Help lifting a log

Just a swag, but I think it's 50% until you get to 45 degrees if the lifting is perpendicular to the ground as you raise the load?

Your intuition is good!

If the log has mass "m", then its weight = W = m*g where g =0.81

Let the length of the log = L

If the log's end points are x and y, with you picking up at x, with force F then:

Sum of moments at point y = 0

Thus F*L = mg*[L/2]....the reason, the force exerted by yourself has the full length of the log as the lever arm, whilst the weight acts at midspan, hence its lever arm is half the length.

Then solving for F = mg/2!, as you already knew.

The length of the log obviously effects the weight of the log i.e. longer log = heavier log

This is correct, assuming the log is uniform in diameter and density along its length. Any real log has a wide and a narrow end, so of course it depends which end you lift!

For a uniform log with ends cut exactly square, the greatest force is indeed required just to barely get the end off the ground. As soon as the end is a significant distance off the ground, a bit of the other end of the log is on the other side of the fulcrum (the corner of the log in contact with the ground), helping to lift.

When you lift the log, length L, you are applying a force perpendicular to the axis of the log, which of course is vertical to start with. This force starts out at half the weight of the log. As you lift it, the log pivots around the other end and your end traverses a circle, as you continue to apply force perpendicular to the log. The center of gravity of the log also traverses a circle of radius L/2. As you raise the log through an angle Θ, you raise the center of gravity L/2*sin(Θ), which represents potential energy. Your force applied is the derivative of potential energy with respect to displacement, that is, F = d(L/2*sin(Θ)) / dΘ = L/2*cos(Θ). As Θ approaches 90 degrees, this force approaches 0.

Interestingly enough, as Θ gets larger, less of your force applied is in the vertical direction, so the pivot is assuming even more load. This should sound familiar to anyone who has moved furniture up a stairway and been at the bottom end.

I am just wondering is this just a problem to work on or does it have an application? If it does may we ask what?

Sure. I was dragging some logs out of the woods with my tractor this past weekend and I was slightly dismayed when the skidding winch I bought worked like a charm pulling 2 logs at the same time up to the tractor, but the 3 point hitch was unable to lift the combined weight of the two logs plus the weight of the winch. I am now trying to calculate the load at the winch to compare to the published capacity of the hitch.

Remember the weight of the logs depends a LOT on the moisture content of the logs. If they are already dead trees at cutting time, they are waaay lighter than recently cut live trees.

Only an engineer would go through all this just to figure out the capacity of the hitch, a lumber jack would just gab one less log.

You're right Labyguy, this is the utmost challenge for an engineer. When I purchased the tractor last year, I had poured over volumes of specs from many tractor companies and set up comparative matrices using excel (how else would you do it?). Lifting capacity was an important selection criteria. If the tractor is not up to spec, ie. 1109 lbs 24 aft of hitch, shouldn't I establish this and have corrective action taken while the tractor is still under warranty?

And don't forget that the butt end is quite a bit heavier than the top end.

50% is right. Essentially, you are using a lever.

The center of mass is the center of the log,

so your initial mechanical advantage is 2:1.

You are also right about the decrease in effort as the

force vectors change with the angle.