Find the force needed to pull a mass of 30kg at constant velocity on a rough incline making 40' angle & uk=.3?
105644
darkmantis, you lost me at "mgsin40, which is 56.7 Newtons".  \nm=30, g=9.8, sin40=0.6428, mgsin40=189.2 Newtons\nso total=67.6+189.2=256.8 Newtons
961494
This is an energy balance situation.  The work done in dragging the object must equal the energy change.  In this case the energy is potential energy, given by m*g*∆h, where ∆h is the change in vertical distance.  The work (energy) required to drag the block is force times distance along the incline.  Equate these two quantities and solve for F.
939535
Sorry, but you'll need to give a little more information. How are the pulleys and weights arranged? Just a little picture would help.\n\nTypically, pulleys, weights, tensions and accelerations are extensions of the block on an incline problem except the pulleys are providing the force vector direction change.\n\nIt might help to think of the problem as a sum of vectors and look for where their vector sum lies.
883912
Let the banking angle be '$'.\nThe reaction force will be perpendicular to the banking surface.\nLet the reaction force be 'R'.\nLet 'm' be the mass of the car.\nvelocity of car (v)=27 mt/s\nradius(r)=60 mt.\n\nThe centrifugal force will be in horizontal direction. And this force will be balanced by a component of the reaction force 'R'.\nR sin$= mv^2/r ---------------------------(1) \n\nThe weight of the car will be vertically downward. This weight will be balanced by Rcos$.\nRcos$=mg  ----------------(2)\n\nNow dividing eqn(1) by eqn(2)\n\nRsin$/Rcos$=mv^2/r /mg\n=>tan$=v^2/rg = 27 x 27/60 x 9.8 = 729/588 = 1.239\n\n$= tan(inverse)(1.239)\n\nsolve above tan(inverse)  function. and find the value of the banking angle
743343
Just like any other weight on an incline problem, draw a right triangle.  However, the lines aren't distances, they're force vectors.  The vertical vector = 17,000.  The horizontal line is irrelevant.  The hypotenuse is the force vector you want.  It represents the force of the truck sliding down the hill.  Basically, find this vector, and that's how much force of friction it takes to prevent sliding (so net force on the truck = 0).\n\nThe angle between the horizontal line of the triagle and the hypotenuse is 4 degrees, 20 minutes.  You want the other (non-right) angle.  SInce all three angles add up to 180 degrees, 180 - 4degs/20min (change that to decimals to save headaches) = the angle you need.\n\nNow, cos(that angle) = 17,000 / (the value of the vector your solving for, all it X).  Rearranging that gives you X = 17000/cos(that angle).\n\nHappy solving.
739982
sin 30= weight at end/ actual weight\nactual weight = 20/ sin 30\n= 40
746185
I can only answer 1 and 3 for you, cause i dont know the others, but here it goes:\n\n             1.) Think of this problem in terms of a right triangle. Draw a horizontal line. This line is the tidal force of 3* (10^3) N. Then draw a line on the end of the line perpendicular to the horizontal one. This line is your wind force of 6* (10^3) N. Connect the two ends of the lines with another line, to make a complete right angle triangle. This is where it gets a little tricky. The angle formed when the slanted line and the horizontal line meet is the angle you are trying to find. To find it, use the tan function, which is defined as opposite side's value (6*10^3 N) over the adjacent side's value (3*10^3 N). Lets use x as a variable for the unknown angle. So...