Is there a mechanical engineer in the house?

[justsomeguy]

I know there's at least one, tkanzler be his name. It's kind of related to the mast raising quandry thread, in which he was so kind to help with the physics. In the attached drawings, a line is threaded through a pulley, and equal tension is present on each side of the pully in each drawing. Horse sense (if I've still got any) tells me that the pulley will feel the most "pull" in the first case, ramping down in the second, even less in the third, to an assumed nil when the lines of force are equally opposed in a straight line. The gist of my question is to find a formula, in layman's terms, to quantify the forces acting on the pulley's mounting surface. Thanks in advance. Tom

 

[Stu Jackson]

In this case, you don't need a mechanical engineer (I are one, anywaze...) but wikipedia or any good block & tackles source like Harken should be your friend. Do a google search on blocks & tackles. What you're looking for is called the "mechanical advantage."

 

[Randy6414]

Your horse sense is correct. The most pull will be felt when the lines are paralleled, and the least when the line is straight with opposing pulls.

As for a formula, the first example is easy--the total force is 2x the force on one leg of the line.

The others can be calculated using failry simple trigonometry. It's been years since I used trig regularly, so the actual calculations escape me at the moment. Stu's ideas of Wikipedia or the Harken reference are both good ones.

Randy

 

[Marc14857]

Stu

I think you are ultimately asking what is the force in the vertical direction that is parallel to the forces in your first picture. If that is indeed what you are asking you just have to break down the forces into the vertical and horizontal components. I can draw a sketch later this evening but the formular is simplly the sine of the angle of the line to the pulley equals the vertical force times the the overall force applied via the line. Remember high school sine equals opposite over hypoteneuse. Just solve the formular Vertical force = overall force applied to the line x Sine of the angle of the line to the pulley.

Marc

 

[tkanzler]

Take the included angle (between the two legs of the line), and divide by two. Now you're dealing with just one side.

For the force component acting upward (towards the top of the page) in your sketch, as would be experienced by the block attachment or sheave pin, multiply the line force by the cosine of the angle (or half-angle, since you divided by two). That will be the force in the upward direction as contributed by that half of the line (one side only is being considered here).

Twice that is the total force in the upward direction, since there are two halves to the line, and since you're assuming (as stated) that there is no friction in the sheave, the two sides will just be mirror images.

If the included angle between the lines is zero degrees, then the half angle is zero, the cosine of zero is 1, and the line force times 1 is the line force. Twice that is the force 'felt' by the sheave or block connection.

At 90 degrees between them, the included half-angle is 45 deg., cosine 45 is .707, for the vertical component is .707 x line force x 2 = 1.41, or 141% of the pull on the line.

If the included angle is 180 degrees, the half angle is 90 degrees, the cosine of 90 is 0, so the vertical force at the block attachment is also 0, times two is still zero. This is consistent with what you stated in that the two lines simply oppose each other, and since the block/sheave is not deflecting the line, there is no force.

Block/sheave reaction force will never be more than twice the line force (with both pulling vertical in the diagram), and never less than zero (both pulling horizontally in the diagram), which makes sense since the cosine of the angle varies between 0 and 1 (not counting when it goes negative).

For the horizontal component of the line force, multiply by the sine of the half angle. The two sides (left and right) will be equal, of course, but since they're acting in opposite directions, they cancel.

HTH

 

[Happy Dolphin]

Try this block and tackle site - http://en.wikipedia.org/wiki/File:Four_pulleys.svg
In the site - you will see the force is equal for a simple pulley, adding additional pulleys (or blocks) will reduce the necessary force to move or lift an object (at the rate indicated)...
As others had mentioned, your question may be more of a TRIG formula. Angles and the force needed to move (or raise) your mast?

 

[Ross]

If the arrows represent the force on the ropes and the lift is vertical then the load on the pulley will be constant but the force on the ropes will increase to approach infinity when the rope approaches horizontal. The typical example is the brick on a clothesline.

 

[justsomeguy]

Stu

I think you are ultimately asking what is the force in the vertical direction that is parallel to the forces in your first picture.
Marc

snip

That's right. I've updated the drawings, but still haven't done the math!

 

[Marc14857]

Tkanzler

Maybe we are saying the same thing but wouldn't the horizontal force on each leg be the sine of 1/2 the angle times the hypotenuse i.e. sine = opposite over hypotenuse?

 

[tkanzler]

Sure is. I'm not sure where the confusion is, but it's probably in my description. So here are some diagrams. Life is much easier with diagrams.

In the first pic, the two lines are shown with magnitude Fline. The sheave resultant force is Fsheave. The half-angle is simply A.

The other diagram is the force vectors broken into horizontal (Fline_x) and vertical (Fline_y) components. It's obvious that the Fline_x component vectors on the left and right are equal and opposite, as you would expect.

