What is the equations for tension?
967107
(tension in crossbar) * (its height) =(half of ladder weight) * (length of ladder * sin(30/2) / 2)
763604
a) The stress in the wire is given by force per unit area = m*g/A\nIf Y is the Young's modulus of the material of the wire, \nY = stress / strain. Therefore \nstrain = stress / Y.\nElongation = strain*original length.\nb) When the object is accelerated upward with an acceleration of a, the force acting in the wire = m*(g+a)\nThe rest of the equations above are valid.\nc) If the elastic limit of the wire is E, the greatest mass that can be accelerated up with an acceleration of a is got from \nE = Mmax*(g+a). Hence\nMmax = E/(g+a)
658774
I would think you would use the force created by compressing the spring.\n\nThis force is given by the equation:\n\nF = k(Li - Lc)\n\nWhere:\nF = force\nk = spring constant\nLi = uncompressed spring length\nLc = compressed spring length\n\nThe below link is for a calculator for this force.\n\nhttp://www.efunda.com/DesignStandards/springs/calc_comp_Fstress.cfm
737967
Draw a free body diagram as shown below: \n\n                [m1]------> T<---[m2]------->89N\n              <--------            <---------\n                f1 =19.6N        f2=29.4N\n\n\nFirst solve for f1 and f2:\n\nf=ukR is the general formula where f is the force of friction, uk is the coefficient of friction, and R is the force normal to the surface. In this problem R is the weight of the block=mg. Now substitute known values\n\nf1=0.2*10*9.8\n   =19.6N to the left\nf2=0.1*30*9.8\n   =29.4N to the left\n\nLooking at the free body diagram:\n\nfor m1:\n\nnet force, f1=T-19.6\n                      \nFrom the formula, f=ma  where f is the net force, m the mass, and a is the acceleration.:\n\nf1=10a\n\nThus,\n\nT-19.6=10a\n         a=(T-19.6)/10\n\nfor m2:\n\nNet force, f2=89-29.4-T\n                  =59.6-T\n\nSimilarly as for m1:\n\n59.6-T=30a\na=(59.6-T)/30\n\nNow a for m1 is also the same a as for m2, therefore:\n\n(T-19.6)10=(59.6-T)/30\nThe zero on each side cancels out. \n3T-58.8=59.6-T\n4T=59.6+58.8\n4T=118.4\n  T=29.6N in the direction shown in the above free body diagram.
632824
Most popular Equation is... given by Newton..\n\nF = M A\nForce = Mass x Acceleration
970605
For your first problem...\n\nYou know that T = Fc = (mv^2) / r\n\nWhere T is the Tension, Fc is the Centripetal Force, m is the mass, v is the velocity, and r is the radius.\n\nYou're given m = 80 kg, r = 0.90 m, and T = 600N.\n\nSo...\n\n600 N = (80kg * v^2) / (0.90 m)\nSolve for v...\nv = +/- 2.598 m/s\n\nSo since -2.598 m/s doesn't make sense (a negative speed), the solution would be ~ 2.6 m/s.\n\nFor your second problem...\n\nIts been a long time since I've done circular motion.  This site has the equation you need to use (its too messy to type here): http://hyperphysics.phy-astr.gsu.edu/hbase/mechanics/carbank.html if you can figure out the angle.  So if you know what the angle should be, that equation will get you the coefficient of friction.\n\nIf you can't figure out the angle, at least attempting to solve it using that equation and some reasonable angle should get you some points I'd imagine.\n\nIf I can think of the correct angle I'll post an edit to this.
806201
In order to solve these types of problems, you should always draw a free-body diagram to determine all the forces acting on the sled.  In this case, there are 3 forces acting on the sled:\n\nWeight force, denote it 'W'\nForce exerted by rope, denote it 'F'\nNormal force exerted by hill, denote it 'N'\n\nWhen you draw the forces, W is directed vertically, F is parallel to the hill incline going up, and N is perpendicular to hill incline.  Now denote the axis to set up Netwon's 2nd law equations.  Since we can create these axes in any way we want, you should set them to make calculations easier.  In this case, let's make the axese coincide with N and F.  We can set the axes in this manner because they are perpendicular.  Let's denote the 'x-axis' parallel with F direction and 'y-axis' parallel with N direction.\n\nNow set up the force equations.  The only force we need to break up is W because this force does not coincide with the x or y-axis.  We need to break up the components parallel to these axes in order to solve the problem.\n\nBy using trigonometry, you see that W*cos(theta) component is parallel with N, going in the opposite direction.  Also, W*sin(theta) component is parallel with F, going in the opposite direction.\n\nNow write Newton's 2nd law equations (sum of forces in one direction = 0)\n\nx-direction ---------> F - W*sin(theta) = 0 -------> F = W*sin(theta)\ny-direction --------> N - W*cos(theta) = 0 ------> N = W*cos(theta)\n\nYou can now solve for both F and N.\n\nForce exerted by rope (i.e. F) = W*sin(theta) = 58*sin(34) --->\n\nF = 32.43 N\n\nForce exerted by hill (i.e. N) = W*cos(theta) = 58*cos(34) ----->\n\nN = 48.08 N\n\n-----------------\n\nHope this helps
281995
Is it elasticity? Try to solve the problem with some formula below:\nYoung's Modulus(Y)=Stress/Strain\n=(F/A)/(dl/L) \nwhere, F=force applied, \nA=cross sectional area, \ndl=leangth increase and L=total lenght.\nBulk Modulus(B)\n=dP/(dV/V) \nwhere P=pressure and V=volume.