\n\ntanx=(6*10^3) / (3*10^3)\nUse inverse of tan now (on calculators, you usually have to press shift or 2nd, then the tan key.) So...\n\nInverse tan ( (6*10^3) / (3*10^3) ) = x\n\ndo that in your calculator, and the value you get is the angle.\n\nFor the net force just subtract (3* 10^3) from (6*10^3) and the result will be your net force.\n\n3.) I dont know if u know this or not, but there is a formula that is part of Newton's second law. It is F=m*a.\n"F" being the Force, while "m" is the mass, and "a" is the acceleration. Just plug in the values you are given so that it comes out....\n\n390=270*a\n\n390/270=a ("a" is in meters per second squared)\n\nHope that helps
888459
v(t)= -gt - ve(ln((m - rt)/m))\n\nx(t) = ∫v(t) dt\n\n= ∫[-gt - ve(ln((m - rt)/m))] dt (Note ∫lnu du = ulnu - u + k)\n\nLet u = (m - rt)/m Thus du = -r/m dt ie dt = -m/r du\n\nSo x(t) = -½gt² - ve*-m/r (u(lnu - 1) + c\n= mve/r [ (m - rt)/m * (ln((m - rt)/m)  - 1] - ½gt²\n= ve(m - rt)/r * (ln((m - rt)/m) - 1) - ½gt²\n\nIf g = 9.8 m/s^2, m = 30000 kg, r = 155 kg/s, and ve = 3000 m/s and t = 1 minute = 60 s\n\nx(60) = 3000 * (30000 - 155 * 60)/155 * (ln (30000 - 155 * 60)/155 - 1) - ½ * 9.8 * 60²\n\n≈ 138390 m\n\n= 138.39km
593602
The answer is to draw an outline of the box and sketch all forces acting upon it.\n\nThen, resolve those forces into their horizontal and vertical components.\n\nThe box is not required to accelerate in order for constant-velocity motion to occur.  So, it is simpler than F = ma.  Just use Sum (F horizontal) = 0, and sum (F vertical) = 0.\n\nSolution:\n\nSum of horizontal forces:\n\nF - 0.3 * N * cos(12) * N * sin (12) = 0\n\nSum of Vertical forces:\n\n- mass * gravity - 0.3 * N * sin(12) + N * cos(12) = 0\n\nBe careful with pounds as a unit of mass.  Better to convert to slugs by dividing by 32.17 ft/s^2 (gravity).\n\nFirst solve for N (the normal force perpendicular to the plane acting upward on the box) using the vertical force eqn:\n\n-(35/32.17) * (32.17) - 0.3 * N * sin(12) + N*cos(12) = 0\n\nThen plug N into the Horizontal Force sum equation to solve for F.\n\nWatch your units!
77860
hey, didnt i answer something very similar to this for you? you need to do F=ma so 43000000N/2100000kg then you get an acceleration, but then you have to subtract gravity from this because thats the acceleration produced by the rocket but there is still gravity pulling with 9.8 m/s^2 in the opposite direction. so just do (43000000/2100000) - 9.8, and that's your answer :)
784311
If M0 is the mass at the origin and My is the mass on the y-axis, and Mx is the mass on the x-axis then\n\nFy = G*M0*My/Ry^2\n\nwhere Ry is the distance between M0 and My. Likewise,\n\nFx = G*M0*My/Rx^2\n\nwhere Rx is the distance between M0 and Mx. The magnitude of the resultant force (F) is then\n\nF = sqrt(Fx^2 + Fy^2)\n\nand the angle (q) can be computed from\n\ntan(q) = Fy/Fx.
131445
Torque is force*distance.\n\nFind the weight of the person (M*g) = 89kg *9.81m/s^2 \nM*g= 873.09 Newtons\n\nthere is a 1:3 lever arm there--> 4cm/12cm\nThe muscle is having to exert a force of 3*873.09N or 2619N\n\nThe torque around the pivot is 2619N @ 4cm (or .04m)\n2619N * .04m = 104.77N*m\n\nYour question says you want to "find the torque" yet you ask for the "forces pushing up".  Make sure you know which one you need.
