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Viscosity Formula: Why Not (mass x length)/(time x area)?
AI Thread Summary
The discussion centers on the confusion surrounding the correct formula for viscosity, specifically questioning why the alternative formula (mass x length) / (time x area) is not valid. Participants emphasize that viscosity should be expressed in terms of mass divided by (length x time), which aligns with the standard definition. There is a disagreement regarding the proposed formula, with some participants asserting it is incorrect. The conversation highlights the importance of understanding the units involved in viscosity calculations. Ultimately, clarity on the correct formula is essential for accurate scientific communication.
Neglect gravity other than that of water; neglect friction.
F1=ρgsh=1000*10*1*(1+4)=50000 N;
F1=ρgsh=1000*10*0.1*4=4000 N;
F3=50000-4000=46000 N?
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}...
I was told forces are vectors so they can be divided into x and y components, but what about the normal force or the force of static friction? do you divide those into x and y components if say you had a block that was lying on an angled surface?
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