Physics problem resolving forces

[MatskiMonk]

The question I'm trying to get my head around is this:

The blue beam is a rigid piece, attached to nothing, but resting on a surface at points A and B. Forces are being applied, and measured at F1 and F2 - but the measurement is the sum of forces being directly applied downwards, and resultant forces from moments in the beam. L1, L2 and L3 are all unequal, and F1 and F2 can be measured as unequal. How would I calculate what the forces at A and B were?

Now, I'm looking at this, and I'm thinking to myself... I'm over thinking this... but for example, F2 exerts a downwards force at B and A using F1 as a pivot. At the same time, F2 exerts an upwards force at A because it also uses B as a pivot. F2 exerts an upwards force on F1, and vice versa...

So if the indicators at both F1 and F2 both read 50N (for example), and L1 and L3 are unequal, then the forces at A and B are going to be different... but how to calculate what they are?

edit: to be clear the force being applied at F1 and F2 are unknown but the total resultant force at F1 and F2 is.

 

[MaxAttack]

I'm out of practice and wasn't very good at mechanics when I was studying it earlier this year, but it appears to me in this situation you have equilibrium.

So with all forces in Y direction equal to 0, I'd think you might be able to calculate the individual turning moments using the given distances, which might lead you in the right direction?

I'm not sure, but once we got into turning moments people seemed to find it very easy not to recognise an equilibrium, and this made answers impossible to find.

Hope this helps, but I could well have missed the point entirely.

 

[Danoff]

Looks like a free-body diagram for an object at rest (ie: sum of accelerations and moments = 0).

Just generate the necessary free body diagram equations and set to zero.

Edit:

Follow this guy's example:

 

[TexRex]

Looks like a tropical bird.

 

[MatskiMonk]

Hmmm... thanks for the help so far, though it doesn't help Will keep trying anyway, perhaps when I post some working it will become clearer. This is a mental block that probably shouldn't be a mental block.

 

[UnkaD]

Hmmm... thanks for the help so far, though it doesn't help Will keep trying anyway, perhaps when I post some working it will become clearer. This is a mental block that probably shouldn't be a mental block.

20200306_135253

• UnkaD
• Mar 6, 2020

I got this done. Sorry if notation might be off; it's been a very long while since I took statics. The first two equal zero, so you can take the moment about "b" and solve it for force of "a" and substitute that into the force of "b" equation. Simplify that one and substitute it back into the force of "a" equation. That should yield two equations for finding "a" and "b" given values of the two downward forces and all three lengths. Hope this helps. I was overthinking it at first too.

 

[Maninashed]

I asked a journalist and they said about 1000000N per support. My cat said try THIS

 

[MatskiMonk]

I got this done. Sorry if notation might be off; it's been a very long while since I took statics. The first two equal zero, so you can take the moment about "b" and solve it for force of "a" and substitute that into the force of "b" equation. Simplify that one and substitute it back into the force of "a" equation. That should yield two equations for finding "a" and "b" given values of the two downward forces and all three lengths. Hope this helps. I was overthinking it at first too.

Thanks, I need to puzzle through this.

To reiterate, the downward forces at F1 and F2 are unknown, only the resultant (compression) of the downward force, and the upward force of the beam at each end (caused by the moment of the F at the other end)...

This is where I got to originally, trying to solve it my way...

Where F1E and F2E are the unknown downward forces, F2R and F1R are the resultant upward forces of F1E and F2E acting about pivots A and B. IF1 and IF2 are the indicated forces on gauges that measure the compression between F1E and F2R, and F2E and F1R. I've chucked in some values for scale. As you can see, I got stuck even before getting to the Force on A or B.

My cat said try THIS

Your cat is wise (mine thought the answer was 'corned beef'), and I indeed tried that site, but I don't know the load, I know the compression that the load and the upward force give, but not the loads alone.

Sorry if I'm being dense here chaps.

 

[TexRex]

Corned beef is always the answer.

11 more days...

 

[UnkaD]

Thanks, I need to puzzle through this.

To reiterate, the downward forces at F1 and F2 are unknown, only the resultant (compression) of the downward force, and the upward force of the beam at each end (caused by the moment of the F at the other end)...

This is where I got to originally, trying to solve it my way...

View attachment 897491

Where F1E and F2E are the unknown downward forces, F2R and F1R are the resultant upward forces of F1E and F2E acting about pivots A and B. IF1 and IF2 are the indicated forces on gauges that measure the compression between F1E and F2R, and F2E and F1R. I've chucked in some values for scale. As you can see, I got stuck even before getting to the Force on A or B.



Your cat is wise (mine thought the answer was 'corned beef'), and I indeed tried that site, but I don't know the load, I know the compression that the load and the upward force give, but not the loads alone.

Sorry if I'm being dense here chaps.

