This isn't a homework problem. I'm asking cause I'm wondering how strong a person would have to be to pull it off. Their arms would be extended, both hands gripping the staff at one end, and the staff is facing directly away from them, parallel to the floor, going through their 'z-axis' as it were (with x being left-right and y being top-bottom).
[Physics] How much force would be necessary to hold a 6-foot steel staff weighing 20 lbs from one end of it, parallel to the floor
forcesstaticstorque
Best Answer
It's a 20-lbs staff. 20-lbs net force are required to hold it up, regardless of its orientation.
If you just apply this force to one end of the staff, though, there would be a net torque.
Instead, you need to use your hands to apply two forces to the staff. One force, exerted by the near hand on the very end of the staff, should be down. The other, exerted by the far hand a few inches down the staff, should be up. You can imagine a see-saw. The far hand is the pivot of the see-saw, which pushes up on the board. The near hand is a very heavy person sitting very near the pivot, and the pole itself is a very light person sitting far from the pivot. (The weight of the pole itself can be considered to be concentrated at the half-way point.)
There are three forces on the pole and hence two distances between them. The ratio of the forces at the extremes needs to be the inverse of the ratio of the distances. That is, if one force is three times further out from the pivot, it should be one-third as large.
Let's say the distance between your hands is four inches, or 1/3 a foot. The distance to the middle of the pole is about 3 feet, or about 10 times as much. So you need to exert a force of about 200 pounds down with your near hand and 220 pounds up with your far hand.