I cannot work out constants a and b --

  • Thread starter Thread starter tosingoodnessamao
  • Start date Start date
  • Tags Tags
    Constants Work
AI Thread Summary
The discussion revolves around solving for constants a and b in the equation v = ay + by^2, given specific velocity values at different distances from a surface. Two equations were established based on the velocities at 1.5 mm and 3.0 mm from the surface, leading to a system of equations with two unknowns. Initially, a mistake was made by incorrectly multiplying b by y instead of y^2. After recognizing this error, the user successfully solved for the unknowns a and b. The conversation emphasizes the importance of correctly setting up equations to find solutions in mathematical problems.
tosingoodnessamao
Messages
3
Reaction score
0
Homework Statement
If the velocity v of the air in a boundary layer having a dynamic viscosity of 18 × 10−6
kg/ms is given in terms of the distance y from the surface by:
v = ay + by^2
where a and b are constants, calculate the surface shear stress if at 1.5 mm from the surface
the velocity is 75 m/s, and at 3.0 mm from the surface it is 105 m/s.

I know that a is the shear strain/rate and i need to times it by the dynamic viscosity but i really cannot work out a and b.

* update - realised that I forgot to square the value for y when multiplying by b.
Relevant Equations
shear stress = (shear strain/rate) x du/dy
75 = (1.5 x 10^-3)a + (1.5 x 10^-3)b
105 = (3 x 10^-3)a + (3 x 10^-3)b
 
Physics news on Phys.org
You are given three things:
(1) v = ay + by^2 where a and b are constants
(2) at 1.5 mm from the surface the velocity is 75 m/s
(3) at 3.0 mm from the surface the velocity is 105 m/s.

From these, you should be able to set up two equations with two unknowns and solve them for a and b. Have you tried that?
 
  • Like
Likes tosingoodnessamao
Two equations with two unknowns. Have you tried solving them?
 
  • Like
Likes tosingoodnessamao
phyzguy said:
You are given three things:
(1) v = ay + by^2 where a and b are constants
(2) at 1.5 mm from the surface the velocity is 75 m/s
(3) at 3.0 mm from the surface the velocity is 105 m/s.

From these, you should be able to set up two equations with two unknowns and solve them for a and b. Have you tried that?
Yes finally found the unnowns a and b because initially I kept multiplying b by y rather than y^2 which was my mistake.
 
Mayhem said:
Two equations with two unknowns. Have you tried solving them?
Yes I have managed to solve them now I recognised my mistake.
 
Thread 'Variable mass system : water sprayed into a moving container'
Starting with the mass considerations #m(t)# is mass of water #M_{c}# mass of container and #M(t)# mass of total system $$M(t) = M_{C} + m(t)$$ $$\Rightarrow \frac{dM(t)}{dt} = \frac{dm(t)}{dt}$$ $$P_i = Mv + u \, dm$$ $$P_f = (M + dm)(v + dv)$$ $$\Delta P = M \, dv + (v - u) \, dm$$ $$F = \frac{dP}{dt} = M \frac{dv}{dt} + (v - u) \frac{dm}{dt}$$ $$F = u \frac{dm}{dt} = \rho A u^2$$ from conservation of momentum , the cannon recoils with the same force which it applies. $$\quad \frac{dm}{dt}...
Thread 'Minimum mass of a block'
Here we know that if block B is going to move up or just be at the verge of moving up ##Mg \sin \theta ## will act downwards and maximum static friction will act downwards ## \mu Mg \cos \theta ## Now what im confused by is how will we know " how quickly" block B reaches its maximum static friction value without any numbers, the suggested solution says that when block A is at its maximum extension, then block B will start to move up but with a certain set of values couldn't block A reach...
Back
Top