I’m replacing the serpentine belt in my 2000 f150, what size is the square driver I need to spring the tensioner i tried a 3/8 and it's to big and a 1/4 is to small.
thanks

 

Try the 3/8" again. Pretty sure thats what size it is. maybee your driver was'nt squared up with the hole, and if you use a ratchet, be sure the ball detent on the ratchet goes in.

 

It's actually a 1/2 inch drive. I don't know why your 3/8 didn't fit. And you'll probably need a breaker bar or a huge ratchet. You have to put a lot of force on it, and a regular ratchet just won't give you the leverage.

 

movement is not the issue i'm not able to get a 3/8 drive ratchet into it.

 

Maybe there's something stuck in it then because it's definitely a 1/2" drive.

And all I was saying is that you'll need a cheater bar or something to help you get leverage once you do get a ratchet in there.

 

http://www.fordf150.net/forums/viewtopic.php?p=340496

 

thanks
i wonder if the 2000 4.6 is different or if something is stuck in it, i bought it used a few years ago.

 

All three engines should be the same. I have a 2002 4.6L and that's how mine worked.

Are you shure you're putting the ratchet into the correct spot?

 

Precisely. OP - I had the exact same truck - I used a long 1/2" breaker bar.

Who knows, though, right?

Pics, maybe - with a 1/2" short extension adjacent to the hole you are using for comparison?

 

Not sure…I assume it would be at the spring portion not at the actual pulley…but I could be wrong.

 

I think it actually is at the pulley. It's been a while though so don't take my word for it.

 

yup....you are correct, i'm all set thanks for your help.