# Boy and box on ice

I have a troubles. Can you help me? I have a task, called "slippery activity".

Boy (he has $m=45kg$) Stands at ice and tries to move a big box ($M=90kg$) with a string (rope). Boy's Slip ratio is 0.4 and box's is 0.3. With which minimum angle to the horizon Boy must pull the rope for move the box?

When I tried to solve this task, I had equation with 2 unknowns.

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Oh, when T is the power, acting on the rope, I had 7*T*sin(a)=900. But there are 2 unknowns –  KupuJIJI Nov 27 '12 at 14:50

Let's denote:

$\mu_1$ (boy's slip ratio)
$\mu_2$ (box's slip ratio)
$m$ (boy's mass)
$M$ (box's mass)

$F$ (pulling force of the boy)
$\alpha$ (the angle of the rope to the horizon)

Then

1) The box will start to move if the following inequality holds:

$$F\cos \alpha \geqslant \mu_2(Mg-F\sin \alpha)$$

2)The boy does not slide along the ice if the following inequality holds:

$$\mu_1(mg+F\sin \alpha)\geqslant F\cos \alpha$$

Combining both inequalities together we get:

$$\frac{\mu_2 Mg}{\cos \alpha +\mu_2\sin \alpha}\leqslant F \leqslant \frac{\mu_1 mg}{\cos \alpha -\mu_1\sin \alpha}$$

At minimum angle the inequality becomes an equation:

$$\frac{\mu_2 Mg}{\cos \alpha +\mu_2\sin \alpha}=\frac{\mu_1 mg}{\cos \alpha -\mu_1\sin \alpha}$$

A solution of this equation:

$$\alpha_{min}=\arctan \left (\frac{\frac{M}{\mu_1}-\frac{m}{\mu_2}}{M+m}\right)=29.05^{\circ}$$

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I concur with 29.05 –  xxx Nov 28 '12 at 18:52