Weighing a package

[jnglmassiv]

It’s common to weigh trucks and train cars one axle at a time. However, this only works since the scale is relatively low and the weight is bearing down on the platform. If the weight were applied at an angle, the load cells flex in a somewhat sideways direction which will give a false reading. A major design consideration for portable axle weighers is to keep them low in profile.

[PastTense]

HoneyBadgerDC:

Frankly if you are shipping any significant number of packages you should get a scale and buy postage online–thus saving time at the post office.

For example you can buy UltraShip 55 lb. Digital Postal Shipping & Kitchen Scale for about $36 at Amazon or elsewhere.

You can buy postage at:
Shipeasy:
https://app.shippingeasy.com/orders
Paypal:

Log in to your PayPal account

U.S. Postal Service:
https://cns.usps.com/labelInformation.shtml

Shipeasy and Paypal have a 5% discount on US. Postal costs for some services. Shipeasy is free up to 50 packages a month (excluding the postal cost) and Paypal is free (excluding the postal cost). They have various differences in what shipping options they offer. So I use a combination.

[HoneyBadgerDC]

@PastTense

HoneyBadgerDC:

Frankly if you are shipping any significant number of packages you should get a scale and buy postage online–thus saving time at the post office.

For example you can buy UltraShip 55 lb. Digital Postal Shipping & Kitchen Scale for about $36 at Amazon or elsewhere.

You can buy postage at:
Shipeasy:
ShippingEasy
Paypal:
Log in to your PayPal account
U.S. Postal Service:
https://cns.usps.com/labelInformation.shtml

Shipeasy and Paypal have a 5% discount on US. Postal costs for some services. Shipeasy is free up to 50 packages a month (excluding the postal cost) and Paypal is free (excluding the postal cost). They have various differences in what shipping options they offer. So I use a combination.

I currently have everything at my house and do my own labels and pay on line. I was actually going back a couple of years but didn’t mention that because I was more concerned with talking about the weighing method and what her and I used to go through.

[ftg]

@HoneyBadgerDC

No it would be more like putting you toes on one end and supporting your head on the other end and then reversing the process.

If the two measurements are not equal (which is extremely likely), then you have more Math to do other than just addition. As well as some more measuring.

This is a standard example in Physics and a good test question.

[Irishman]

@HoneyBadgerDC

On a side note just for the sake of math and nothing practical lets suppose that you had a six foot long 4" box and the balance point was 12" toward one side. For some reason you were only allowed to weigh on one end. Basing it on the balance point you could determine at what point to support one end either at the hand or the scale or the angle to hold it at that would make both sides equal. What would be the formula for figuring either of these two ways?

Using sum of forces and sum of moments equations. If you hold the package level, it is simpler.

Let Fm = Force measured on scale
Fh = Force holding with hand
L = length of box (6 ft in your example)
s = scale support width
d = offset of center of weight from center of length (12 inch in your example)
h = where you hold it (measured from far end of box, not scale end)
Align box end on scale with edge of scale.
Fm + Fh = 0, ergo Fh = -Fm

Fm (s/2) + Fh (L - h) = 0

x = L/2 - d - h

substituting

Fh = - (1/2)Fm / (L - h)

If you hold at an angle, there will be some trigonometry involved.

[HoneyBadgerDC]

@Irishman

Using sum of forces and sum of moments equations. If you hold the package level, it is simpler.

Let Fm = Force measured on scale
Fh = Force holding with hand
L = length of box (6 ft in your example)
s = scale support width
d = offset of center of weight from center of length (12 inch in your example)
h = where you hold it (measured from far end of box, not scale end)
Align box end on scale with edge of scale.
Fm + Fh = 0, ergo Fh = -Fm

Fm (s/2) + Fh (L - h) = 0

x = L/2 - d - h

substituting

Fh = - (1/2)Fm / (L - h)

If you hold at an angle, there will be some trigonometry involved.

Great answer!