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A tank (which its area of its corners is negligble) with mass M, moves on a surface with coefficients of friction between the surface and the tank [tex] \mu_{s} \mu_{k}[/tex] (static and kinetic).Rain drops are dropping on the tank with angle [tex]\alpha[/tex] with speed V, the water comes to the tank with constant dm/dt=Q, in t=0 the tank is at rest.

1.what is the condition that the tank will move from its initial place?

2.assume the codnition in 1 is satisfied find the velocity of the tank as a function of time?

3.how much time will elapse until the tank is stopped?

attempt at solution

1. I think the condition is that: [tex]\frac{dp_x}{dt}=\mu_kMg[/tex]

and [tex]\frac{dp_y}{dt}=0[/tex]

2. now when i open these terms i get: [tex]Q*(Vcos(\alpha)-v_x)+(M+Qt)\frac{dv_x}{dt}=\mu_kMg;-Q(Vsin(\alpha)-v_y)+(M+Qt)\frac{dv_y}{dt}=0[/tex] now after integration i get that:

[tex]v_y=(Vsin(\alpha))*(1-(Qt+M)/M);v_x=\frac{(\mu_kMg+QVcos(\alpha))(1-M/(M+Qt))}{Q}[/tex] but i dont see how to get from this when will the tank stop, according to these equations i get that it will not stop, so i guess something is wrong in my reasoning, any hints ,tips, encouragements, are welcomed. (-:

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# Homework Help: A tank of water.

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