melanie1235 said:
Can somebody pleas help me, I'm desperate.
Here is more help then I probably should provide (purists forgive me), but it seems the most practical way to assist here. Sorry if this is a bit long but I've tried to explain in detail.
To keep the working less cluttered, we can (without loss of generality) say G = 1 force-unit. Then P = 1.26 force-units. We then don’t need ‘P’ or ‘G’.
##F_A## is the compression or tension in AC. Similarly for ##F_B## and BC.
Block-A can slip left or right. Same for block-B. Let’s consider one case in detail: block-A about to slip left.
To make block-A slip left, AC must be in compression. So the vertical force on block-A from AC is ##F_Asin(30º)=0.5F_A## downwards. Since weight = 1 unit, the total downwards force on block-A is ##1 + 0.5F_A## and the limiting frictional force is ##0.3(1 + 0.5F_A)## = ##0.3 + 0.15F_A## (to the right)
The horizontal component of ##F_A## is ##F_Acos(30º)## = ##0.866F_A## (to the left).
When about to slip:
##0.866F_A## = ##0.3 + 0.15F_A##
##F_A## = ##0.419## (force-units)
Now the tricky bit. The 3 forces acting at point C are ##F_A##, ##F_B## and P. We could try analysing these forces using components, but we soon get into a terrible mess (I tried it!). But we note the rods are perpendicular and we can easily use the force-triangle for these 3 forces in equilibrium at C.
There are 2 cases to consider: BC might be in compression or it might be in tension. I’ll consider BC in compression (but see note below). In this case the 3 forces at point C are all directed inwards, to point C. I can’t do a drawing easily but I hope this diagram is enough. The force-triangle is ΔPQR:
...Q
P
...R
where ##\vec{PQ}## = ##F_A##, ##\vec{RP}## = ##F_B## and ##\vec{QR}## = P
You should draw this force triangle for yourself anyway to make sure you understand how to do it.
QR is θ to horizontal. PR is 60º to horizontal. So ∠R = θ - 60º. Since ∠P = 90º:
sinR = PQ/QR = |##F_A##|/ |##P## |
sin(θ - 60º) = 0.419/1.26
θ = 79.4º
So one of the required angles (α₁, α₂, α₃ or α₄) is 79.4º. (But you need to check all my arithmetic.)
Note: If BC were in tension, this would be a different situation. To be rigorous you should repeat the above process to check whether or not BC in tension gives a feasible solution. and if so, if it gives a value of θ in the allowed range.
Now repeat the whole process for block-A slipping right, for block-B slipping left and for block-B slipping right! Once you understand the method and are comfortable using force triangles, it's tedious but not difficult.
