Torque Equilibrium Question

[stitchattack]

Q: A 0.5 kg uniform meter stick is suspended by a single string at the 30 cm mark. A 0.2 kg mass hangs at the 80 cm mark. What mass hung at the 10 cm mark will produce equilibrium?

Set up is suppose to be: (mg)(20cm) = (0.2 kg)g (50 cm) + (0.5 kg)g(20cm) and you're suppose to find the m.

What I don't understand is the set up. I understand the left side which is the unknown mass located 20 cm from the suspension and I understand the 0.2 kg mass which is hanging 50 cm from the suspension. I don't understand the set up for the uniform meter stick.

Will someone please explain why the distance from the suspension for the meter stick is 20 cm and why it is on the right side of the equation?

Thanks!

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[Hadi7183]

For a "uniform" meter stick, the center of mass coincides the geometrical center, i.e. the center of stick (50 cm). Since the fulcrum is located at 30 cm, the distance is 20 cm.

 

[stitchattack]

For a "uniform" meter stick, the center of mass coincides the geometrical center, i.e. the center of stick (50 cm). Since the fulcrum is located at 30 cm, the distance is 20 cm.

Maybe I'm missing something but how did you figure the center of stick is 50 cm. I thought the entire meter stick is the 30 cm + 50 cm = 80 cm so center is 40 cm?

 

[Hadi7183]

But it's called "meter" stick, which means it's one meter long

 

[stitchattack]

But it's called "meter" stick, which means it's one meter long

Ok I just facepalmed myself. That completely went over my head lol Thanks for the clarification!

 

[Hadi7183]

haha... No problem!