Rope and Block Question #1
 

A load on the line is 1000 lbs.

You have just lead a line through a block, around
an angle of 180 degrees. What is the load on the
sheave? [1 pt]

For an angle of 90 degrees, what would be the load on
the sheave? [1 pt]

At what angle does the sheave load match the line load?
[1 pt]

Rope and Block Question #1
 

Bart wrote:

A load on the line is 1000 lbs.

You have just lead a line through a block, around
an angle of 180 degrees. What is the load on the
sheave? [1 pt]

2000 lbs.

For an angle of 90 degrees, what would be the load on
the sheave? [1 pt]

2 x load x cos(angle/2) = 2000 cos(45) = 2000(.707) = 1,414 pounds

At what angle does the sheave load match the line load?
[1 pt]

Load = 2 x load x cos(angle/2)

cos(angle/2) = 1/2

angle = 2 inv_cos(1/2) = 2 * 60 = 120 degrees or pi/3 radians.


//Walt

Rope and Block Question #1
 

Bart wrote:

A load on the line is 1000 lbs.

You have just lead a line through a block, around
an angle of 180 degrees. What is the load on the
sheave? [1 pt]

2000 lbs

For an angle of 90 degrees, what would be the load on
the sheave? [1 pt]

2000*sin(45)=1414 1bs @ 45 degrees.

At what angle does the sheave load match the line load?
[1 pt]

arcsin(1000/2000)=30 degrees (actual angle of line-block-line= 150
degrees)

Cheers
Marty

Rope and Block Question #1
 

Martin Baxter wrote:

Bart wrote:

A load on the line is 1000 lbs.

You have just lead a line through a block, around
an angle of 180 degrees. What is the load on the
sheave? [1 pt]

2000 lbs

For an angle of 90 degrees, what would be the load on
the sheave? [1 pt]

2000*sin(45)=1414 1bs @ 45 degrees. (Note: angle of line-block-line = 135 degrees)

At what angle does the sheave load match the line load?
[1 pt]

arcsin(1000/2000)=30 degrees (actual angle of line-block-line= 150
degrees)

Cheers
Marty

Rope and Block Question #1
 

Martin Baxter wrote:

Bart wrote:
At what angle does the sheave load match the line load?
[1 pt]

arcsin(1000/2000)=30 degrees (actual angle of line-block-line= 150
degrees)

Due to symmetry, there's a 30 degree angle on both sides, so, angle of
line-block-line = 120.

//Walt

Rope and Block Question #1
 

At what angle does the sheave load match the line load?
[1 pt]


Martin Baxter wrote:
arcsin(1000/2000)=30 degrees (actual angle of line-block-line= 150
degrees)


30 degrees from what?

Seems to me that the running part would have to be above
horizontal for the load on the sheave to match the load on
the line.

DSK

Rope and Block Question #1
 

DSK wrote:
At what angle does the sheave load match the line load?
[1 pt]


Martin Baxter wrote:
arcsin(1000/2000)=30 degrees (actual angle of line-block-line= 150
degrees)


30 degrees from what?

30 degrees above the horizontal. That's the point where the vector
parallelograms form nice equilateral triangles.



Seems to me that the running part would have to be above horizontal for
the load on the sheave to match the load on the line.

DSK

Rope and Block Question #1
 

DSK wrote:

At what angle does the sheave load match the line load?
[1 pt]

Seems to me that the running part would have to be above horizontal for
the load on the sheave to match the load on the line.

Huh? Imagine a horizontal line with 1000 lbs of tension on it. It runs
through a block, but the block doesn't deflect the line at all. What's
the force on the block? Zero, right?

Now pull up on the block. As the angle of deflection increases, so does
the load on the block. Eventually it becomes twice the tension (assuming
the tension doesn't change as you change the delfection angle).
Somewhere in between 0 and 2T it equals T.

BTW, I love topological proofs, in case you haven't noticed already.

Or just imagine a bridle with a traveler block. If the angle of the
bridle is 120 degrees, the force on the block is equal to the tension on
the bridle.

//Walt

Rope and Block Question #1
 

3 points to Walt.

My answer was 60 degrees on the third one, but that's
120 looking at it from the other side.


"Walt" wrote
Bart wrote:

A load on the line is 1000 lbs.

You have just lead a line through a block, around
an angle of 180 degrees. What is the load on the
sheave? [1 pt]

2000 lbs.

For an angle of 90 degrees, what would be the load on
the sheave? [1 pt]

2 x load x cos(angle/2) = 2000 cos(45) = 2000(.707) = 1,414 pounds

At what angle does the sheave load match the line load?
[1 pt]

Load = 2 x load x cos(angle/2)

cos(angle/2) = 1/2

angle = 2 inv_cos(1/2) = 2 * 60 = 120 degrees or pi/3 radians.


//Walt

Rope and Block Question #1
 

Did you study Mechanical Engineering Walt?

Rope and Block Question #1
 

Bart wrote:

Did you study Mechanical Engineering Walt?

No. My degrees are in mathemetics, but I used to teach the budding gear
heads & cement heads this stuff. Or try to anyway - most of them seemed
to just want to memorize formulas.


//Walt

Rope and Block Question #1
 

Seems to me that the running part would have to be above horizontal
for the load on the sheave to match the load on the line.


Walt wrote:
Huh? Imagine a horizontal line with 1000 lbs of tension on it.

Sure. It wouldn't be straight though, it would be a catenary ;)


.... It runs
through a block, but the block doesn't deflect the line at all. What's
the force on the block? Zero, right?



Now pull up on the block. As the angle of deflection increases, so does
the load on the block. Eventually it becomes twice the tension (assuming
the tension doesn't change as you change the delfection angle).
Somewhere in between 0 and 2T it equals T.

BTW, I love topological proofs, in case you haven't noticed already.


But that doesn't prove anything, because that's not the
proposed situation. In your proof, if the angle is less than
90, then I'm right.

If the runninag part of the line is below the horizontal,
then the load on the sheave is going to be greater than 1000
lbs because the block is not only supporting the weight but
also opposing the pull of the running part.

Try your proof the other way around, in a way that is
analogous to the situation Bart proposes: Imagine a 1000 lb
weight hanging straight down on a line. Tension = 1000 lb.
Now put a pulley on the line, with no deflection. Load on
the sheave is zero. Now slowly move the pulley sideways & up
such that the angle increases from 0 to 180. Tension on both
parts of the line will always be 1000 lbs but load on the
block will slowly increase from 0 to 2000 lbs (2*T).

You already showed that at 90, load on the block is greater
than 1T; therefor the angle at which load on the block
equals T must be less than 90 ie running part above horizontal.




Or just imagine a bridle with a traveler block. If the angle of the
bridle is 120 degrees, the force on the block is equal to the tension on
the bridle.


And what is the leach tension? Does it affect the jib luff?

DSK