Torque, have answer, need explanation

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    Explanation Torque
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The torque calculated for both the origin and the point (2.0m, 0, -3.0m) is the same at (-1.5)i - (4.0)j - (1.0)k Nm due to the specific alignment of the force and the position vector. The position vector from the point to the force is parallel to the force vector, which results in the torque being unaffected by the change in position. Generally, position does influence torque, but in this instance, the parallel nature of the vectors leads to a unique case where the torque remains constant. Understanding this relationship clarifies that while position typically matters, specific orientations can yield identical torque values. This example illustrates the nuances of torque calculations in physics.
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Force F= (2.0N)i - (3.0N) k acts on a pebble with position vector r= (0.50m)j - (2.0m)k relative to the origin. In unit vector notation, what is the resulting torque about a) the origin and b) the point (2.0m, 0, -3.0m)?

Okay, I got the answers and their right. For both (-1.5)i - (4.0)j -(1.0) k Nm. My question is what is the relation here? Why is the torque the same about the origin and a point? Is it because the position doesn't affect the torque, but the force does? What am I missing because I want to know the relationship here or if it's pure coincidence.
 
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gvcalamike said:
Force F= (2.0N)i - (3.0N) k acts on a pebble with position vector r= (0.50m)j - (2.0m)k relative to the origin. In unit vector notation, what is the resulting torque about a) the origin and b) the point (2.0m, 0, -3.0m)?

Okay, I got the answers and their right. For both (-1.5)i - (4.0)j -(1.0) k Nm. My question is what is the relation here? Why is the torque the same about the origin and a point? Is it because the position doesn't affect the torque, but the force does? What am I missing because I want to know the relationship here or if it's pure coincidence.

It happens to work that way because P=[2m,0,-3m] happens to be parallel to the force F=[2N,0,-3N]. So Fx(r-P)=Fxr-FxP and FxP is zero since F and P are parallel. So Fx(r-P)=Fxr. Position generally does affect the torque. But not in this case.
 
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Thank you sir.
 
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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