Is -16kN*m Equivalent to 16kN*m CW?

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The discussion revolves around the confusion regarding the sign convention for moments in mechanics. A 10kN force applied at point A creates a moment about point O, which is questioned whether it should be represented as -16kN*m due to clockwise rotation. The right-hand rule indicates that clockwise moments are negative, yet the solution manual states it as 16kN*m CW without a negative sign. Clarification is sought on whether "CW" implies that the negative sign can be omitted. Ultimately, the comparison made suggests that -16kN*m is equivalent to 16kN*m CW, similar to directional velocity representations.
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The 10kN force is applied at point A. Determine the moment of F about point O.


I attached the image. I have a question regarding the sign. The direction of rotation is clockwise, wouldn't that mean the moment is (-)? The right-hand rule also points your thumb into the page which means it's negative. The book&solution manual, however, is saying that it is 16kN*m CW. Is the "CW" (clockwise) basically means you don't have to put the (-) sign? I'm confused. Does -16kN*m=16kN*m CW?
 

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pyroknife said:
The 10kN force is applied at point A. Determine the moment of F about point O.


I attached the image. I have a question regarding the sign. The direction of rotation is clockwise, wouldn't that mean the moment is (-)? The right-hand rule also points your thumb into the page which means it's negative. The book&solution manual, however, is saying that it is 16kN*m CW. Is the "CW" (clockwise) basically means you don't have to put the (-) sign? I'm confused. Does -16kN*m=16kN*m CW?

It is the samething as going 20MPH west equals going -20MPH east.
 
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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