# 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

I'm assuming one of your two unknowns is the tension in the rope. Consider applying a constraint such as "The tension is large enough that the boy is on the verge of slipping." Such a condition, along with the boy's force balance in the plane of the ice, will allow you to solve for the tension in terms of the given masses and friction coefficients and the unknown angle. Then you can carry on solving for the angle by analyzing the forces on the box.

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