Question

How do I find Torque, such as what the last column is asking?

ase 2(a) Total Mass (g) Moment arms (cm) Torque ig cm t (CCW) x1 Xo mu m2 m, Show your calculation here:

Answer 1

Hi,

Hope you are doing well.

In this case you can find the torque by multiplying moment arm by total mass at that point,

$\mathbf{ie;}\;\tau=m.d$

Where,

$m$ is the mass hung at a point (g)

$d$ is the moment/distance from fulcrum (x₀) to the point (cm).

NOTE: Mass hung on left side (m₁) will produce a torque (T₁) in counter clockwise direction which can balance the other two clockwise torques (T₂ and T₃) produced by masses m₁ and m₂ respectively.

In short, AT equilibrium (the rod is balanced by the masses),$\tau_(ccw)$ should be equal to $\tau_(cw)$.

$\tau_(ccw)=\tau_(cw)$

Let me show you some quick calculations,

Torque on the left side (ccw),

$\tau_(ccw)=\tau_{1}=m_{1}.d_{1}=265.8\;g\times25\;cm=6645\;g.cm$

Torques on the right side (together cw),

$\tau_{2}=m_{2}.d_{2}=165.8\;g\times10\;cm=1658\;g.cm$

$\tau_{3}=m_{3}.d_{3}=115.8\;g\times35\;cm=4053\;g.cm$

$\therefore \tau_(cw)=\tau_{2}+\tau_{3}=1658\;g.cm+4053\;g.cm=5711\;g.cm$

We have, AT equilibrium (the rod is balanced by the masses),$\tau_(ccw)$ should be equal to $\tau_(cw)$,But in your case the readings didn't match. Try again. Make sure it is completely balanced.

Hope this helped for your studies. Keep learning. Have a good day.

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Thank you. :)