Two blocks on a pulley, finding magnitudes, tension, and rotational acceleration...?
970512
According to energy conservation law: m2*g*h=m1*g*h+m1*v^2/2+m2*v^2/2+I*w^2/2.\nm2*g*h is initial potential energy of the system;\nm1*g*h is final potential energy of m1;\nm1*v^2/2 is kinetic energy of m1;\nm2*v^2/2 is kinetic energy of m2;\nI*w^2/2 is kinetic energy of pulley with inertia moment of I & final rotary speed w rad/s^2.\nNow m1=0.46kg, m2=0.5kg, h=0.75m, r=0.05m, t=5s, g=9.8m/s^2.\nAs h=a*t^2/2, then a=2*h/t^2 – linear acceleration of m1 and m2. \nAs v=a*t, then v=2*h/t – linear speed of m1 and m2.\nAngular acceleration of pulley w’=a/r; angular speed w=v/r.\nThus 2*(m2-m1)*g*h=(m1+m2)*v^2+I*(v/r)^2, so I=(extract it please and plug values in).\nT1=m1*(g+a), T2=m2*(g-a). Don’t worry, be happy!
857685
forces on M : grav. force Mg (vertically down) tension 2T(two strings horizontal, one on pulley) tension T (vertically down due to string on pulley) Normal reaction on M at lower surface N (vertically up) Normal reaction on M due to m R (to the left) Friction due to m on M f1=u1 times R (vertically down)(sliding is going on) friction in M at lower surface f2=u2 times N (to the left) (sliding is going on)\nnet force on M vertically up :ZERO\nto the right M times A\nforces on m  : grav force mg (vertically down) reaction due to M R (to the right) tension T (vertically up) friction due to M on m f1=u1 times R(vertically up)\nnet force on m to the right m times A\nnet force on m vertically down m times a\nunknowns N,R,a,A,T,f1,f2,\nwrite a constraint eqn for relating accns a and A. a=2A \nsolve them!!!!
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
524636
If the forces are both acting in the same direction, add them together. = 65 Newtons
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.
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.
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
918781
Where is your freebody diagram? There are other forces acting on the block than just gravity...\n\nGood Luck
1129609
A pulley is a chain, belt or rope wrapped around a wheel. The mechanical advantage of a pulley system is approximately equal to the amount of supporting ropes or strands. \n\nTherefor, if you had a mass of 60kg and wanted to lift it using two supporting ropes, you would have mechanical advantage (MA) of 2. The mass will feel like one half of what it really is. When lifted with the help of the pulley system your 60kg would only feel like 30kg. Thus the effort force equals 30kg. \n\nHere are a couple Of websites that explain this more.\n\nhttp://www.technologystudent.com/gears1/pulley7.htm\n\nhttp://www.swe.org/iac/lp/pulley_03.html
964900
If i understand your scenario, the force is pushing downward on the block.  The horizontal component of the force is 15 N times cosine of 40 degrees.  The vertical component of the force is 15 N times sine of 40 degrees.  However, there is also the force of gravity, which is (3.5 kg * 9.8 m/s2)  (units are N, like the other forces).  So if you add the force of gravity to the vertical component, you have the total downward force.  Multiply this by the coeficient of friction, and you have the force of friction.  \n\nThe force of friction is directed against the direction of motion.  Subtract the frictional force from the horizontal component, and you have the total horizontal force.  Divide the horizontal force by the mass and you should get the acceleration.\n\nfriction force = [(15*sine 40)+(3.5*9.8)]*.25\nacceleration = [(15*cosine 40)-friction force]/3.5
856335
Here are the key things you need to know:\n\nForces acting on the small box:\n- gravity attracting its mass (pulling it downward)\n- large box (pushing it upward)\n- friction force between the two boxes (pushing it forward)\n\nForces acting on the large box:\n- gravity attracting its mass (pulling it downward)\n- the weight of the small box (= the force of gravity on the small box) (pushing it downward)\n- friction force between the two boxes (pushing it backward)\n- friction force between the large box and the underlying surface (pushing it backward)\n- the horizontal force (pushing it forward)\n\nFirst, treat the two boxes as a single mass, resting on the surface and being accelerated by the horizontal force. Calculate the force of friction between the large box and the surface (remembering that the weight of both boxes is acting to push it against the underlying surface). Calculate the force needed to both overcome this friction and to accelerate the mass of the two boxes at the specified rate (the second part of this requires the formula F = ma). That is the horizontal force.\n\nNow calculate the amount of force required to accelerate the small box at this rate. (Use F = ma again.) Divide this force by the weight of the small box to find the minimum coefficient of friction that will allow the large box to apply this force to the small box.\n\nIf you understand all this, you should be able to draw the two diagrams and answer the questions.\nGood luck!
856854
The acceleration of the system can be found using F=ma\nor a=F/m\na=65/(12+24+31)\n=.97 m/s^2\n\nfor the tensions, use the same equation considering the forces on each string.\n\nfor m1 the force is what is pulling the mass so that it accelerates at .97m/s^2\nF=12*.97\n=11.64 N\n\nfor the string between m2 and m3\nF=(12+24)*.97\n=34.92\n\nj
943484
I expect that you have the data on stretch as a result of tension. \na) The object is just at rest. Tension, T = W = m*g.\n\nb) T = W + m*a\n\nc) Tmax = W + m*a
908729
(a) ω = ωo + αt\n0 = 150 + α * 132\nα = -150/132\n≈ -1.136 rev/s²\n\n(b) θ = ωot + ½αt²\n= 150 *132 - ½*1.136*132²\n≈ 9903 revolutions (nearest rev)\n\n(c) See (b)\n\n(d) 75 rpm = 75*2π rad/min\nα = rω²\n= 50 *(75*2π)²\n≈ 11103300 cm/min²\n= 111033 m/min²\n= 111033/3600 m/s²\n≈ 30.84 m/s²\n\n(e) ???? I do not understand what it is you are seeking.
769883
147 Newtons = 2*Tension*sin(8)\nTension = 147/2/sin(8)\nThe tension in each rope is 528 N.\nYou probably have the tension in both ropes together, each rope only does half of the holding.\nThanks for the picture, by the way.
905915
You're going to have to resolve this two-dimensional problem into two one-dimensional problems.  I'd suggest using vertical and horizontal.  On the vertical side, we have gravity pulling down, and the normal force and one component of the tension from the rope pulling up.  This, by the way, is why the normal force is not equal to the weight - by pulling on the rope, you're partially lifting the sled off the snow.\n\nYou know that there is no acceleration vertically, so the net force vertically must be zero.  Knowing this, you can get the normal force as a function of the tension as the weight of the sled minus the vertical component of the tension.  Knowing the normal force and the coefficient of friction will give you the force necessary to pull the sled at constant velocity.  (If the velocity of the sled is constant, what can you say about the net force horizontally?)  Since the friction is only opposed by the horizontal component of the tension, you can find that.  Once you have that, you'll find that the only variable you have left is the tension in the rope, and you can solve for that.\n\nHope that helps.
846820
The formulae for linear acceleration and angular acceleration are very similar:\n\nF=ma\nT = I a  (the T and a here are usually written Tau & alpha in greek)\n\nYour 'torque' of 60N can't be right though, a torque value would be in N-m. Assuming its 60N-m then\n\na = T/I = 60 /12 = 5 rad/s^2