# Thread: Work, distance, and time

1. ## Work, distance, and time

I have to lift a 3300 lbs cate 25 feet using a horse and a compound pulley with a mechanical advantage of 4. I know W=FD and 1 horse power is 550 ft lbs per sec. I also know the the pulley will make the lift lighter but the horse will have to pull further. I need to find how long it takes for the horse to lift the load. If I haven't made myself clear enough, leave a comment as I will check back frequently this weekend. Thank you!

2. Originally Posted by Shrub
I have to lift a 3300 lbs cate 25 feet using a horse and a compound pulley with a mechanical advantage of 4. I know W=FD and 1 horse power is 550 ft lbs per sec. I also know the the pulley will make the lift lighter but the horse will have to pull further. I need to find how long it takes for the horse to lift the load. If I haven't made myself clear enough, leave a comment as I will check back frequently this weekend. Thank you!
time required = (total work to be done)/(1 horse power)

3. when finding total work, do I use the weight of the load, of do I figure in the mechanical advantage of the pully?

4. Originally Posted by Shrub
when finding total work, do I use the weight of the load, of do I figure in the mechanical advantage of the pully?
you can ignore the MA of the pulley ... the work done is the work done, no matter what machine you use to complete the job.

total work to be done = (3300 lbs)(25 ft) = 82500 ft-lbs

5. You have been extremely helpful.