Can someone check if I did this physics problem (involving a circular pulley, tension, and a mass) correctly?
918983
This doesn't LOOK like a real problem.  So, I'm not going to play your game.  Sorry.
747946
It's a bit hard to tell without the drawing, but I think I can see how this is set up.  Start by calculating the applied force, based on what you know about the spring.  F = -kx, so the applied force is 0.053 m, the compression of the spring, multiplied by 470 N/m, the spring constant; the negative sign can be ignored in this case, because we're dealing with magnitudes, not vectors.  Now, if the block doesn't slip, the upwards friction force must equal the downwards gravitational force.  Call the spring force F_spring.  The friction force, then, would be F_friction = u*F_spring, where u is the unknown coefficient of static friction.  The force of gravity, the weight, would be W = mg, where m is the 1.9 kg mass of the block and g is gravity, 9.8 m/s^2.  So F_friction = W ==> u*F_spring = mg ==> ukx = mg, and u is your only unknown.
784797
Suppose there is load and two men are pulling it in one direction each one giving a force of 5N.\n\nIn the language of vector, two vectors of magnitude 5 are acting in the same direction,\n\nThe resultant is a vector of magnitude 10N in the same direction. \n\nIf they are pulling it in opposite direction the load will not move.\n\nIn this case the two vectors of equal magnitude are acting in opposite direction.\n\nThe resultant of these two vectors is zero.
909402
Well, it's Wednesday morning.  I hope this answer still has some value to you.\n\nI started by summing the torques at the hinge.  There are three:\nT(light)=5*9.8*7.5*cos(37)\nT(cable)=-tension*6.31*sin(37)\nT(pole)=8*97.5/2)*cos(37)\n\nThese all sum to zero.\n\nI computed tension=140.5\n\nThe sum of the horizontal forces must be zero.\nThese are the tension in the cable and the horizontal reaction force at the hing.  So the horizontal component is 140.5\nThe vertical forces are the weights of the pole and the light offset by the vertical reaction force at the hinge\nthe vertical component is (8+5)*9.8\n=127.4\n\nI checked the answer by computing the magnitude of the reaction force two ways.  Using Pythagorean theorem \nsqrt(140.5^2+127.4^2)\n=189\n\nI then summed the forces parallel to the pole:\n5*9.8*sin(37)\n8*9.8*sin(37)\n140.5*cos(37)\n=189\n\nj
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"::
239996
Work done = force x distance moved by object i thinkin a straight line ,so ithink some conversion maybe needed.
574012
I've alread answered two of your questiions in my last answer....\n\nIn this one, the guys above are correct - however, I'd try working it out for myself if I was you. Your tutor will soon realise something is wrong if you are giving answers to a subject you don't understand. Read the chapter, ask you teacher if you need more help. Learn Newtonian mechanics for yourself.
914856
OK as the ball is swung around in a horizontal circle the scale inside the tube is measuring the tension on the string. By knowing the time to complete one revolution  linear velocity can be found from v=r*omega(angular velocity). in the second part the v is given and needs to be converted into radians/second by using 3.7=.5*omega where omega=3.7/5=.74rad/s then using the mass radius and angular velocity you can find the tension in the string. With a= acceleration (into center) omega^2*r=.74^2*.5=.2738m/s^2and using f=ma the tension in the string will equal f which equals .2*.2738=.0548newtons Unless i have a silly math mistake somewhere.
1119730
ahh, good theory problem...\nlet's see, there's energy, friction, force (normal, tension, as well as the other ones mentioned), gravitation, mass, pendulum motion, gotta be more that I've missed, but no doubt someone'll be here to correct me.
893876
Simply keep in mind that summation of forces equal to zero.\n\nSum (Fx)=0\nSum (Fy)=0\nlet th\nF1- force o string supporting W1 (weight on the left)\nF2- force o string supporting W2 (weight on the right)\nF3- force o string between the weights\n\nFx=0=F1cos(54) - F2cos(68) +F3cos (13)\nFy=0= W1+W2-F1sin(54) - F2sin(68) -F3sin (13)\n\nwe have 2 equations and three unknowns.\nSince the dimensions or length of the ropes ar enot given we can't employ moments.\n\nlet's use superposition and solve one side at a time\n\nW1=F1sin(54) then F1=W1/sin(54)=37.73/sin(54)=46.6N\nsimilarly\nF2=W2/sin(68)=90.944/sin(68)=98.08\n\nsince \nF1cos(54) - F2cos(68) +F3cos (13)\n\nthen  F3=(F1cos(54) - F2cos(68))/cos (13)\n\nF3=(46.6cos (54)-98.08cos(68))/cos(13)= - 9.6 N\n\n(Negative sign means that the originally assumed 'positive' direction to the left was wrong. F3 is pointing to the right.)
