# Box-on-incline physics problem

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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andrevdh
Homework Helper
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!

HallsofIvy