The vertical components both act in the same direction, and are therefore additive. They're balanced by Fsheave, which must also be equal and opposite to the sum of the two Fline_y vectors (red and green) for static equilibrium to exist (nothing's accelerating - moving maybe, but not accelerating).

To calculate the vertical or horizontal component of the red or green Fline vector, just multiply Fline by the sine or cosine of the angle A as it's drawn, as appropriate. Since they're both (red and green) the same magnitude, you simply double Fline_y to get the total vertical reaction at Fsheave.

Example 1:

A = 60 deg.
Fline = 100 lb

Fline_y = 100 lb x cosine 60 deg = 50 lb
Fsheave = 2 x 50 lb = 100 lb

Example 2:

A = 80 deg
Fline = 100 lb

Fline_y = 100 lb x cosine 80 deg = 17 lb
Fsheave = 34 lb (since there are two vectors, equal in magnitude, and in the same direction)
Fline_x = 100 lb x sine 80 deg = 98 lb (but negated by the Fline_x on the other side)

Conversely, to calculate the line pull for a given sheave force (brick on a clothesline example), divide the sheave force by two and apply

Fline = Fline_y / cosine A
and
Fline = Fline_x / sine A

Fline_y = 50 lb (half a 100 lb brick)
A = 80 degrees

So Fline = Fline_y / cosine 80 deg = 288 lb

Obviously, as the angle A gets larger (line gets flatter, a la clothesline), the resulting force in the line grows rapidly. Eventually, at 90 deg, you're dividing by zero, which means the force becomes infinite (apologies to the mathematicians out there).

In the bad old days, we'd use a slide rule or trig tables to look up values for sine and cosine. Don't really miss them.

 

[justsomeguy]

Many thanks to all of you

 

[Marc14857]

tkanzler

You are right and the confusion is on my part. I inappropriately described the angle that I was referring to. In my force diagram the angle was the angle between the line and the horizontal plane which made it a sine function but was the complement to the angle you described. OK now I feel better

Marc

 

[Hermit Scott]

The sum of the forces are zero (assuming the block is stationary). So the two forces going in the Y component are equal to the force in the down (-Y) direction. If you have angles to the two ropes it is only the Y component of the force that is counted.
The simplest example of the sum of forces is a person sitting on a chair. The chair pushes up at exactly the same force as the persons weight pushing down. IF not, the chair would start to rise or sink in to the floor.
I went to school for mech engineering but dropped out ( but still tell everyone I am a mech engineer)

 

[Stu Jackson]

I went to school for mech engineering but dropped out ( but still tell everyone I am a mech engineer)

Geez, NOW ya tell me! Think of all the time in school I coulda saved!

 

[Johnb]

Now you know what the case is for having a little slack in life lines - that permits the line to balance a vertical load with a much lower tension in the line.

 

[Winkfish]

My brain hurts.

 

[Bill Roosa]

OK does the force on the line go to zero or become infinite?
My engineering sense tells me youall are considering two entirely different problems using the same diagram. In one you hold the tension in the rope constant and in the other you hold the force on the sheave constant.
They both get answered with the same math BTW but if you insist on not defining the problem you are going to get wrong answers.

If you let Ft be the tension in the line, Fs be the force on the sheave, and theta be the smaller angle between the two lines then you can say;

2.0*Ft*sin(theta/2)=Fs
with Fs acting on a line that intersects the sheave center and bisects theta and is directed away side theta is on.

or solving Ft,

0.5*Fs/sin(theta/2)=Ft IN ALL CASES.

The force never goes to infinity because that would violate numerous physical and quantum laws. For the engineers among us there is no such thing as a physically real rigid body. EVERYTHING bends so theta cannot ever be pi radians. The other way of looking at it is you cannot place an infinite tension on a line as it will break well before you get anywhere close to that loading.

Engineering is taking the math and using it in the real world. Yes 1/0 is mathematically possible with rigid bodies used for the sake of discussion but you cannot do it in real life. Stuff will break (you can engineer where that happens BTW), bend (that too), twist (...), rupture, crack........ a whole list of things but it will never sustain a load in excess of what its material properties are capable of. The engineering is designing the structure so it will handle the load or limiting the load so as to not exceed the capabilities of the structure. Without knowledge of the magnitude of the loads you are not doing engineering.

 

[Hermit Scott]

I went to school for mech engineering but dropped out ( but still tell everyone I am a mech engineer)

Geez, NOW ya tell me! Think of all the time in school I coulda saved!

And Money.

 

[Ross]

Bill, As I said as the rope approaches a straight line the force will approach infinity. You don't need an BSME to know that, just build a farm fence or string a banjo. No matter how you try to tighten the wire you can still bend it.

 

[Bill Roosa]

Yea Ross I agree.
I just get concerned when folks talk about "goes to infintiy" stuff. Long before "infinity" reality sets in and you have to account for it. Free body diagrams and rigid bodies are only the beginning. Strenght of meterials is where the real fun begins. Ahh the days of plastic hinges and statically indeterminate structures , those were the days.