861484
1) The angle of the rope needs to be considered to calculate horizontal and vertical components of the clerk's force. \nFh = 188*cos(25) = 170 N This is what pulls the box.\nFv = 188*sin(25) = 79.5 N This actually lifts the box slightly.\n\nThe symbol for coefficient of kinetic friction is µk. A lower number makes it easier to move the box - if you spilled some olive oil on the floor µk would perhaps go to 0.1. N is normal force. If this was on a slope, normal would be different from weight, but the floor here is level so N = W? \n\nActually no. Because the rope pulls up on the box slightly, the normal force pressing the box to the floor is less than the box's full weight. The box's weight = m*g = 36.9 kg*9.8 m/s^2 = 362 N. The vertical component of the tension in the rope relieves 79.5 N of that so N = 362 - 79.5 N = 282 N. \n\nSo Ff = .450*282 N = 127 N is the force that friction fights with. The horizontal portion of the clerk's pull is 170 N. So there is 43 N left over to accelerate the box. Newton's 2nd law says\nF = m*a\n43 N = 36.9 kg*a\nso a = 1.17 m/s^2\n\n2) Here the force is directed down at an angle so the vertical component of the force will add to the normal force used in calculating the friction.\nFh = 435 N*cos(35) = 356 N\nFv = 435 N*sin(35) = 250 N\n\nN = the weight of the box + Fv = 259 N + 250 N = 509 N\n\na) Ff = µk*N = .57* 509 N = 290 N This is not enough to hold the box stationary so 356 - 290 N = 66 N are available to accelerate the box. Newton's 2nd law again:\nBut first we need the box's mass. W = m*g\nm = W/g = 259/9.8 N = 26.4 kg. Now\nF = m*a\n66 N = 26.4 kg*a\na = 66/26.4 m/s^2 = 2.5 m/s^2\n\nThis is getting complicated. To move 9 m will take t seconds:\nx = (1/2)*a*t^2 Solve for t.\n\nb) Ff = µk*N = .75* 509 N = 381 N This is more than the horizontal portion of the clerk's push, so the box will resist. The answer is 0.
792677
F - Ff = m*a\n\nThis is because the force you're applying is in one direction, while the frictional force is ALWAYS in the opposite direction. Frictional forces act to prevent the applied movement.\n\nBecause you're moving at constant velocity, you therefore have no acceleration, so:\n\nF - Ff = 0\n-or-\nF = Ff\n\nFf is the force due to friction which:\nFf = u * Fn\n\nWhere "u" (it's actually the greek letter mu, but this will do) is equal to the coefficient of friction, and Fn is equal to the perpendicular force acting on the box. On flat ground, the normal force is equal to the gravitational force, but on inclines you have to break it up into its vector components.\n\nF = u*Fg\nF = (0.20)*(25kg)*(9.8m/s^2)\nF = 49N\n\nYou see that the velocity at which you push it has no bearing on the answer because neither your applied force nor frictional forces are due to constant velocities.
456850
Do your home work. \n\nStill I tempt to answer this question.\n\nW= weight of the box\nmu= coefficient of friction.\n\nWe assume there is no acceleration. So force to upward movement is Fup = Wcos(20)*mu + Wsin(20),\n\nsimilarly Fdown = Wcos(20)*mu - wsin(20),\n\nso\nWcos(20)*mu + wsin(20) = 4Wcos(20)*mu - 4wsin(20)\n5Wsin(20) = 3Wcos(20)*mu\nmu = 5sin(20)/3cos(20)\nmu= 5/3 tan (20)\nmu = 0.606
848443
when u will draw ur inclined plane and mass on it there r different forces acting in it \nfirst force will be directly perpendicular to the ground level i.e. Fg=mg\nsecond force acting is mgcos(theta) which is perpendicular to hypotenuse of the inclined and makes up Fn=mgcos(theta)\nthen another component parallel to the hyp. of the inclined \nwhich equals mgsin(theta) which goes down the ramp\nand  then u can\nequal the force going down with the frictional force\nmgsin(theta)-(mew)mgcos(theta)=ma\n(as we know frictional force is (mew)Fn=(mew)mgcos(theta) )\nsolve by putting values\nmew=0.146    ans
617628
Ignore the answerer above. Learning physics can be tough.\nSolve as below:\n\nu=30m/s\nv=0\na = -6m/s^2\n\nusing v=u+at\nt=(v-u)/a\nt=(0-30)/(-6)\nt=5 seconds is the answer