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i guess that must be the case thankyou for all your help its being most appreciated. Could i trouble you to look at my other problem. It is the orifice question, although i just need confirming that i have done the right thing.

ok so now we have the 9cm correct what about the 12.8cm From what you have said, i should just take the Force acting on AB and when calculated I am out by 40nmm which is a little too much, i mean it could be right but i feel i may have made a mistake. So here's what i did using the same...

ive recalculated the force for 9cm i get 9cm, using the formula, h/2 * h * w * g * p = F where h =height, w = width g= 9.81 and p = 1000 so i get 0.09/2 * 0.09 * 0.075 * 9810 = 2.98N Acting at 2/3 from top of water so 30mm So the level arm turns out to be 172mm Putting into the...

So as I am just taking moments about O, then the force applied by the mass should be equal to the force applied by the force on ab. Would i then simply take 12.8cm and use it in the equation 0.128/2 * .128*.075 * 9.81 * 1000 = 6.02 N and would act at 159mm from 0. then the moment...

Are you suggesting that the pressure applied to the curved outersurface is equal to the pressure applied to the line AB. Obviously times by the leaver arm which is 202mm. Also how would i calculate where the Resultant Force F would act, as for a straight line it would be 2/3 of the height...

The problem was find the theoretical difference in head pressure before and after orifice. Then compare with experimental difference in head pressure. No diagram wasnt given one to begin with. sorry

could really use some help?

Using the picture added Q4, work out the hydrostatic force applied to AB when the water level is at 9cm and 12.8cm. 50g of weight was added to balance the setup. And 260g and 460g were used respectively. width of container was 75mm Also need the moments Heres what I've done F...

Orifice problem I've been given a list of quantities Pipe before orifice head = 255mm Diameter = 31.75mm Pipe after orifice head = 230mm Diameter= 20mm Time for 10litres to fill in collection tank 67.09seconds Assuming g= 9.81 and density = 1000 from the experiment difference in...

Got it, the length should be root 3 instead of 1. so it should be (-10 x (root3x2) - +5x1)/2=Ha

All forces are in KN

ok so the new calculated resultant forces would be He= 15 ( (-10 x 2) - (+5x1) + He=0) Ha= -20 (+15 + 5 - ha= 0 therefore Ha= -20) Va= -10 (+10 -Va=0) But now checking my lecturers answers he gives 14.82 for He And a resultant force for a 22.2 acting 26.46...

Im desperate and really need someones help urgently.

To add to it i have just tried sum MA= 0 Therefore he x 2 - 5x1 - 10x2 =0 he =12.5 sum of horizontal 5+12.5 =17.5 therefore ha = -17.5 sum of the vertical Va -10 x 1 =0 so Va = 10 Is that right?

The problem is attached to this post, basically i need the exterior reactions and how to work them out. i know i need to take moments I am just not sure how to work it out. Basically what i worked out is i assumed horizontal members are equally to 1 sum of the horizontal = 0...

reaction at C 8x V2 -(30x8x4) - 50x2 -20x10=0 8V2 = 1260 V2 = 157.5 Reaction at A 8V1 - (30x8x4) -50x6 +2 x 20 8V1= 1220 V1 = 152.5 Is that not right?

Is the max bending moment equal to the of moment about D i.e. 2x V2 -40 +m = 0 where V2 is the reaction at C i.e. 2 x 157.5 Giving me 275. Is this correct?

Ive drawn the beam and worked out the reactions, where do i go from there to calculate the max bending moment?

Homework Statement The beam ABCD in question is 10m long with supports at A and C. At B there is a 50KN point load and another 20Kn point load at D. The is a uniform load between A and c of 30KN/m. A is 8m from C, A is also 2m from b and d is 2m from c. I need to find the max bending...