the bending moment in timber baem M, at a point at a distance X from one end of the beam is given by formula:-
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2M = wx(L-x)\n\n(2M)/w = x(L-x)\n\n2(50N/m) / (5N/m) = x(20m-x)\n\n(100N/m) / (5N/m) =\n\n(20 N/m) = \n\nsorry i couldn't get an answer\nthat is as far as i can get\nit might help a little bit though
914639
T = mg.\n\nThe length of the cable is irrelevant in this example. The beam is hanging off the cable.\n\nThe tension in the cable therefore is m*g, which is 2450 N.
790576
You have to find basic wind speed, K1,K2,K3 factors, Design wind pressure, Gust factors, exposed area & solidity ratio, drag co-efficient.\nWind load on panel = ExpArea*Drag*Gust*Design Wind Pr.\nPanel load is distributed to nodes & you can analyse and find the axial forces (Comp & ten) treating these are truss members.\nThen design of these members is as explained in IS802.
101115
It is depended on the moment and the surface moments if inertia of cross section of cantilever.\n\nMoment is force time distance : M=r*f\n\nand I is used to show surface moments if inertia.\n\nthe more the moment more is the deflection.\nthe more the I less is the deflection.
254925
I'm assuming that the pivot point will land between the 15 and 30 Kg loads because it seems more logical. If you try putting the X distance between the 20 and 15 loads, you'd probably get a negative distance. \n\nActually, the weights have to be multiplied by the pull of gravity (9.81m/ss) to make them actual forces but then again, they would all just cancel out.\n\ni hope you understand the crude diagram\n\n|-------10------|       - distance where weight of beam is applied\n|---------x-------|    - pivot point dist from left\n ____________________  - beam (20 m)\n|_________|__________|\n20_______15_________30 - applied forces (weights)\n\n20kg(X) + 15kg(X-10m)  = 30kg(20m-X)\n20kgX + 15kgX - 150kgm = 600kgm - 30kgX\n20kgX + 15kgX + 30kgX  = 600kgm + 150kgm\n                65kgX  = 750kgm\n                    X  = 750kgm/65kg\n                    X  = 11.53846 m from left end
896679
A mature tree trunk may support a greater force than a fine steel wire but intuitively we feel that steel is stronger than wood. To predict the forces which a structure of any size, made of any material, can sustain without failing, we need the concept of mechanical stress.\n\nThe definition of stress originates in two observations of the behaviour of a one-dimensional body subject to uniaxial loading, for example, a steel wire under uniaxial tension:\n\n   1. When the wire is pulled, it stretches. Up to a certain load, the stretch ratio (current length / initial length) is proportional to the load divided by the cross-sectional area of the wire. We therefore define the stress as σ = F/A.\n   2. Failure occurs when the load exceeds a critical value for the material, the tensile strength multiplied by the cross-sectional area of the wire, Fc = σt A.\n\nThese observations suggest that the quantity that affects the deformation and failure of materials is stress.\n\nIn the more general setting of continuum mechanics, stress is a measure of the internal distribution of force per unit area that balances and reacts to the external loads or boundary conditions applied to a body. Stress is an example of a type of quantity called a second-order tensor. In three dimensions, a second-order tensor can be represented by a 3×3 square matrix (containing nine components). However, in the absence of body moments, the stress tensor is symmetric and can be fully specified by six components. These six components are sometimes written in Voigt notation as a 6×1 column matrix which is often, misleadingly, called the stress vector.
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Extreme position! \nBecause acceleration is directly propotional to the displacement from the mean position and directed opposite to it.\n\na=-(ω^2)x
715400
Angle of line from horizontal position.
229194
At a local airport, a powerful light is placed on top of a 45-foot tower. If the tower is at one end of the runway, what is the angle of depression needed so that the beam from the light extends to the end of the 2200 foot runway?\n\ntan x = height of tower/length of the runway\ntan x = 45/2200 = 0.02\nx = 1.17 deg
317313
If you need an exact balance, you will need too weights (combined).  You will need 3/4 the weight of the beam at the end of the short end to balance the beam (unless you can neglect that).  You then need to balance the weight (w) at the end of the long end with a counter weight CW) at the end of the short end.  A formula that woorks should look like this:\n\nW x 3ft = CW x 1ft\n\nCW = (W x 3ft)/1ft\n\nCW = 3W (plus 3/4 beam weight)
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Well, a clear question would help. \n\nAre you trying to calculate the power that a certain material can withstand?   Or are you trying to calculate strength?
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length   882.75 feet\nbreadth (beam)  92.5 feet
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2t, where t = arctan(Electrostatic force/Weight);\nThe electrostatic force will act horizontal, and the weight vertical, and these two must compose to a force along the suspending thread.\n\nSubstitute the numerical values in and find the angle.
592688
assuming the board is also of uniform dimensions the CM is at 2.325m from left end
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The CG is between the two scales. let us say it is at a distance of x1 from scale 1 and x2 from scale 2. (assume the plank remains horizontal)\n\nThen, x1*Fg1 = x2*Fg2 (the moments must be equal)\n\n>> x1/x2 = Fg2/Fg1\nand x1 + x2 = 2\n>>\nx1 = 2 * (310/(450+310)) ~ 0.816 m\n\nNow tell me where the feet of the woman are.
185355
Axial or Normal Stress (denoted by sigma) is equivalent to the force on the beam divided by the cross sectional area that the force is perpendicular to.\n\nThere are other stresses related to different loadings such as shearing stress, stresses from bending moments and a myriad of others. Your question could be more specific.
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The stress in the cable is m*g/A.  The Young's modulus of steel is about 200*10^9 N/m^2 (pascal).  Therefore the strain will be\n\nm*g*/(A*200*10^9).\n\nThe elongation is strain*length.\n\nFor the second part, replace g above with (g+a), where a =2.7m/s^2\n\nFor the last, compute the mass from S = m*(g+a)/A, The elastic limit is Sm, so the max mass is m = Sm*A/(g+a)\n\nNote: keep the units straight.  Use kg, m, sec, N in all calcs then convert to the asked-for units last.