- FlyVirgin744
**Posts:**1282**Joined:**

Hey guys, I've been stuck on this problem for awhile. Maybe someone can help me.

Figure 11-43 shows two blocks, each of mass m = 3.2 kg, suspended from the ends of a rigid massless rod of length L1 + L2, with L1 = 0.20 m and L2 = 1.8 m. The rod is held horizontally on the fulcrum and then released. What are the magnitudes of the initial accelerations of (a) the block closer to the fulcrum and (b) the other block?

Now the answer to this situation is

0.956 m/s^2 and 8.60 m/s^2. But next time I do it I will get new numbers.

Can anyone help?

Figure 11-43 shows two blocks, each of mass m = 3.2 kg, suspended from the ends of a rigid massless rod of length L1 + L2, with L1 = 0.20 m and L2 = 1.8 m. The rod is held horizontally on the fulcrum and then released. What are the magnitudes of the initial accelerations of (a) the block closer to the fulcrum and (b) the other block?

Now the answer to this situation is

0.956 m/s^2 and 8.60 m/s^2. But next time I do it I will get new numbers.

Can anyone help?

Sometimes I go about in pity for myself and all the while a great wind carries me across the sky.

I've just done a quick calculation, and I think the answers are

(a) 0.16 m/s^2

(b) 1.44 m/s^2

I will post again if I come up with a different answer.

airways1

(a) 0.16 m/s^2

(b) 1.44 m/s^2

I will post again if I come up with a different answer.

airways1

- JetService
**Posts:**4611**Joined:**

Hi,

First calculate the torque around the fulcrum.

T=L2*F2-L1*F1

Then use Newton's second law for rotation

T=I*a/r (r=L in this case) where I=I1+I2 and I1 and I2 are the moment of inertia for each of the two blocks added together (I=mr^2)

This gives a1=(T*L1)/I and a2=(T*L2)/I

I got the answers you posted above doing it this way. Hope it helps!

Staffan

First calculate the torque around the fulcrum.

T=L2*F2-L1*F1

Then use Newton's second law for rotation

T=I*a/r (r=L in this case) where I=I1+I2 and I1 and I2 are the moment of inertia for each of the two blocks added together (I=mr^2)

This gives a1=(T*L1)/I and a2=(T*L2)/I

I got the answers you posted above doing it this way. Hope it helps!

Staffan

I seem to have the same problem as you - I get a different answer each time.

Now I have (taking g=10N/kg)

(a) 0.99 m/s^2

(b) 8.89 m/s^2

I thought my previous answer seemed a bit low...

airways1

Now I have (taking g=10N/kg)

(a) 0.99 m/s^2

(b) 8.89 m/s^2

I thought my previous answer seemed a bit low...

airways1

- FlyVirgin744
**Posts:**1282**Joined:**

That was it Staffan, thanks a lot

Sometimes I go about in pity for myself and all the while a great wind carries me across the sky.

http://panda.unm.edu/courses/finley/P160/HW/HW31Sol.html

JetService..your a math expert. Can you ramble off some of your bullshit errr.. math knowledge?

- JetService
**Posts:**4611**Joined:**

Flight152, I was going to see if someone else could help first. I hate to be a know-it-all, but after seeing the responses, I guess I have to.

FlyVirgin's first problem is his polyfractals are all askew. He doesn't realize (as most of the population, except me) that whenever you have two opposing unilinial lengths on a reflective plane, it immediately puts the fulcrum in an opposing mass state. This basically inverts your polyfractals incrementally, thus skewing them. To compensate when calculating the overall static basis, all you have to do is place your variables (L) in the same matrix as the subvortex. Otherwise you end up with a coefficient leak. And everyone know that a coefficient leak, while durinominal, is only a raw element in a saturated array. Hope that helps.

FlyVirgin's first problem is his polyfractals are all askew. He doesn't realize (as most of the population, except me) that whenever you have two opposing unilinial lengths on a reflective plane, it immediately puts the fulcrum in an opposing mass state. This basically inverts your polyfractals incrementally, thus skewing them. To compensate when calculating the overall static basis, all you have to do is place your variables (L) in the same matrix as the subvortex. Otherwise you end up with a coefficient leak. And everyone know that a coefficient leak, while durinominal, is only a raw element in a saturated array. Hope that helps.

"Shaddap you!"