How Do They Calculate the 9 Degrees in the Force Diagram?

AI Thread Summary
The discussion focuses on understanding how the 9 degrees in the force diagram is calculated, specifically through the use of the inverse tangent function. The calculation involves determining the angle with the x-axis using arctan(30/190), which confirms the 9 degrees. There is confusion regarding the 19N force value, with participants suggesting that while it may not seem accurate, the components can still help derive the correct magnitude. The conversation also highlights the importance of using a calculator or online tools for precise calculations. Clarification is sought regarding the origin of the 130 value in the force calculations.
circuscircus
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Homework Statement


http://img61.imageshack.us/img61/4585/8m1cdhx1.gif

I understood every except, How do they get the 9 degrees in part c?

Homework Equations





The Attempt at a Solution

 
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Look at the components of the resultant of the three forces: using an inverse tangent you should be able to find the angle it makes with the x axis. (Draw a right triangle.)
 
So then is arctan(30/190) 9 degrees?

(I don't have access to a radian calculator right now), also how do they get the 19N?
 
circuscircus said:
So then is arctan(30/190) 9 degrees?
Yes.
(I don't have access to a radian calculator right now), also how do they get the 19N?
You don't need a radian calculator. (If ever you don't have a calculator handy, Google has one.)

The 19 N makes no sense. But the components are close enough, so you should be able to figure the magnitude properly from that.
 
So the real F is sqrt(130^2+30^2) am I correct?
 
circuscircus said:
So the real F is sqrt(130^2+30^2) am I correct?
Where did the 130 come from?
 
Thread 'Variable mass system : water sprayed into a moving container'

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}...
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