Engineering - Support Reactions

Do you have the question?

There’s not really a question, it’s just an example of how to find R1 and R2 in terms of P, w and L. Attached the slide in my OP.

THere was nothing before defining P and w? What did it do for moments about R2?
Gut feeling is its a typo but the "w/unit length" isnt very clear on the diagram.

Nope. For moments about R2 it did:
-R1 x L + P x (L/2) + w(L/2) x (L/4 + L/2) = 0.

But if I was to do it the way I thought was right I’d do moments about 2:
(R1 x L) + (-P x L/2) + (-w x L/2) = 0.
So for R2 again I’m not sure where they’ve got that last bit of (L/4 + L/2).

Honestly the slides for these lectures are so bad, there’s just random numbers everywhere with no explanation of where they came from (and they’re badly formatted too lmao).

If w was the weight of a uniform beam Id agree with you. However w seems to be something related to weight per unit length so wL would be the weight. However, that doesnt explain the L/4 and 3L/4 (though they do sum to L). Id really need to properly understand how w (and P) were defined.

Okay so in another PowerPoint I found a diagram and it says P = a point load and w = a distributed load (load/unit length).

Okay so in another PowerPoint I found a diagram and it says P = a point load and w = a distributed load (load/unit length).

Actually looking at the original diagram, w only applies (uniform) to the first 1/2 of the length so the load is wL/2 and the com is L/4 from R1 and 3L/4 from R2. Its zero in the second 1/2 of the length.

So the L/4 and 3L/4 have come from the mid point of where the force is acting? Is that right or am I being thick?

THats right. There is a uniform force/load acting from 0..L/2 (on the left) so that means it acts like a point force/load at L/4 from the left and 3L/4 from the right.

Okay thanks, it’s clicked now! Makes sense. Appreciate it