How Do I Find h-min in a Box-on-Incline Physics Problem?

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The discussion centers on solving a box-on-incline physics problem involving static equilibrium and the forces acting on the box, including gravity, normal force, friction, and a horizontal force "h." The goal is to determine the minimum horizontal force "h-min" needed to prevent the box from slipping when the incline exceeds a critical angle "theta-s." The user has derived several equations relating these forces but struggles to isolate "h-min." A suggested approach involves substituting previously established equations to simplify and ultimately solve for "h-min." The conversation also touches on a light-hearted comment regarding the spelling of "voila."
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I'm stuck on a seemingly easy box-on-incline physics problem. It's a typical incline with a surface with friction and one box; gravity, the normal force ("n"), the frictional force ("f"), and another force "h" that acts on the box. The "h" force is applied horizontally on the box, from the left. The box is in static equilibrium.

The problem asks us to show that "h-min" = m*g*tan(theta - "theta-s"). "theta-s" is the maximum angle at which the box can remain still; if the "h" force did not exist and the angle of incline goes beyond this point, the box will slip. The "h" force is supposed to keep this box in place, and "h-min" is the minimum amount of force necessary to keep the box from slipping when the angle of incline is beyond "theta-s".

What I've proved so far is that

"h-min" * sin(theta) + m*g*cos(theta) = n

and

"h-min" * cos(theta) + mu*(n) = m*g*sin(theta)

and

mu = tan(theta-s).

But I can't seem to find "h-min" -- all those darned sines and cosines and tangents are getting in my way. What I've figured out so far is that

m*g*tan(theta - "theta-s") = m*g*(tan(theta) - tan("theta-s"))/(1 + tan(theta)*tan("theta-s"))

which is just an application of a simple trig expansion for tangent. I know that I can go on from there, simplifying things and such, but every path I've taken always leads me to a seemingly dead end - I can't find anything left to simplify, but I still haven't proved anything. Can someone help me please?
 
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jahz said:
What I've proved so far is that

"h-min" * sin(theta) + m*g*cos(theta) = n

and

"h-min" * cos(theta) + mu*(n) = m*g*sin(theta)

and

mu = tan(theta-s).
Substituting your first and third equations into the second should bring you to
h_{min}=w\frac{\sin\theta - \tan\theta_s\cos \theta}{\cos\theta+\tan\theta_s\sin\theta}
multiply with
\cos\theta_s
top and bottom and use
-\sin\theta=\sin-\theta
and viola, the demon have been concered - with some help though!
 
andrevh- this is off the point and I hope I'm not being condescending but whenever I see "viola" (or occasionally "wallah") I can't help but wonder whether it is an intentional joke or the person is simply misspelling "voila"!
 
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