).

The pulley radius needs to be 0.0027 ft (= 19.5 / 7132) or a diameter of 0.065" to produce the required force and speed. It is of course a ridiculous size, which means you need a mechanical advantage of some sort like gears or pulleys. If your actual pulley diameter is 1.95", then you need to slow down the motion with a pulley ratio of 30:1 (= 1.95 / 0.065) between the shaft pulley and your final pulley where the cable is connected.

- #5

SevenToFive

- 56

- 3

Again, thanks for all of the replies.

- #6

SevenToFive

- 56

- 3

Wouldn't mg be equal to 8000lbs x 32.2ft/s, so 22964.6lbs of force? Which would be the load on the winch correct?

For power the customer is already using a 1HP motor with a 60:1 gear reducer along with two 3" diameter pulleys. These pulleys have 5 wraps around them, which should give some mechanical advantage. How do I calculate the mechanical advantage, and whether or not it can pull up the load?

Thanks to those who reply, your help is greatly appreciated.

- #7

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No. 'lb' is already a unit of force. In the US system, we measure aWouldn't mg be equal to 8000lbs x 32.2ft/s, so 22964.6lbs of force? Which would be the load on the winch correct?

By definition, 1 lbf = 1 slug.ft/s². So the correct way to use ##W = mg## with your numbers is to find the mass in slug which is 8000 lbf / (32.2 ft/s²) = 248.45 slug.

By definition, 1 lbf = 32.2 lbm.ft/s². This means that 1 slug = 32.2 lbm, such that 248.45 slug X 32.2 lbm/slug = 8000 lbm. So you can see that 8000 lbf is equivalent to 8000 lbm under the Earth acceleration. This way, there is no need to do a conversion (at least numerically).

As drawn, this configuration doesn't gives a mechanical advantage (if both pulleys are of the same diameter). It only reduces the chance of belt slippage and/or breakage.These pulleys have 5 wraps around them, which should give some mechanical advantage. How do I calculate the mechanical advantage, and whether or not it can pull up the load?

Your mechanical advantage is in the gear reducer only, thus 60:1.

So - if I correct the error in my previous post where I used 80 000 lb instead of 8 000 lb - you need 713 lb. With 1 hp, you can pull at speed v = 550 X 1 hp / 713 lb = 0.77 ft/s. It will probably be a little less as there will be some power losses in the transmission components.

The relationship between velocity (in ft/s) and rpm is:

[tex]v = \frac{rpm \times d}{229.2}[/tex]

Where ##d## is the diameter of the pulley in inches.

So the rpm of 3" pulley must be 58.8 rpm (= 229.2 * 0.77 / 3). Multiply by your mechanical advantage to get the motor rpm and you get 3528 rpm. That should be your motor rpm.

To check, the motor torque should be 1.49 lb.ft (= 5252 * 1 hp / 3528 rpm) or 17.86 lb.in. After the gear reducer, you get 1071.6 lb.in (= 60* 17.86). With the 3" pulley (1.5" radius) you get a pulling force 714 lb (= 1071.6 / 1.5). The error is just due to rounding errors.

3528 rpm sounds about twice as much as a typical electric motor. Either I made a mistake in my conversion, or you have a supplemental mechanical of 2:1 somewhere. From your drawing, it is still not clear how you transform the rotation into linear motion. What is the diameter of the pulley at #2 where the cable winds up?

I assumed it was 3" like the ones with 5 wraps around them. If it was 6", it would reduce the rpm by 2 (1765 rpm). Which implies that the pulley torque is also twice as much.

- #8

SevenToFive

- 56

- 3

I assumed it was 3" like the ones with 5 wraps around them. If it was 6", it would reduce the rpm by 2 (1765 rpm). Which implies that the pulley torque is also twice as much.[/QUOTE]

Correct both pulleys are 3" in diameter, the cable starts on pulley 2 and then goes back and forth to pulley 1 for a total of 5 wraps. At the last wrap of pulley 1 the cable goes to the cart, but rides over pulley 2 not under in my drawing. Thanks for all of your time and effort helping me out with this, it is greatly appreciated.

- #9

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So basically, the two 3" pulleys make a «flat» drum to store the cable as it winds up. In that case, you cannot say you have «5 wraps», as you will have more and more as it winds up (or less and less as it unwinds). The number of wraps is then irrelevant to mechanical advantage and my previous calculations would stand. The key equations are:

At the motor output:

[tex]P = \frac{T \times rpm}{5252.11}[/tex]

After the gear reducer:

[tex]T_{out} = T_{in} \times GR[/tex]

[tex]rpm_{out} = \frac{rpm_{in}} {GR}[/tex]

At the pulley:

[tex]v = \frac{rpm \times d}{229.183}[/tex]

[tex]F = \frac{24 T}{d}[/tex]

[tex]P = \frac{Fv}{550}[/tex]

Where:

If the tension in the cable is greater than the load it pulls, then the cart will accelerate.

At the motor output:

[tex]P = \frac{T \times rpm}{5252.11}[/tex]

After the gear reducer:

[tex]T_{out} = T_{in} \times GR[/tex]

[tex]rpm_{out} = \frac{rpm_{in}} {GR}[/tex]

At the pulley:

[tex]v = \frac{rpm \times d}{229.183}[/tex]

[tex]F = \frac{24 T}{d}[/tex]

[tex]P = \frac{Fv}{550}[/tex]

Where:

- ##P## is power in hp;
- ##T## is torque in lb.ft;
- ##rpm## is angular velocity in rpm;
- ##GR## is the gear ratio of the gear reducer;
- ##v## is the linear velocity of the cable in ft/s;
- ##d## is the diameter of the pulley in inches;
- ##F## is the tension in the cable in lb.

If the tension in the cable is greater than the load it pulls, then the cart will accelerate.

Last edited:

- #10

SevenToFive

- 56

- 3

Thank you very much Jack Action for all of your time explaining this to me.

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