How to calculate force to 'stand up' an object?

Yes! That DOES make sense. A lot more sense now solving a real world problem than 40 years ago in college.

I knew, intuitively, that it wasn't as heavy as lifting the whole thing off the ground but your "the ground supports the other half" offers a realistic "handle" for future problems.

So, to erect a 1 ton pole from 15 degrees the most I need to lift is 966 lbs. which decreases to zero as it reaches vertical. Cool!

Thank you very much for your quick, clearly stated, freely given knowledge. This attitude is 50% of what makes the internet so great and bodes well for mankind.

Ralph Chandler

Dir Prototype Development

As we later found out, it was not a student's homework! :)

In any case, nice clear answer!

Worked out for the best this time; I'll have to be slightly more cautious in the future!

No! You did good. Please, don't be cautious or hold back your knowledge and experience because of some imagined or possible cheating. People are basically good not bad.

I think killowatt0's comment was snide, rude, suppressive and uncalled for. He probably routinely thinks people are bad and tries to stop them.

You (and Hendrik) think people are good and try to help them.

Had it been a student he'd have had the force formula in his books and you, in effect, simply referred him back to the formula with some clarification.

Thanks again for your help.

RC

What a nice clear answer! Kudos.

I think, there is something wrong.

Please correct me if I am wrong.

Assumption

- The force is applied in vertical direction

- C.G. of the pole is in centre

Equating the moments -

F X l cos α = W X 0.5 X l cos α

So, F = W X 0.5 ------------ 1

or dF = dW X 0.5

W = mgh (Potential energy of mass 'm' at height 'h')

= mg (0.5 X l X sinα)

Since 'α' is changing as you lift the load,

dW = mg (0.5 X l X cosα)

d α

dF = mg (0.5 X l X cosα) X 0.5-----------from equation 1

= mg (0.25 X l X cosα)

"W = mgh (Potential energy of mass 'm' at height 'h')"

Here's the error. Weight is not potential energy; it's just mass times gravity.

Thanks for the spot-check though - you guys help keep me honest.

wt is not potential energy.

There is the mistake .

I should have equated the energy.

E1 = Wgh (Potential energy of mass 'W' at height 'h')

= Wg (0.5 X l X sinα)

Since 'α' is changing as you lift the load,

dE1/dα = Wg (0.5 X l X cosα)--------------------- 1

Similarly,

If E2 is energy used for lifting the load,

dE2/dα = dFg/dα X l -----------------------------2 (thanks for suggestion 'dkwarner')

equating 1 & 2

dFg/dα X l = Wg (0.5 X l X cosα)

So, dF/dα X l = W (0.5 X cosα)

If you follow your assumption in doing a real job, you are working too hard. Any component of force not perpendicular to the centerline of the object ( or maybe the bottom edge of the object would be more accurate) is wasted effort. You are rotating the object. Rotation is caused by a torque - a force times its lever, where the lever is perpendicular to the force. At 45°, the effective lever of a vertical force is only 0.7 of the length of the object.

To say the same thing another way, at 45° only 0.7 of your effort is pointing in the direction required (perpendicular to the length of the object) to rotate the object. The other 0.7 of your effort is pulling along the length of the object, causing no rotation and therefore wasted. Remember that at 45°, 0.7+0.7=1.0 (vector addition)

With the force remaining vertical, at 85° you are exerting 11.5 times more force than necessary! You may as well just lift the thing!

CSM -

Helping a student to grasp the problem is not a bad thing. Mr Chan must just use it in the right way.

If he just copy the work to the reply sheet then it may actually end up in a bridge collapsing or other mishap.

If Mr Chan however read the answer step by step (says mmmm) and store it away in the grey stuff. You have actually achieved something positive.

If Mr Chan now comes back and ask what will happen if the pole is tapered from thick to thiner at the top more students (and part of CR4) may actually learn from it.

Assume you apply the force tangentally. Standing the pole up, you are raising the center of gravity, which is in the center of the pole and traces out a circle with a radius 1/2 the length of the pole. The height of the center of gravity represents the potential energy, and equals 1/2 the length of the pole times the sine of the angle. The force applied is the derivitive of the potential energy (height of the CG) with respect to displacement (angle). Therefore the force is 1/2 * (2.2 lb) * cos(Θ).

Yes. This is what really happens.

You don't keep lifting straight up, you lift tangentially on the top of the pole and walk forward as the top of the pole moves through the arch toward vertical until you're not lifting but pushing.

RC

Thank you all for your kind words.

But, Kilowatt0 is correct. Luckily, it was a legitimate question this time, but I would have done my profession and trade a disservice by blindly answering a student looking for the easy answer. While I will always freely share the fruits of my experience on this forum (I consider this a form of "giving back" to my profession) and with whomever asks, I'll consider this a lesson learned. As I'm sure you are all aware, it can be a thrill to be the first to post an answer to a technical question, especially among this gathering of peers, and I fell into that trap. I will be more aware in the future and make efforts to "qualify" the inquiry before spouting an answer. "A little knowledge is a dangerous thing."

Again, thank you all for your input and support.

Thank you to all for your help.

I have all the answer I need.

RC

Some cautions:

1. Weight is not equivalent to Mass. Sometimes, you can ignore this fact and get away with it. Other times you cannot. For example, you cannot move this pole without accelerating it. Therefore, if you were planning to move the pole with a crane, and you wondered it the crane's cable would break, then you need to considered how quickly the crane can accelerate the mass of the pole. This acceleration in a real-world crane could easily be 1g (or 9.8M/S/S), which would double the instantaneous force. CSM's excellent post gives you the force involved in a static situation. In the real world, things are often not static. There is a difference between supporting a pole in a particular position, and moving a pole through a particular position.

As I'm sure you know, your mass is the same on earth as it is on the moon. Your weights are dramatically different. Closer to earth, the differences become significant in many contexts, an obvious one being flying, in which a utility class airplane (typical Cessna) is certified to fly at 3.8 G, to allow for steep turns, turbulence, mishandling, and the simplest aerobatics. So, if you were designing a mechanism that would have to work on an airplane, you would need to design for forces much higher that those for a mechanism that sat on the ground all day.

2. In the real world, force may not be normal to the pole. The force required to right the pole has been calculated here (by CSM, for example) as a force normal to pole, at its end. In your drawing in which you show pole at 80 degrees, you show the force at about 40 degrees, so that force would not be normal to the pole. In practice, the direction of force may vary quite a lot. In this pic, one arm is very nearly normal, others are not.

In the real world, the force would rarely be normal. (Consider where your arms are when you walk a ladder up.) In the case of the crane I mentioned before, the crane would need to keep backing away from the pole to keep the cable normal.

All very important and accurate qualifications to the original answer. "Score 1 for Good Answer!"

It is maximum force Fmax at zero degree angle.

Fmax = 0.5*P*cos0. (Default object's centre of gravity is mid point of object)

Fmax = 0.5 *1 *1= 0.5 kg.