PLEASE HELP! Phyics Problem with Force?
927827
K so...I haven't done any Grav froces recently...but I'll try\n\nIm guessing use  Net force = ma\nThe net force is the Gravitational force, Gm1m2 / r^2 is = m2 4 (pie)^2 r/ T^2\n\nfor the 4 (pie)^2 r/ T^2    r is the distance form the planet's center, to the moon, for T, find how many second 40 days is.\n\nand the other r is the planet's radius\n\nm2 cancels\n\nsolve for m1, and yo uhave the mass of the planet.\n\nI guess plug in numbers after that, into the equaltion Fg = Gm1 m2  /  r^2\n\nm2 would be the 70 kg\n\nhope thats right
210676
Use free body diagram to balance the forces and u can derive the equation of motion \nIf u want i can give u full derivation.\n"::
67664
friction is F=uN where u is the coefficient and N is the normal force.  In this case 260N is the normal force. So:\nu*260N=150N* cos 40\nYou must subtract off (150 * sin 40), also because you are counteracting part of the normal force.\nSo the final eqn should be\nu = (150N* cos 40) / (260N - (150* sin 40))\nLet me know how it goes\nemail: ke7alw@yahoo.com
966154
The magnitude of a force is measured in newtons(N)
494540
Haha whats with all these answers, if you know basic physics, One of Newtons laws of motion is that:\n\n"for an object that is stationary or travelling at constant velocity, no force acts."\n\nso you ask what is the force?\nthere is no resultant force, the driving force and resistance are equal and cancel each other out.\n\n\nAnd additionally the basic equation to find force is:\nF=ma (force = mass x accelleration), this is evidence that an accelleration needs to be taking place for a force to act.\n\nLets say you want the force for a 50kg object accellerating at 3ms-2. F=ma  ->  F=50x3 ->  F=150N.\n\nNow lets say you want the force for a 50kg object travelling at constant velocity, no accelleration.\nF=ma  ->  F=50x0  ->  F=0N\n\n\nThere you go, wasnt that easy
127000
This problem is solved with the application of work and energy principle which states that total work done on a moving particle is equal to the change of its kinematics energy:\nW=T2-T1\nHere the only work done on the particle is dot to gravitation force and is equal to -m*g*h:\n\n-m*10*0.45 = (1/2)*m*(1)^2-(1/2)*m*(v)^2 --->\nm*v^2=(4.5+0.5)*m*2=10*m --->\nv=sqrt(10) m/s\n\nv is equal to sqare root of 10.(approximately 3.162 m/s)
1014316
would you like to supply me with some formulas? im out of grade 12 physics and i forgot most of the formulas. trust me from experience, just plug numbers into a formula and things will work\n\nF=mg = ma\nFk=μkN\na= vf^2 - vi^2 / 2d
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
812295
Er, gravity really has nothing to do with this question. Fk = coefficient*Fn.  Fn = mg.  -19.6 is your Normal Force.  But yea, the answer is correct.
879813
you need to identify that there're two parts of the force, one acting horizontally, the other acting vertically.\nSo:\n1) the vertical part is:\n70sin30\n2) tje horizontal part is:\n70cos30
771534
The problem is about conservation of angular momentum.  The tricky part is (b). To solve this use this definition of angular momentum and draw a picture to find out the value of r sin θ\n\nThe angular momentum of a particle of mass m with respect to a chosen origin is given by vector product L = r x p= mvr sin θ
617970
I am terribly afraid of the equilibrium problem.. the first one on the free response!!! well, good luck!
152239
AIR RESISTANCE CANCELS OUT THE FORCE DUE TO GRAVITY , THERE IS NO ACCELERATION BECAUSE THE OPPOISTE FORCE IN THE SAME.\nSO YOU CAN SAY 0
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.
932992
it is a much easier problem than it seems..\nthe key point is the fact that the body is moving at constant speed. meaning that the net force on it is zero --> no acceleration\n\nthere are two forces. friction and the horizontal component of the force that is exerted at 40 degrees.\n\nso simply ...\n\n\nFx = Friction==>\n976 cos40 = coeff. *N *g\n\n\n now there is equilibrium at the vertical axis as well.\n\nmeaning that \n\nN+Fy = m*g =>\n\nN = (mg- fy)\n\nso we mix them together and hve\n\n976cos40 = 0.667 (mg- 976sin40) =>\nm=50.31kg\n\nin this problem you use that there is equilibrium in both of the x and y axis.
969486
I just helped someone else with their physics question, so I'm alittle tired of it.  But, I suggest you make an equation sheet...physics is all about equations.  So, if you have a piece of paper with your equations, all you have to do is look at the sheet and plug things in...
663213
I am sorry I can not provide you with an answer to your question. However it might help you get more answers if you put this question in again but put it in the Science and Mathmatics section. Under Physics. I think there are more people willing to help you out there!