How to determine What are the forces acting on bars and wires? (photo of exercise in comment)

Question

Answer 1

So Gladidi K.,

Look like you haven't had a lot of practice in wading through statics physics problems?

Let's look at the first one; all the rest are broken down in the same way, just the details differ!

1) First principle: this is a statics problem; forces must all balance out! (nothing is accelerating)

2) When forces at a point balance, they do so in both the x and y directions, and you can calculate those more-or-less independently.

So now: what is generating force in the first problem? It's that 5 kN force, and it is acting straight down (in the (-) y direction) at the junction of bar and cable.

And what holds it (assume it is a suspended mass) up in the air? Only the bar S1. But the bar is at a tan^-1(3) angle, so it can only supply part of its force in the (+)y direction, namely, 3/(10)^0.5 proportion (look at the triangle until you see that; the straighter up the bar, the more of its force is as (+)y vector). Hence, to supply 5 kN of pure (+)y force, it must push with 5 *10^0.5 / 3 kN total force. The other 1/(10)^0.5 kN of force is still directed in the (+)x direction!

OK, now let's get rid of that bit of (+)x force. That happens by the action of the cable S2. S2 is angled at 45 degrees (don't forget to add both the 2m and the 1m on the ground!). So, since it is only (2)^^0.5/2 effective in each of x and y directions, it must pull with 1/(10)^0.5 * (2)^0.5 kN = 1/5^0.5 kN. That is its tension.

Now, you aren't still QUITE finished yet -- do you see why? When you pulled on cable S2, it ALSO is exerting a (-)y component where it hits the junction with the bar. So you balanced out the x-components, but you still didn't account for that extra bit of y-component force. It then transfers into the bar. How to calculate: the amount of (-)y force transferred is the same as the (-)x force, since the cable is pulling at 45 degrees angle. So that is 1*2^0.5/10^0.5 or 1/5^0.5 kN. The bar is fairly efficient at transmitting as (-)y namely 3/10^0.5 proportionally, so the total compression uses the inverse of that factor, so 1/5^0.5 * 10^0.5 /3 = 2^0.5 / 3 kN (additionally to what the initial force of 5kN caused).

So let's sum up: the bar has compressive force of (5 * 10^0.5 / 3 )+ (2^0.5 /3) kN and the cable has tension 1/5^0.5 kN (standard form = 5^0.5 /5 kN).

This may seem incredibly roundabout to calculate to you now, but once you do it a few times, it will get better. You just have to trace where the forces come from (or, in other problems, where they go), and not miss any conversions due to the geometry of the triangles.

-- Cheers, --Mr. d.