Recent content by Adam Rabe

ok i understand now, the mass of the ball at the bottom already has buoyant force accounted for and so if it was out of the water it would have that weight force added back to it. thanks a lot for helping me with my stupidity if i may ask how would you go about this without calculating...

hello sorry for the late reply... oh i see. so you need to add the bouyant force? you then get 1 + 0.5 = 1.5N 1.5 N ---> 0.15 kg 0.15/0.00005 = 3000 thanks a lot! but i don't get why you added, ill have to think about it

Homework Statement Homework Equations F = ma F = pVg p = m/V 3. The Attempt at a Solution [/B] First i tried finding the volume of displaced water which i did by... (50/1000) * 10 = 1000 * 10 * V ---> V = 0.00005 m^3 Next as the metal sits on the bottom its weight force minus the bouyant...

Hello sorry for the late reply. i think i see where i went wrong... Using your advice, i added the pressure change due to height and subtracted the pressure difference between the two speeds (as fluid speeds up pressure must drop) 5*10^5 + (1000 *10) - ((0.5 * 1000 * 30^2) - (0.5 * 1000 *...

Homework Statement Homework Equations AV = AV P + 1/2*p*v^2 + pgh = P + 1/2*p*v^2 + pgh The Attempt at a Solution For a.) ... divide 3 by 0.1 because of (AV = AV) = 30 ms-1. answer is right. For b.) I am stuck on this one. I tried using the 2nd relevant equation above but i can't seem to...

Hi sorry this is taking up a lot of your time. I really appreciate your help Does this image make sense? its essentially just like the previous drawing i did

Say two support forces from left to right are A and B. I meant that when 'Left is Fulcrum'. left support force (A) is the pivot point of system. Right is fulcrum, right support force (B) is the pivot point of system

Then the distance would of the left side would be longer... just like on my drawing so point A from weight force would be 1m and point B from weight force would still be 0.5m Left as fulcrum: Torque = 0 = (-200 * 1m) + (Fb * 1.5) => Fb = 200/1.5 Right as fulcrum: Torque = 0 = (-200 *0.5) + (Fa...

I don't quite understand what you mean by about that axis. Do you mean like making the weight force the centre of the two support forces? So... have the weight force in between support force A and B Left as fulcrum: Torque = 0 = (-200 * 0.5) + (Fb * 1) <-- support force by other side of axis (B)...

I understand. So how would you go about calculating these shifts in distances and size of support forces? I wasn't really taught how to do so, my lecturers kinda just threw me the torque formula. Is there some algebraic method i need to do?

Hey is this easier to follow? Am i on the right track

Yes it does. Try balancing a pen on two places and lifting one end up, the other end is still supported.

Hello sorry for the late reply. Ok i see now that there are two support forces. For the rod to lift off net force upwards must be greater than downwards. The two support forces must still be providing 100 each (is this right?) If i lift one end, the other end would still be providing a support...

Homework Statement Homework Equations T = fd f = mg The Attempt at a Solution I appear to be suffering from autism. Heres my working out... F(bar) = 200 Two support forces of equal length, each 0.5 m from fulcrum (given). Therefore torque magnitude is the same. 100 N on both sides = counters...

omfg i got it. thanks a lot i derped out with the log rules.