- #1
physiker99
- 36
- 0
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.)
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.)
Last edited by a moderator: