A 1 kg mass on a frictionless inclined plane is connected by a pulley to a hanging 5.0 kg mass,?
957436
By drawing a sketch diagram, you can work out the system in question can be represented as a right angle triangle, with the hypotenuse equal to 1 (kg) and the opposite side = 0.5 (kg).  the angle you are trying to calculate is between the hypotenuse and adjacent, and is given by the formula \n\nsin a = opp / hyp\n\ntherefore sin a = 0.5 / 1\nsin a = 0.5\n\na = arcsin(0.5)\n\na  = 30 degrees.\n\nAs both weigths are acting in opposition to each other, the angle will be positive, so your answer is option 2, 30 degrees.\n\nHope this helps!
313833
So, this looks like a homework problem, but this is how you would set it up.  And, it looks like the beam is massless...\n\nFirst, you know that the beam's not going anywhere, so the sum of the forces on it must be equal to zero.  That is, the sum of the masses must be equal to the sum of the forces exerted by the two supports-- 5N+80N+30N=s1+s2, where s1, s2 are the two forces.\n\nNext, you know that the beam's not rotating, so the sum of the torques on it  around any point must be zero.  To make things simpler, I'll set up the equation with the torques about end A--  5N*1m+80N*2.5m+30N*1.25m-s2*4m=0.  Note the difference in sign of the torque produced by s2 relative to the other forces.\n\nNow you see that the second equation has only one unknown-- s2.  Solve for s2, then plug that value into the first equation.  Then solve for s1.
865118
Force acting along the inclined plane[30.1]\n=mgsin30.1\nForce acting normal to the inclined plane\n=mg cos30.1\nForce of kinetic friction opposing the motion\n=.242*mgcos30.1\nWhat is the net acceleration?\n=gsin30.1-.242*gcos30.1\n=g[.5015-.242*.8651]\n=.2922g\nA]If we let the crate roll, it will accelerate due to gravity with out any external force being applied. To make it move in uniform motion we have to apply force in opposite direction\n=.2922*9.81*30.3\n=86.85 newtons\nB]work done by the force\n=86.85*4.75\n=412.55 newton-meters\n=412.55 joules\nC]Force of friction*4.75\n=30.3*9.81*cos30.1*.242*4.75\n=295.60 joules\nD]zero, since it acts perpendicular to direction of motion\nE]Work done by gravity\n=mgsin30.1*4.75\n=30.3*9.81*.5015*4.75\n=708.06 joules\nF]as in B]
202303
i will assume that you just asked how to find the mass of an object using a meter rule \nif that was the question it is simple\ntake the meter rule and hang it in such a way that it is in equilibrium i.e the scale is hanging horizontally\n\nknown hang the known mass on one side and hang the other mass on the other side\narrange in such a way that it is in equilibrium\nnow find the ditance of the known and the unknown mass from the position where the meter scale is tied and let it be a & b\n\nnow using the Principle Of Moments\n\nM*a=m*b\n\nhope it was helpful
269311
assuming that force is applied in the same direction that displacement....\nwork = force * displacement * cos(angle between force and displacement)\nwork = (10N)*(5.0m)*cos(0)\nwork = 50 Joule
754118
E1=M*g*h, M=(your mb)=32kg, - pot energy of the system at rest,\nh=2m and g=9.8m/s^2.\nE2=kinetic of M=32kg\nE3=kinetic of m=20kg\nE4=kinetic of pulley u=5kg\nE5=potential energy of m at the height h, E5=m*g*h\nThus energy conservation: E1=E2+E3+E4+E5.\nE2+E3=(M+m)*V^2/2, V to be found.\nE4=I*W^2/2, angular speed being W=V/r, r=0.1m, and I=u*r^2/2 for your pulley.\nHence 2*M*g*h=(M+m)*V^2+(u*r^2/2)*(V*r)^2+2*m*g*h,\n(M+m+u/2)*V^2=2*(M-m)*g*h, - does not depend on r.\nV=2.08m/s, don’t believe & check it! – good picture gal – a piece of art.
545558
Force = mass * acceleration\nForce = 1300 kg * 2.6 m/s2\nForce = 3380 Newtons
173724
If it is a fixed pulley, you need the same force as the weight of the load which is 9.8*0.2 Newtons (9.8 is the acc. due to gravity and 0.2kg is the mass of the load) or simply 200gf (where gf is gram force). Now you can tell one half of this number.\n\nAlternately, if you have a movable pulley, the force required is half the weight of the load, i.e., 100gf.  You can tell one half of this number too.\n\nIf you need to find the difference between fixed and movable pulleys, look up a physics text (or science text), I am not totally familiar with the schooling system here.   Otherwise, Google should show you pictures of fixed and movable pulleys.
796334
Consider the forces and torque in this problem. Setting up so that the x-axis is positioned parallel to the incline\n\nSFx = m*a = m*g*sin(q) - Fs\n\nwhere m is the mass of the cylinder, a is the acceleration, g is the acceleration due to gravity, q is the angle of the incline and Fs is the statis friction force. For the torque\n\nST = I*a/r = r*Fs\n\nso that we have for the statice friction force\n\nFs = I*a/r^2.\n\nPlugging this into the first equation\n\nm*a = m*g*sin(q) - I*a/r^2\n\nand solving for the acceleration\n\na = m*g*sin(q)/(m + I/r^2)\n\nNow, finding the time to travel a length L can be done in several ways. The first, and perhaps simplest is\n\nL = 0.5*a*t^2\n\nso that\n\nt = sqrt(2*L/a)\n\nAlso, you can consider how many revolutions of the cylinder is takes to achieve the same distance. This mean dividing the distance L by the circumference of the cylinder. Multiplying the result by 2*pi gives the angular travel so that the time can be determined by\n\n2*pi*L/(2*pi*r) = 0.5*alpha*t^2\n\nwhere alpha is the angular acceleration and is equal to a/r. \n\nConservation of energy can be employed as well\n\nm*g*h = 0.5*m*v^2 + I*w^2\n\nwhere h (= L*sin(q)) is the height, and w (= v/r) is the angular speed. That is to say that the potential energy at the start is equal to the kinetic energy after traveling through the distance L along the incline. Solving for v yields\n\nv = sqrt(2*m*g*L*sin(q)/(m + I/r^2))\n\nwhich is equal to the acceleration times the time traveled. From this \n\nt = sqrt(2*m*g*L*sin(q)/(m + I/r^2))/a
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.
746965
You don't need to find the mass if use conservation of energy to solve the problem.\n\nThe kinetic energy of the can initially is:\n1/2*m*v^2\n=.5*m*2^2\n=2*m\nand when it comes to rest it is zero\nThe energy was lost to the work done by friction.\n\nthe force of friction times displacement is the work:\n\nu*g*m*d\n=.2*9.8*m*d\n=19.6*m*d\nnow set the energy of the system equal to zero:\n2*m-19.6*m*d=0\ndivide out the mass and solve for d:\nd=19.6/2\n=9.8m\n\nj
725904
this is the amount of force.  Force equals mass times acceleration.  The mass is ur bodyweight in kilos and the acceleration is gravity 9.8m/s/s
702358
Why did you post this question twice?\n\nYou need to be careful when you draw the force diagram.  When I read the question, I drew the pulley directly above the box, but when you do that, the angle turns out to be arctan .620 = 31.8 degrees, independent of the weight.\n\nTo get the diagram right, the pulley must be placed "up the incline" so that the 45N force acts parallel to the inclined plane.\n\nHaving said that, here we go.  Let the angle of inclination be x.  The vertical force on the box is 130N.  Resolve that into two components -- one normal (perpendicular) to the plane (Fn) and one parallel to the plane (Fp).  These forces are\n\nFn = 130 cos x  and  Fp = 130 sin x\n\nThe frictional force, directed up the plane, is Fn times the coefficient of static friction 0.620.  You also have the 45N force directed up the plane.\n\nPut all this together to get the forces up and down the inclined plane:\n\nFp = 0.620 Fn + 45 \n\n130 sin x = 0.620 (130 cos x) + 45\n\n130 sin x = 80.6 cos x + 45\n\nThis is not an easy equation to solve (even though we know the answer).  I think we need a numeric solution using trial & error (successive approximations).  To jump ahead, let's try the teacher's answer:\n\n130 sin 49 = 80.6 cos 49 + 45\n98.11 = 97.88\n\nClose.  Suppose we don't know that 49 degrees is the right answer.  Let's start by defining a function\n\nf(x) = 130 sin x - 80.6 cos x - 45\n\nand try to find f(x) = 0.  Here goes:\n\nf(45) = (130 - 80.6) sin 45 - 45 = -10.07\n(because sin 45 = cos 45)\n\n(If I were using calculus, I'd probably take the derivative now to get my next approximation.  But I won't do that since you may be taking a physics course that does not use calculus.)\n\nFor our next guess, let's try x = 50:\n\nf(50) = 130 sin 50 - 80.6 cos 50 - 45 = 2.78\n\nWe have f(45) = -10.07 and f(50) = 2.78.  To do linear interpolation, we need the equation of the line through those points.  The slope is m = (2.78 + 10.07)/(50 - 45) = 2.57.\n\nInterpolation works like this ... y - y0 = m(x - x0) ... we want the value of x when y=0, so we set y=0 and solve for x.  When you do that, you get x = x0 - y0/m.  Our next approximation is\n\nx = 45 - (-10.07)/2.57 = 48.9 degrees\n\nf(48.9) = 130 sin 48.9 - 80.6 cos 48.9 - 45 = -0.02\n\nWow!!!  That's a better answer than 49 degrees.  Go with it.  To see that 48.9 is best, compare these three results:\n\nf(48.8) = 130 sin 48.8 - 80.6 cos 48.8 - 45 = -0.28\nf(48.9) = -0.02\nf(49.0) = 130 sin 49 - 80.6 cos 49 - 45 = 0.23\n\nand we were trying to get f(x) = 0.  So your best answer is 48.9 degrees.\n\nBTW, this was a good problem because of the need to get a numerical solution.  We used linear interpolation, which worked fine.
905447
The net force is 0 Newtons. The force of friction is 100N.\n\nThe key here is that\n1. No acceleration.\n2. Horizontal force only.\n\nThis means that the forces are balanced in the horizontal direction. The object isn't accelerating so net F = ma = 0.
193176
well F = kx. im assuming you have to find out the spring constant in this. So the force is the force of gravity that is acting on the mass which is 9.8*0.55kg which is 5.39N. and the x is the amoutn it stretches, which is 36cm or 0.36m. so then\n5.39N = k (0.36m)\nk = 5.39N / 0.36m\nk = 14.972222222222222222222222 N/m. \n\nhope that helps. let me know fi u need more help. later
701530
Plug the numbers into the Newtons Equation :\n\nF =  (GMm) / r^2\n\nF = Gravitational Force / weight  \nG = Universal gravitational constant   6.673*10^-11 (note the - sign on the exponent)\nM = mass of earth (kg)                 5.98*10^24 kg\nm = mass of satellite (kg)             150kg\nR = radius of earth    (m)             6,378,137 m\nh = height above earth (m)             12,800,000m\nr   = h+R\n\nThe answer will come out as the weight being @ 162.67N\n\nNot weightless, although someone standing on the satellite would feel weightless as its in continuous free fall.\n\nEver been in an elevator and felt lighter as it falls or been weighless on the big dipper as it plummets?\n\nThe gravitational force (weight) of 162.67N is what keeps it in orbit and prevents it from flying off at a tangent. Basically it pulls the satellite towards the earth but the satellite is travelling so fast it misses and continues round in its orbit.\n\nAdded:\n\nWeight is measured in Newtons and is the attractive force between the two masses. (M & m)\n\nIn this case the earth and the satellite. A 1 Kg mass on the surface of the earth will be pulled towards the earth's centre with a force of 9.8N, therefore its weight at ground level is 9.8N. If you plug take m as 1kg, h as 0 and plug them into Newton's formula should get the answer 9.8N, which is where we get the constant g from. \n\nAs you get further away from the centre of the mass the gravity field decreases in inverse proportion to the distance squared. Hence the divide by r^2 on Newtons fornula. Therefore the weight of a 1Kg mass at 12,800km is 1.08N.\n\nBecause the satellite is in a stable orbit the opposing forces of the satellite trying to escape orbit and the gravitational pull towards the earth are equal so yes the weight is 162.7N and the centrefugal force is 162.7N. This is the correct answer to the problem. \n\nAs an aside; for simplicicities case in this scenario the mass of the satellite can be regarded as constant due to its relatively low velocity compared to the speed of light. \n\nHowever as you come closer to the speed of light mass increases according to Einstein's theory of relativity, eventually reaching infinity at light speed.
975661
Depending on which reference you go by, it is in the vicinity of 6.0 X 10^24 kilograms.  \n\nSince, 1 kilogram = 0.00110231131 short tons we can say that the earth weighs approximately 6.61386787 x 10^21 short tons.