498153
Think of it like a line of people with their arms linked. Surface tension is just a measure of how hard it is to break through the surface. So, arms linked tighter? Harder to break through the line.
121693
cmon...dont u know basic physics...just calculate the resultant of these 2 forces\nanswer= [(132)^2 + (181)^2]^1/2\nif this makes sense!!
102434
Part (a): \n   Given three values (Mrock = 1.2 kg, Xrock = 5 cm, Xstand = 14 cm), and we need to solve for one unknown (Mstick).  To clarify, Mrock means mass of rock, and Xrock means position of the rock on the meter stick.\n   Since you're being tested on this concept, I'll assume you know the equation for torque: T = Force x Lever arm.  We will also use the concept of Static Equilibrium, which is a fancy term that means that nothing is moving.  Since this system is in static equilibrium, we can use the equation T(clockwise) = T(counterclockwise), using the stand as our pivot point (14 cm).\n   To find clockwise torque, we need to find the Lever arm from the center of mass to the pivot point.  The center of mass is located at 50 cm, and the pivot point is at 14 cm, so our lever arm is 50 cm - 14 cm = 36 cm.  T(clockwise) = Mstick * g * 36 cm  (It's okay to use units of cm in this problem because it cancels out in the end.)\n   If the rock is hung from a massless string, we can assume that the force on the meter stick from the string, which is the tension in the string, is equal to Mrock * g.\n   Counterclockwise torque: Lever arm = 14 cm - 5 cm = 9 cm.  Torque(counterclockwise) = Mrock * g * 9 cm (plug in numbers later to simplify calculation)\n   Torque(counterclockwise) = Torque(clockwise) ==>  Mrock * g * 9 cm = Mstick * g * 36 cm  ==> g's cancel, manipulate equation to get: Mstick = (9 cm / 36 cm) * 1.2 kg = 0.3 kg\n\nPart (b):\n   Since the stand is not causing the meter stick to accelerate upwards, it must be exerting a force upward equal to the force the meter stick exerts downward.  The total force from the meter stick is equal to (Mstick * g) + (Mrock * g) ===>  (Mstick + Mrock) * g ===>  1.5 kg * 9.8 m/s^2  ===> 14.7 N\n   I hope this helps (I'm a physics major...this should be a reliable answer)
816666
Let A = ten of rope at 28 deg angle from ceiling\n      B = ten of rope at 71 deg angle from ceiling\n \nSum forces in x direction\n\nA cos 28 = B cos 71\n \nsum forces in y direction\n\nA sin 28 + B sin 71 = 254.8\n \nSolving for A & B\n\nB = 254.8/ (tan28 x cos 71 + sin71)\n \nA = B cos 71/ cos28\n\nyou crunch the numbers.
383953
In what domain? \nWith what force?\nWith what resistances?
852384
I can't help you with the first question, but to figure out the second one you just need to use Hooke's law (I'm assuming the spring to be almost perfectly elastic.) In equation form the law looks like: F=k*x\nWhere F is the force applied, k is the spring constant, and x is the deformation.\nThe force applied will be the objects weight, so 9.8*.5=4.9N, and the deformation is .04m.\nSo 4.9N=k*.04m\nk=4.9N/.04m\nk=122.5N/m
760545
If a weight is hanging from a V formed by two ropes, the angle of the V cannot be 180 deg, unless the weight is 0.\n\nThis is because the Tensions in the two ropes must compose to a force equal and opposite to the weight, which is acting vertically down.\n\nIf the angle is 0, then each tension is half the weight = 500/2 = 250 N\n\nFor any other angle, we need to know the relative lengths of the ropes.  If they are equal, the tensions are equal and\n2Tcos(a/2) = weight.\nwhere a is the "V" angle.
224079
try this site... it shows step by step how to go about sloving these equations, good luck:
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