Oh, I thought there were no values and they just wanted you to solve for specific variables. So the given value is the combined force of F1 and F2 and everything else is unknown? That changes quite a bit. hmm...

 

[Bealdor]

Thanks, I need to puzzle through this.

To reiterate, the downward forces at F1 and F2 are unknown, only the resultant (compression) of the downward force, and the upward force of the beam at each end (caused by the moment of the F at the other end)...

This is where I got to originally, trying to solve it my way...

View attachment 897491

Where F1E and F2E are the unknown downward forces, F2R and F1R are the resultant upward forces of F1E and F2E acting about pivots A and B. IF1 and IF2 are the indicated forces on gauges that measure the compression between F1E and F2R, and F2E and F1R. I've chucked in some values for scale. As you can see, I got stuck even before getting to the Force on A or B.

You just need to solve the moments about points A and B.

MA = 0 = IF2 x (L2+L3) - FB x L2 --> FB = 500 x (270+60) / 270 = 611.1

The counter rotating component of F1E x L1 is already included in IF2, that's why you can ignore that here. Now let's calculate FA.

MB = 0 = IF1 x (L1+L2) - FA x L2 --> FA = 490 x (50+270) / 270 = 580.7

Again, you can ignore the counter rotating component of F2E x L3 because it's already included in IF1.

Since we know both reaction forces we know that: F1E + F2E = FA + FB = 1191.8

As a result, we can subsitute F1E and F2E with each other as needed: F1E = 1191.8 - F2E and F2E = 1191.8 - F1E

And now you just need to solve the moments about points A and B again, just with F1E and F2E:

MA = 0 = F2E x (L2+L3) - FB x L2 - F1E x L1 --> Substitute F1E with 1191.8 - F2E

MA = 0 = F2E x (L2+L3) - FB x L2 - (1191.8 - F2E) x L1
MA = 0 = F2E x (270+60) - 611.1 x 270 - (1191.8 - F2E) x 50 --> Now solve for F2E
MA = 0 = F2E x 330 - 164997 - 59590 + 50 x F2E

F2E = (164997 + 59590) / (330 + 50) = 591.0

Now you can solve F1E:

F1E = 1191.8 - F2E = 600.8

Proof:

F2E x (L2+L3) - FB x L2 - F1E x L1 = 0
591.0 x (270+60) - 611.1 x 270 - 600.8 x 50 ~ 0 (the result is not exactly 0 due to small rounding errors)

F1E x (L1+L2) - FA x L2 - F2E x L3 = 0
600.8 x (50+270) - 580.7 x 270 - 591.0 x 60 ~ 0 (again, the result is not exactly 0 due to small rounding errors)

Edit: Made a quick Excel sheet for you.

 

[MatskiMonk]

You just need to solve the moments about points A and B.

MA = 0 = IF2 x (L2+L3) - FB x L2 --> FB = 500 x (270+60) / 270 = 611.1

The counter rotating component of F1E x L1 is already included in IF2, that's why you can ignore that here. Now let's calculate FA.

MB = 0 = IF1 x (L1+L2) - FA x L2 --> FA = 490 x (50+270) / 270 = 580.7

Again, you can ignore the counter rotating component of F2E x L3 because it's already included in IF1.

Since we know both reaction forces we know that: F1E + F2E = FA + FB = 1191.8

As a result, we can subsitute F1E and F2E with each other as needed: F1E = 1191.8 - F2E and F2E = 1191.8 - F1E

And now you just need to solve the moments about points A and B again, just with F1E and F2E:

MA = 0 = F2E x (L2+L3) - FB x L2 - F1E x L1 --> Substitute F1E with 1191.8 - F2E

MA = 0 = F2E x (L2+L3) - FB x L2 - (1191.8 - F2E) x L1
MA = 0 = F2E x (270+60) - 611.1 x 270 - (1191.8 - F2E) x 50 --> Now solve for F2E
MA = 0 = F2E x 330 - 164997 - 59590 + 50 x F2E

F2E = (164997 + 59590) / (330 + 50) = 591.0

Now you can solve F1E:

F1E = 1191.8 - F2E = 600.8

Proof:

F2E x (L2+L3) - FB x L2 - F1E x L1 = 0
591.0 x (270+60) - 611.1 x 270 - 600.8 x 50 ~ 0 (the result is not exactly 0 due to small rounding errors)

F1E x (L1+L2) - FA x L2 - F2E x L3 = 0
600.8 x (50+270) - 580.7 x 270 - 591.0 x 60 ~ 0 (again, the result is not exactly 0 due to small rounding errors)

Edit: Made a quick Excel sheet for you.

Well, I need to read this with a clearer head, but looking at the numbers the spreadsheet outputs it seems to demonstrate at very least the rough magnitude of the issue when comparing a best and worst case issue for L1, L2 and L3.

Thanks for everyone's efforts.