Pushing a cylinder over a lip

[bry1]

I am attempting to calculate the force required to roll a large roll of paper over a lip. The roll of paper is resting stationary on the floor and is "chocked" against a raised lip that I am assuming to be square. I realize that I huge factor in this problem is the firmness of the roll but I am going to assume that it does not deflect for now. Any help would be appreciated!
Givens:
Paper roll weight=6000 lbs
Paper roll diameter=50"
Lip height=1"

 

[bigTomHanks]

sum moment = 0

F(49")-W(10")=0

F approximately = 1225 lbs

 

[JStephen]

Sum moments about the lip, assuming cylinder is just lifting off floor. Cylinder weight is known, horizontal force to be found.

If you are actually pushing this with something that is not free to rotate, you would also get some vertical downward force at the push point due to friction.

 

[Cockroach]

Actually this is a very easy problem, typically found in first year static textbooks.

The lip adds a reactionary force to the free body diagram consisting of weight and say a force pulling the body at some angle, not necessarily horizontal. This is the most general of cases.

So use the free body diagram and sum forces in the horizontal and vertical directions. Just prior to movement, sum of the vertical forces go to zero. This enables one to compute the reactionary force of the lip. In the horizontal direction therefore, you can solve for that unknown force required to move the piece. This is essentially solving two equations in two unknowns.

Therefore consider your situation: circular cross section body wedged up to a chalk. A force at angle theta is used on the right hand side to LIFT that body over the lip. The lip causes a reactionary force that resists any attempt of motion, that is the function of the chalk in the first place. Weight of course, acts in the vertical plane downwards.

Compute the angle of action through the centroid of the body as a result of the chalk. Given that the chalk is "h" high and the body to be moves is of circular cross section "r", then the angle of action is "beta".

cos(beta) = (r-h) / r which gives us angle beta.

Imagine a free body diagram with your applied force "F" acting at angle theta from the right hand side, weight "W" vertically downwards at 270 degrees. The reactionary force "R" of the lip acts at beta to the left of weight. Sum the forces.

Sum Fx = F cos(theta) + R cos(90+beta)
Sum Fy = F sin(theta) + R sin(90+beta) - W

Simplification using compound angular relationships in circular measure therefore give:

Sum Fx = F cos(theta) - R sin(beta)
Sum Fy = F sin(theta) - R cos(beta) - W

At motion impending, Sum Fy = 0 so that solving for R:

R = [W - F sin(theta)]/cos(beta) reaction force of chalk

Substitute into sum of horizontal forces that also equal zero at motion impending, the simplifiy:

F cos(theta) - R sin(beta) = 0 for R given above,
F cos(theta) cos(beta) + F sin(theta) sin(beta) = W sin(beta),
F = [W sin(beta)]/[cos(theta) cos(beta) + sin(theta)sin(beta)],
where beta = arccos{(r-h)/r}, r=radius of body, h=chalk height.

In your case you want the force to act strictly horizontal, therefore angle theta is zero. The above equation reduces to F = W tan(beta).

And there you have it.

Kenneth J Hueston, PEng
Principal
Sturni-Hueston Engineering Inc
Edmonton, Alberta Canada