Keeping a uniform bar in horizontal equilibrium

[ogodwin]

Homework Statement

A 2.35kg uniform bar of length 1.3m is held in a horizontal position by three vertical springs as in Figure P8.83. The two lower springs are compressed and exert upward forces on the bar of magnitude F1 = 6.80N and F2 = 9.50N, respectively. Find the force exerted by the top spring on the bar, and the location of the upper spring that will keep the bar in equilibrium.

Relevant Equations

Sum of Forces = 0
Sum of Torque = 0?
F = -kx?

I'm able to get the force of the top spring (6.75N upwards) by setting the sum of all forces equal to 0 as everything is stationary. Where I'm stuck is starting the second part of the problem. I initially tried setting the sum of torques equal to 0 using the top spring as a lever arm but unless I missed a variable I'm not really sure how to start the second part.

 

[kuruman]

Will you please post the figure? We cannot help you without it. Also, please be more specific about what you tried already by posting your actual work.

 

[jbriggs444]

Will you please post the figure?

Google gave a hit (from a homework cheating site, of course).

 

[kuruman]

Google gave a hit (from a homework cheating site, of course).

View attachment 324118

Excellent sleuthing job! Now we wait for OP's attempt in more detail.

 

[Freyja]

Please do not take my word as gospel, but I would approach this problem as an "inverted" seesaw.

Hints:

- Springs work linearly.
- Top spring is your fulcrum.

Somebody correct me if I'm wrong, please?

 

[jbriggs444]

- Springs work linearly.
- Top spring is your fulcrum.

Somebody correct me if I'm wrong, please?

Looks like a static setup to me. The linear behavior of springs will not enter in. In any case, we are waiting on OP to contribute. Meanwhile, OP seems to be a one-post wonder doing a drive by.

 

[Freyja]

Looks like a static setup to me. The linear behavior of springs will not enter in. In any case, we are waiting on OP to contribute. Meanwhile, OP seems to be a one-post wonder doing a drive by.

I hope I didn't break any rules, I didn't mean to post any answers before OP showing his/her attempts to solve.

It's just, this problem clicked right away, since I just had to perform a mechanical aptitude test. No harm intended :P

 

[kuruman]

Please do not take my word as gospel, but I would approach this problem as an "inverted" seesaw.

Why inverted? Compare what you have here with a regular seesaw

Two forces at each end directed down. Check.
One force at the fulcrum directed up. Check.
The weight of the bar directed down. Not shown but check.

 

[jbriggs444]

Why inverted? Compare what you have here with a regular seesaw

Two forces at each end directed down. Check.
One force at the fulcrum directed up. Check.
The weight of the bar directed down. Not shown but check.

Inverted because the springs in the situation at hand are under compression:

The two lower springs are compressed and exert upward forces on the bar

Of course, it is the same algebra either way. Only the signs on some of the parameters get changed.

Hmmm. The algebra is also unchanged if we squint our eyes a bit and consider the Earth as the seesaw beam and the beam as the anchor to which it is affixed. Now the springs on the bottom of the drawing exert a downward force and the spring on the top of the drawing has the sign of its force inverted.