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]

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

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

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

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

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

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

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

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

Did you study Mechanical Engineering Walt?