Recent content by Bolts

If there is no load then w = 0 therefore v = 0, meaning there is no deflection.

Umm, don't forget to in include the weight either T = F.R . Which would be the "torque required to rotate a supported vertical load" as described in the first thread. Where F is the weight of the item and R is the horizontal radius of the load from the pivot position (without acceleration).

Yes, that is the correct connection.

The equations I presented are for each support. Given that the total load is 200kg then the sum of the vertical forces at the supports has to equal 200kg (100+100) and the vertical component on each support is therefore 100kg as use in my calculations. Sum of the vertical forces equals zero...

If the sag in member x and y is 10deg from horizontal and the vertical component of that force is 100kg then the horizontal component will be 100/TAN(10). When I look at the new drawing I can see that the tension in members x and y is 575.9kg(100/SIN(10)). The horizontal component will be 567kg...

If members X and Y had 10 degrees of sag then the force on each anchor would be 567kg (100/TAN(10)).

Not sure if this helps but you will have a primary coil and a secondary coil the ratio of the voltages is the same as the ratio of turn in each coil. V1/V2 = N1/N2. So if you know the number of turns in each coil you can calculate the voltage. I am pretty sure that the current will be in the...

In regards to the original question. If you assume that the structure is pin jointed or tension cables then: With the system you have drawn the tension in member F-C and F-B will be 141.4kg (100/COS(45)) and the tension in member D-E will be 100kg (100/COS(45)). With the supports being...

To lift 350lb at 10ft/s will require the use of at least a 6.4hp motor. At that power with 12V the current required will be at least 400AMPS. Sounds a bit impractical. Perhaps go for lower speed or higher voltage.

Normally you would put the dampener in the middle of the spring if possible. The center of force for the spring should be the same for the dampener unless there is other mechanical guides in place that eliminate unwanted vibration modes.

One way for a warning system would be to setup a diode with a resistor (or trim pot) combination that switches at 11.8V.

After working with old cars for many years I have found that steel tanks don't rust (internally) unless they get water in them (which they all do) that is with conventional gasolines/petrols where the water and petrol do not mix, in this case the water sinks to the bottom of the tank and causes...

The high pressure of 10.2 MPa combined with the low flow rate of 2m3/hr might create to small a pressure difference across the orifice to solve. Try increasing the flow rate and see if it solves and make a conclusion based upon that result and what you are asking to solve in the first case.