I went ahead and solved it. There is actually one other step I used prior to summing moments about point C. You still have to isolate part of the structure as noted above which is the most important idea for this problem. I also found your vertical reaction at D looks like it is going in the...
Use virtual work. Set the deflections equal for each case and solve for P. I would guess this is how your instructor wants it to be solved and not just by manipulating existing tables.
It is just making a free body diagram with the cable detached from the support. That is how I found the force in the cable. There are 3 steps; sum moments about A with the cable cut to get the cable force (note the cable is conveniently perpendicular to the platform), then sum vertical forces...
I got it now, I had left out a cosine. Make a cut through the cable up at where it attaches to the top support. Then sum moments about point A using the vertical weights given in order to find the tension in the cable.
The question wants you to minimize the force in the cable "F". So in order to do that, you would want to find the angle and magnitude of the resultant of forces given already (like it says to do) then make "F" parallel with this force (at the same angle). So first break apart the given forces...
It is a cantilever beam, cantilevered on each side of the fulcrum. But don't try to equate the first class lever to a single cantilever of the entire lever length with the combined load on the end. That is why you got 4 times the moment the first time (8674).
"The original lever I calculated as a resistance of (about) 310N a distance 7mm from the fulcrum with an applied force of 85N a distance of 26mm from the fulcrum."
For a second class lever, this is close enough to right.
"After this I have multiplied each lengh of the side of the lever by...
Sorry, I tried to rearrange my last post to the end before you posted again.
For a first class lever;
http://en.wikipedia.org/wiki/Lever
The max moment will occur at the fulcrum (not like in second class), so now Mmax=effort*26mm which is the same as the resistance*7mm. Now that you have...
I understood your opener to be a total of 26mm long. I don't understand how you are adding the 26 and 7 to get 33. What you should have is this:
http://en.wikipedia.org/wiki/Image:C...ualization.png [Broken]
Where the distance from the force to the first support (or resistance in your...
No, adding the force onto each bar will not equal the original force.
Each component part of your equation is right, but I do not understand why you are trying to add the components. When you see a force written as F=#i+#j, the 'i' and 'j' are 90 degrees apart.
Get the section modulus (S) of the lever then solve for the bending stress, fb=M/S, this is the stress to compare to the yield stress of the material (for practical purposes). Most likely, yielding of a metal bottle opening lever from bending would be the failure mode and not shear. You would...
The box he is sitting on has only one cable at the left end so the box would rotate under his weight, unless there is some other information you are not giving. Is the box supposed to be in the air or sitting on the ground?
There are no zero force members for this load arrangement. JD would be a zero force member if there was not a load applied at D.
There has to be equilibrium at each end of member BI. HI is balancing joint I and BC is working to balance joint B
I believe that is the same as what I am calling the integration method:
Either way, the assumption has to be made that the shear deformation is negligible compared to flexural deformation, or else you would need additional beam properties to get the correct solution.
There are a couple of ways to solve an indeterminate beam problem. If you were not given the moment of inertia or modulus of elasticity, then the "flexibility" or "integration" method would be the one I would use to give you the approximate answers, since the E and I for this problem are likely...
If "A" is a roller and has one reaction, and "B" is fixed for 2 translations and 1 rotation, then that is an indeterminate problem. You will need additional beam properties to be able to solve it.
To make sure I understand what you have; a constant rectangular cross section with a uniform horizontal force applied only along the top surface of the section and perpendiular to the longitudinal axis of the beam, and the beam has one end fixed and one end free.
Your deflection can be...