How do you find tension when no lengths are given?
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.
57888
the tension is just the force of gravity(since there is a constant velocity, the forces going up must equal the forces going down.)\nso its just the mass times gravity so 79 x 9.8
967107
(tension in crossbar) * (its height) =(half of ladder weight) * (length of ladder * sin(30/2) / 2)
576519
tensile strength = (2*P)/(pi*t*d)\nwhere; P =maximum load you get from the machine(kPa); pi = 3.14; t = thickness/height of specimen (m) and d = diameter of specimen(m)
415302
You need to use Pythagorean theorem.\nMeasure the lenght of the lotus plant which is above the water.(X)\nBend the plant to any one side until it is just submerged under the surface of water.\nMeasure the Dist. between the initial and final positions of the plant(Y)(i.e. the initial position and final position where it is submerged)\n\nLet the actual lenght be L.\nThen (L x L) = (Y x Y) + (L - X)(L - X)
317039
Imagine the stretching of the surface of the drum as the stretching of a spring. use hooke's law to do calculation. u will need to know about the strength of material and etc. very very very complicated
866574
Mental picture for you.  Take a belt by the ends.  Now pull as hard as you can, you have now created tension in the belt.   Get it now?  a kind of stress that's built up.
774792
Let the length of side "b" be x ft.\nTherefore side a=2x and side c=2x+3\nAccording to the problem,\nx+2x+2x+3=93\n=>5x+3=93\n=>5x=93-3=90\n=>x=90/5=18\nTherefore.side a=2X18=36 and side c=36+3=39\nAnswer a=36ft. b=18 ft and c=39 ft
937470
Getting started, have to overcome static friction.  Therefore, force applied = static friction force just before movement\n\nF = umg\nF = 75N\nm = 20kg\ng = gravity\nfind u, plug and play\n\nsame problem for sliding friction
914303
read stretching scientifically.... you might be able to download it\nmichael\nwww.bandconsulting.com
706663
The idea is that all forces at the nodes (connecting points) are balanced. Draw the two free body diagrams.\n\nLeft point:\nFx\n0=-T1cos53+T2cos11\nFy\n0=T1sin53-T2sin11-57\nRight point:\nFx\n0=-T2cos11+T3cos42\nFy\n0=T2sin11+T3sin42-W2\n\nBy inspection we see only the first 2 eqns are required. Solve the system:\n0=-T1cos53+T2cos11\n0=T1sin53-T2sin11-57\n\nT1=83.62 N\n{=}8.5 kg
973341
A)\n\nTo find tension in the wire, we'll calculate the moment about the base of the beam:\n\nweight of beam = 145 N  acting at l=1.7m/2=0.85 m\nweight of sign = 235 N acting at  l=1.7m\n\nonly the vertical component of force in the wire will be used \nbecause the horizontal component passes through the base\n\ntherefore Tvert=Tsin35 acting 1.35 m from the base\n\nso here's the Moment eq\n\n-235x1.7 (down) - 145x0.85 ( down ) + Tsin35 x 1.35 = 0\nT=(235x1.7 + 145x0.85)/(1.35xsin35)\n  =675.1N (up and to the right )\n\nB)  from above the T_vert=Tsin35 = 387.2 N (upwards)\n    and T_horiz=Tcos35=553.0 N (inward)\n\nNow we can find vert and hor forces at the base\n\nHorizontal====>       horizontal Force + 553 N (inward ) = 0\n\nhorizontal force = -553 N (wall is pushing outward ) \n\nVert====>Sign + beam + T_vert + vert_base_force = 0\n              -235 (downward ) -145 (downward ) + 387.2 (up) +vert_base_force=0 \nvert_base_force = 235 +145 - 387.2= -7.2 N ( down )
770512
The total force on the block is mg sin (theta). The mg cos (theta) and force normal cancel out. Since there is no friction, the only force, again, is only mg sin (theta). \n\nThis is obvious if you consider that there is a force mg, and break it down into components, the solution should seem obvious.
481398
sum the forces in the direction that they are applied and it will be equal for a static object. \n\nRule of thumb on this one is that you have 100N down ward, then you need 100 N's up ward.\n\nGivens, legs are equal lengths, same material, and no angles to calculate.\n\n4 legs - 4 forces - each at 25 Newtons.  \n\nNow if you want to get fancy, break out the strain gauges and let us see what we really find.  But it helps if you have preformed a few material tests and experimenters first.
179742
Various springs (flat or helical) deflect linearly with load and resistance of certain wires (nichrome, etc.) varies linearly with length.  If your experiment allows or expects large deflections you may be able to combine the spring/resistors into a circuit that calibrates force/deflection/resistance-reading suitable for measurement on an ohmmeter or perhaps a strip chart recorder.  Good luck.
857685
forces on M : grav. force Mg (vertically down) tension 2T(two strings horizontal, one on pulley) tension T (vertically down due to string on pulley) Normal reaction on M at lower surface N (vertically up) Normal reaction on M due to m R (to the left) Friction due to m on M f1=u1 times R (vertically down)(sliding is going on) friction in M at lower surface f2=u2 times N (to the left) (sliding is going on)\nnet force on M vertically up :ZERO\nto the right M times A\nforces on m  : grav force mg (vertically down) reaction due to M R (to the right) tension T (vertically up) friction due to M on m f1=u1 times R(vertically up)\nnet force on m to the right m times A\nnet force on m vertically down m times a\nunknowns N,R,a,A,T,f1,f2,\nwrite a constraint eqn for relating accns a and A. a=2A \nsolve them!!!!
905049
First assume that we neglecting any losses from air resistance or internal friction.\n\nWith that simplification, imagine two states:\n1) you are on the bridge 100m above the water and at rest\n2) you are suspended by the spring at its point of maximum extension, and therefore with zero velocity\n\nSince we assumed that there were no losses, the energy of both of these states is equal. Where is that energy located?\n -->1: your gravitational potential energy (no kinetic energy, no stored elastic energy)\n--> 2: your remaining gravitational potential energy, and the stored elastic energy in the spring (by definition, no kinetic energy)\n\nSo what has happened is that the spring has stored elastic potential energy that is equal to your loss of gravitiational potential energy. That's the only way the energies of the two states are the same.\n\nNow you can set up some equations. Let's say that you start at height H0 and end up at height H1.\n\nchange in PEg= m * g * (H1-H0)\nchange in PEk = 1/2 * k * [(H1-H0)-30m]^2\nwhere we have accounted for the 30m long unstretched length of the spring.\n\nNow by observing that -PEg = PEk, and plugging in the given values for m and k and the known value for g, you can readily solve for the change in height (H1-H0) (it should be < 0) It's a quadratic, so you expect to find one answer that is not physically realizable - reject that one, and select the other one.\n\nDoes that help?