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A torque question

  • Thread starter toqp
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Homework Statement



http://img21.imageshack.us/img21/7103/fig1jpg.png [Broken]

The rods in the system are rigid and massless. The only mass in the system is that of the weight (mass m).

The system as a whole is pivoted at a point A. The lower rod is pivoted at a point C.

If the system is set correctly, will the free point B go up?

The Attempt at a Solution



Of course there's first the torque T of the lower rod. It is supposed to be in balance so the sum must be zero. If the gravitational force acting on the mass is mg, then there is some force F acting on the point where the upper and lower rods meet.

[tex]\sum T = mgr_b - Fr_c = 0 \Rightarrow F=mg\frac{r_b}{r_c}[/tex]

(It seems my latex code is not working, so..)

T = mgrb - Frc = 0 --> F=mg (rb / rc)

When considering the whole system, we can think of the mass m as acting at the point C, so that it tries to rotate the whole system down.

[tex]T_{down}=mgr_a[/tex]

Tdown=mgra

On the other hand, if the lower rod is kept stationary, the there must be a force equal to F acting at the point r_a + r_c.

[tex]T_{up}=F(r_a + r_c)[/tex]

Tup=F(ra+rc)

So yes... the point B can move upwards. But this seems so counter-intuitive to me. I surely have missed something or done something wrong.
 
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Answers and Replies

  • #2
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Edit:

Of course... I've completely forgotten translational conditions. The pivotal point C is not translationally at rest.
 
Last edited:

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