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

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SUMMARY

The discussion focuses on calculating the minimum horizontal force, "h-min", required to keep a box stationary on an inclined plane with friction. The established formula for "h-min" is derived as h-min = w * (sin(theta) - tan(theta-s) * cos(theta)) / (cos(theta) + tan(theta-s) * sin(theta)), where "theta-s" is the maximum angle of static friction. The user has successfully formulated equations involving the normal force and friction but struggles to simplify them to find "h-min". The conversation highlights the application of trigonometric identities in solving static equilibrium problems.

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  • Concept of friction and its coefficients
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jahz
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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
[tex]h_{min}=w\frac{\sin\theta - \tan\theta_s\cos \theta}{\cos\theta+\tan\theta_s\sin\theta}[/tex]
multiply with
[tex]\cos\theta_s[/tex]
top and bottom and use
[tex]-\sin\theta=\sin-\theta[/tex]
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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