Inclined Plane Forces - with a twist

[Spencer25]

Homework Statement

Please see attached photo

Relevant Equations

Please see attached photo

Hi Smart People, I have spent a lot of time on this and have looked in Reed's and Wallker's applied mechanics but can't find a clear explanation. Calculating the force required when pulling up an inclined plane whether horizontally to the plane or at an angle seems pretty straight forward. But, if dragging the object down the plane how would the answer change...not sure what to do with the friction angle. Does the formula shown work in this case...also, not sure if the "absolute" sign is required. Thanks a bunch everyone.

ps...what if the friction angle is greater than the ramp angle?

 

[Spencer25]

Photo attached

 

[haruspex]

Looks like the angle φ is the angle the rope makes to the slope. That was not clear.

I don't get the sign switch in the denominator. Seems to me it is always the difference between φ and α. According to the given answer, when pulling downslope the force needed is minimised when φ is -α, which makes no sense.

Regarding the absolute sign, consider the frictionless case. Does the given formula make sense?

And I would never suggest memorizing such a formula. Too specialised for my taste. But then, I always found it much easier to derive formulas as necessary than to remember them.