481598
There are three objects; 5kg, 9kg, and a cable of negligible mass.\n\nWhen the three move, they must have the SAME acceleration.\n\nTherefore, consider the three as a single object of mass 5 + 9 = 14 kg, with acceleration ‘a’.\n\nThe only force that is pulling the three is the force given by the 9kg object. (The gravitational pull is only on this 9kg.)\n\nThis force is 9 x 9.8 N = F.\n\nSince F = M a,   9 x 9.8 = 14 x a.\n\nFrom this a = 6.3 m/s^2.\n\nAll the three objects are moving with an acceleration 6.3 m/s^2.\n\nIf the body 5kg is moving with an acceleration of 6.3 m/s^2, then what is the force acting on it?\n\nWe know that f = ma; hence, f = 5 x 6.3 = 31.5 Newton.\n\nSince the tension of the rope is the pulling force, tension =31.5 Newton.\n\nThe tension can be calculated using the forces acting on the other end of the cable also.\n\nThe downward force on 9kg is 9 x 9.8 N; the tension is the upward force T. \n\nThe net force acting on 9kg is (9 x 9.8 - 9 x 6.3) = 9 x 3.5 = 31.5 N = T.
38713
well whatever force that is being exerted on the rope, that is going to be the tension. that means that all you have to do is calculate the force.  you just have to realize that not only is the gymnist accelerating, producing force, she is also overcoming the force of gravity. therefore, if you were to draw your free body diagram, you would have 2 forces pointing down, the force of gravity(mass times gravity), and this acceleration force(which would just be mass times the acceleration.  if you add these two forces together, you get the total force being exerted on the rope which is equal to the tension. here is the equation:\n\nT=F + Fg  (where T is tension, F is the force caused by the acceleration, and Fg is the force due to grabity.)\nT= ma + mg\nT= (74)(.11) + (74)(9.8)\nT= 733.3N.\n\nthis can also be done by just adding the acceleration to the acceleration due to gravity then multiplying this new acceleration by the gymnist's mass. either way you will get T=733.3N
918781
Where is your freebody diagram? There are other forces acting on the block than just gravity...\n\nGood Luck
351440
well ok. for part a, the acceleration is in the horizontal direction, because its moving horizontally. so the force is just mass x acceleration which is 70*6=420N. but thatz only in the horizontal direction.\n\nfor part b, you have to account for gravity. so gravity is pulling down at 9.45m/s^2, but youre accelerating upwards at 6m/s^2. so that means that youre overcoming the gravity and accelerating at 6m/s^2 MORE than the gravitational acceleration.\nso 9.45+6 = 15.45. this is your actual acceleration.\nso now you just do F=ma again and you get 70*15.45=1081.5N. \n\nhope this helps. let me know if you need more help. and this isnt thermodynamics.
833368
find the force of her weight. \nthat should be the answer
252925
The two simplest pulley systems demonstrate the principles of their operation.  First if there is a single pulley overhead and you pull down to lift a load, you must pull with the same force as the weight of the load because oonly one rope pulls up on the load.  100 pound load = 100 pound pull.\n\nSecond, if the load is below you and you pull up on a rope connected to your own level through a pulley, you pull with a force one half the weight of the load because two ropes are pulling now up on the load.  100 pound load = 50 pounds pull + 50 pounds on fixed rope.\n\nThe overhead pulley required you to pull the rope say 10 feet to lift the load 10 feet; but the pulley below you required you to pull the rope 20 feet to lift the load 10 feet.\n\nYou can calculate more complicated pulley systems either by counting the number of ropes that are under tension that actually lift the load and dividing the load by that number or you can determine the length of rope you must pull to raise the load a given distance and use that ratio to determine your force vs the load weight.  you can do this by making sketches rather than actually setting up real pulleys, ropes and loads.  Good luck.
806029
1st link \n\nb) F = 40 N\n\nc) F = 60 N\n\n2nd link\n\nb) linear velocity = 0.59 m/s\n\nc) Taking g = 9.8 m/s^2\n\nv = 0.83 m/s\n\nd) It would not effect the answer as the expression for v which is sqrt(mu * r * g) is independent of the mass. \n\n3rd link\n\na) Rotate the setup, note the reading on the balance. Find the time required for 1 circuit.\n\nb) No work is done\n\nc) T = 5.48 N\n\nd) 5.9 %\n\ne) If the cord is exactly horizontal, it won't touch the rod and hence cannot be rotated.