Solve dp/dt Velocity Problem: rho, v, g, f, u

[physiker99]

Object’s above area is S. Dust particles with density rho and velocity v fall over it. Fallen particles on object leave object and fall to the ground. Gravity is g. What is the velocity of the object, u? (Friction coefficient between ground and the object is f)

http://img411.imageshack.us/img411/4336/adszqv0.jpg

Solution is as since there is no acceleration, net force is zero.
Then G+N+ F(friction) + dp/dt = 0

Then x component of dp/dt is calculated as followed: (dpx/dt)

Ffriction = fN = dpx/dt = J * v(relative) = J * (v. Cos(theta) – u) (A)

(dpy/dt > y component of dp/dt)

N = mg + dpy/dt = J * (v*sin(theta)) (B)

J (mass of dusts fall in unit time) = S * rho * v(sin(theta))

Then solver makes derives simple equations from A and B and finds u.
My questioon is, how dp/dt is used. I guess it must be derived, but it is not derived and m*v is directly written.

If anyone understood the solution, please let me know. (Solution is correct by the way, it is taken from a book.)

 

[Redbelly98]

Object’s above area is S. Dust particles with density rho and velocity v fall over it. Fallen particles on object leave object and fall to the ground. Gravity is g. What is the velocity of the object, u? (Friction coefficient between ground and the object is f)

http://img411.imageshack.us/img411/4336/adszqv0.jpg

Solution is as since there is no acceleration, net force is zero.
Then G+N+ F(friction) + dp/dt = 0

I am guessing these terms were written as vectors in the book you got it from, rather than as simple scalars as shown here?

Also, "G" is mg, the weight of the object?

Then x component of dp/dt is calculated as followed: (dpx/dt)

Ffriction = fN = dpx/dt = J * v(relative) = J * (v. Cos(theta) – u) (A)

(dpy/dt > y component of dp/dt)

N = mg + dpy/dt = J * (v*sin(theta)) (B)

J (mass of dusts fall in unit time) = S * rho * v(sin(theta))

Then solver makes derives simple equations from A and B and finds u.
My questioon is, how dp/dt is used. I guess it must be derived, but it is not derived and m*v is directly written.

The velocity change can be calculated from:

• The dust has velocity v, at angle θ, before "colliding" with the object.
• After collison, the dust has the same velocity as the object.

Besides the velocity change, you need to know how much mass of dust has collided with the top surface S per unit time. This is determined by density and volume. The volume is determined from S and v, θ (the before-collision velocity)

If anyone understood the solution, please let me know. (Solution is correct by the way, it is taken from a book.)

If you have more questions, feel free to ask.

 

[physiker99]

I am guessing these terms were written as vectors in the book

You're right, those were written as vectors. Excuse me.

Also, "G" is mg, the weight of the object?

Right, G is the mg (as far a i remember)

Besides the velocity change, you need to know how much mass of dust has collided with the top surface S per unit time. This is determined by density and volume. The volume is determined from S and v, θ (the before-collision velocity)

How can we use velocity component to calculate mass. Can you explain to me how you use velocity and S to calculate volume? I have difficulty to understand it.

 

[Redbelly98]

How can we use velocity component to calculate mass. Can you explain to me how you use velocity and S to calculate volume? I have difficulty to understand it.

You'll have to use some geometry, and figure out where dust particles are located that would hit the surface within a time Δt later. That will define a volume of space.