Calculating the translation force on a cam follower

[T=FxR]

Hi all,

I have been trying to solve this problem, I basically have a cam that's pushing up a follower and I was wondering if someone could check my work

So from the diagram below I have a force acting on the follower at a 77.6 deg angle.

So I did Fy = F x Sin(12.4deg) = 1.012 lbf, Then I subtract the weight of the follower from the froce netF= .027lbf - 1.012lbf = .986 lbf

Heres a FBD
http://imageshack.us/a/img594/1393/90053966.jpg

hopefully this makes sense, thanks for checking my work its been a long time since I've had a problem like this so my memory isn't so good. Not sure if I posted this in the right section, thanks again.

 

[ZapperZ]

This has been moved to the HW/Coursework forum even though it isn't formally a HW/Coursework forum. This is because it is of the same TYPE of problem, and there are many excellent people here that will be able to help you with that.

Zz.

 

[aralbrec]

A cam follower follows a prescribed kinematic motion so any net force on the follower is going to have to take that into account.

For example, the vertical acceleration of the follower is determined by the angular rotation and shape of the cam pushing the follower upward.

I'm not an ME so you'll have to think on that and see if it rings any bells but what you've drawn is a follower with a certain net force upward that will cause an upward acceleration that I don't see (from your diagram) will match the cam shape.

 

[CWatters]

Fy = F x Sin(12.4deg) = 1.012 lbf

Looks ok.