Finding the Distance for Static Equilibrium of a Suspended Rod

[flash2]

A uniform .122kg rod of .90 m length is used to suspend two masses as shown below. At what distance x should the .20kg mass be place to achieve static equilibrium?

Okay so I got the Torque of the 0.2kg mass

T= (0.2)(9.8)(x) with x being the length I hafta find.

Then I got the Torque of the 0.50kg mass

T= (0.5)(9.8)(0.2)

Then I plugged it into this...

sum of Torques CW= Sum of Torques CCW and the answer was 0.5m and I got that but originally for the 0.50 mass, I put in (0.25) for the length because that's what it is but I didn't get teh answer so I did trial and error with the numbers I had and I figured 0.2m was the right length to get the right answer. So I'm wondering why it's 0.2 m and if the mass of the rod matters.. cause I dind't use it.

 

[factor]

Well the answer I'm arriving at is .503 meters, or just .5 if you want to keep it simple. When they give you mass of the rod, that usually means it matters, and they make it simple when they tell you it's uniform. Since the mass-density is uniform you can assume that the rod applies all it's weight as though it were concentrated at the length of the rod divided by two, or .45. Since .45 is to the left of the support point, you can say the rod applies a torque of (.122*9.81*.2)

So the equation becomes (.2*9.81*x) + (.122*9.81*.2) = (.5*9.81*.25)
Solving for x yields .986886/(.2*9.81) = .503

When we plug this value into our original summation of torques equation we arrive at 1.22625 = 1.22625 so the bar must now be in static equilibrium.

 

[daniel_i_l]

The mass of the bar matters because even without the weights the bar will tilt in the direction of the longer side.

 

[flash2]

Oh man I get it so much more now. So is there a torque from the rod exerted on the right side of the balance too? Or does taht even matter?

I know this is prbly really easy but it's extremely hard when your teacher doesn't teach the lesson.

Thanks!

-f

 

[OlderDan]

Oh man I get it so much more now. So is there a torque from the rod exerted on the right side of the balance too? Or does taht even matter?

I know this is prbly really easy but it's extremely hard when your teacher doesn't teach the lesson.

Thanks!

-f

You only need on torque from the rod. You may take all the mass of the rod as if it were located at the center of the rod (assuming the rod has uniform density).