"Tension of Mass on Pole: Does it Change?

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The tension of a mass on a vertical pole changes as the mass is moved towards the end due to the distribution of forces along the pole. Tension is defined as the force exerted by the mass, calculated using T=mg, where m is mass and g is gravitational acceleration. The geometry of the setup significantly influences the tension experienced by different sections of the pole. A diagram would clarify the specifics of the tension distribution. Understanding these principles is essential for analyzing the forces acting on the pole.
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Homework Statement



how does the tension of a mass on a pole change as you move the mass towards the end?

Homework Equations



T=mg

The Attempt at a Solution



i thought tension was just mass times acceleration (gravity) so I am not sure
 
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Tension is how much force [thank you LawrenceC - slip] the pole is under.
The answer depends on the geometry.

If the pole is vertical, for eg. then consider which parts of the pole are under tension.
 
Last edited:
kthejohnster said:

Homework Statement



how does the tension of a mass on a pole change as you move the mass towards the end?

Homework Equations



T=mg

The Attempt at a Solution



i thought tension was just mass times acceleration (gravity) so I am not sure

Your question is vague without some sort of diagram. Tension is force. Stress is force per unit area.
 
Simon Bridge said:
Tension is how much force [thank you LawrenceC - slip] the pole is under.
The answer depends on the geometry.

If the pole is vertical, for eg. then consider which parts of the pole are under tension.

yes the pole is vertical overhang off a table
 
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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