Calculating Tank Velocity and Stopping Time Due to Rain Drops

[MathematicalPhysicist]

Problem stated
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 $$\mu_{s} \mu_{k}$$ (static and kinetic).Rain drops are dropping on the tank with angle $$\alpha$$ 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: $$\frac{dp_x}{dt}=\mu_kMg$$
and $$\frac{dp_y}{dt}=0$$
2. now when i open these terms i get: $$Q*(Vcos(\alpha)-v_x)+(M+Qt)\frac{dv_x}{dt}=\mu_kMg;-Q(Vsin(\alpha)-v_y)+(M+Qt)\frac{dv_y}{dt}=0$$ now after integration i get that:
$$v_y=(Vsin(\alpha))*(1-(Qt+M)/M);v_x=\frac{(\mu_kMg+QVcos(\alpha))(1-M/(M+Qt))}{Q}$$ but i don't 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. (-:

 

[Astronuc]

Problem stated
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 $$\mu_{s} \mu_{k}$$ (static and kinetic).Rain drops are dropping on the tank with angle $$\alpha$$ 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: $$\frac{dp_x}{dt}=\mu_kMg$$
and $$\frac{dp_y}{dt}=0$$
2. now when i open these terms i get: $$Q*(Vcos(\alpha)-v_x)+(M+Qt)\frac{dv_x}{dt}=\mu_kMg;-Q(Vsin(\alpha)-v_y)+(M+Qt)\frac{dv_y}{dt}=0$$ now after integration i get that:
$$v_y=(Vsin(\alpha))*(1-(Qt+M)/M);v_x=\frac{(\mu_kMg+QVcos(\alpha))(1-M/(M+Qt))}{Q}$$ but i don't 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. (-:

The tank will move when the x-component of the momentum exceeds the static friction force on the tank, which is initially at rest $$\mu_sMg$$.

$$\frac{dp_x}{dt} > \mu_sMg$$ and one must find that in terms of Q, V and $\alpha$

Then as it rains 'constantly', the applied force is constant (verify this), BUT the tank is accumulating mass at a rate Q.

Then as the tank accumulates mass, the dynamic friction force is increasing
$$\mu_k(M+Qt)g$$

 

[MathematicalPhysicist]

you mean greater or equals.
so the equation should be, for the change in momentum in the x direction instead of $$\mu_kMg$$ it should be $$\mu_k(M+{Qt)g$$, correct?
btw, what with the change of momentum in the y direction? because of the y component of the speed of the rain drops?

 

[Astronuc]

you mean greater or equals.
so the equation should be, for the change in momentum in the x direction

In the case of starting to move, the force applied must be greater than the static friction force. If the equal to the friction force, the tank still does not move.

instead of $$\mu_kMg$$ it should be $$\mu_k(M+Qt)g$$, correct?

M(t) = M + Qt, so apply M(t) where appropriate.

Starting at rest, M(t) = M.

btw, what with the change of momentum in the y direction? because of the y component of the speed of the rain drops?

Oviously, he momentum of the tank does not change in the y-direction, but the momentum imparted by the rain does produce a vertical force on the tank, which would be as one treats the weight with respect to friction.

 

[MathematicalPhysicist]

so it should be dp_y/dt=M(t)g?

 

[Astronuc]

so it should be dp_y/dt=M(t)g?

No, the tank does not gain momentum in the y-direction, assuming it is on a level (horizontal) surface. The rain imparts a constant force related to QV, where Q = dm/dt. So the normal force on the tank is (M(t) + QV)g to which friction is proportional.

On a horizontal surface v_y of the tank is zero.

 

[MathematicalPhysicist]

so what is the equation of motion?
shouldn't it be: (M(t)g+QV)?
so the equation of motion should be something like this:
$$\mu_k(M(t)g+QVsin(\alpha))=Q(Vcos(\alpha)-v_x)+M(t)\frac{dv}{dt}$$ where $$M(t)=M0+Qt$$, correct or wrong again?