physics. forces and acceleration, a man pulling himself up. need help w/ a problem.?
354313
Ok first of all know about the term :Machanical advantage.\nMechanical advantage of a pulley is equal to the number of supporting ropes.\n   As per your description yours is a single fixed pulley.\n\n   A fixed pulley is a pulley in which the wheel does not move. Fixed pulleys change the direction of the effort force. It does not increase the size of the effort force. The effort force is equal to the resistance force in a fixed pulley; therefore, the mechanical advantage (MA) of a fixed pulley is equal to 1.\n\n       But the actual defination of mechanical advantage is as follows:it is th ratio of  \nF(out) = force exerted on the load by the machine and \nF(in) = force you exert on the machine\n\nHere the load is 950N and the friction is 0.Thus by the above formulae:\n\nMA=F(out)/F(in)\nor\nMA=Load/Effort\n1=load/effort\nBut load=950N\n1=950/effort\neffort * 1=950N       <By cross mul>\neffort=950N           <Answer>
761342
We did something like this in my physics lab with an Atwood machine.  It was a pully with weights hanging from it.  I can't exactly remember how it worked, but maybe this will inspire you.  We also did something with an air track and these two masses colliding and we'd see which one would go in which diretcion.  You need a nearly frictionless surface, like an air hockey table or ice slab.
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.
514076
It's true - gravity pulls harder (i.e. more force) on a more massive (i.e. heavier) object. BUT that object is harder to accelerate - it has more interia. The two effects cancel each other out.\n\nClearly you don't like equations, but putting it in numbers can help:\n\nIf a 1kg object is pulled towards the earth, what will be the force, and what will be the acceleration?\nForce = mass * earth's gravitational field strength\nso Force = 1kg * 9.81N/kg\nForce = 9.81N (or 10N, roughly)\n\nAcceleration = Force / Mass\nSo Acc'n = 9.81N / 1kg = 9.81 m/s^2\n\nNow consider the same for a 10kg object:\n\nForce is now 98.1N (or 100N, roughly)\nSo Acc'n = 98.1N / 10kg = 9.81 m/s^2\n\nSo the acceleration is the same.
461030
For the first problem, you need a law of motion. (final velocity)^2=(initial velocity)^2+2*acceleration*displacement. Substituting, you get 0=30^2+2*a*20m and a=22.5 m/s^2. Now, we need to write a force equation: ma=Tension force-weight. To find the mass of the elevator and its load, divide its weight by gravity (9.8 m/s^2) to get about 16.3 kg. And you already calculated the net acceleration. Our force equation can now be rewritten as (16.3 kg)*(22.5 m/s^2)=Tension force-160N. The tension force is about 207N.\n\nFor each of the weight scenarios in the second problem, you know the net acceleration to be 30 m/s^2. You can also find the mass of the passenger by dividing his/her weight by gravity. With the force equation ma=(force of feet up)-(weight) with respect to the passenger, you can find the force of his/her feet. Here is the work for the first scenario.\n\n16N/(9.8 m/s^2)=1.63 kg\n1.63 kg*30 m/s^2=(force of feet)-16N\nForce of feet=64.9N\n\nAnswers for the second and third scenario:\nb) 2600N\nc) 260N
1006189
Explain Newtons laws of motion.
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
853308
The net force on the man is the sum of the force of gravity and the force on the ground. You've already calculated the net force, so simply subtract the man's weight from it. Remember that gravity is pulling him down, so you'll be subtracting a negative number (i.e. the final result should be greater, not less, than the net force).
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.
585763
When the lift descending,the acceleration is increasing.Relatively,body and the lift is moving down with a gradually increase speed.The acceleration will increase until its value is the same as the value of the acceleration due to gravity.\n\nLet R = normal reaction of the floor of the lift on the body, \n   mg = the weight of the body inside the lift\n    g = acceleration due to gravity\n      = 9.81 meter per second square (unit SI)\n    F = resultant force of the whole system\nusing this equation:\n\n     F = ma     \n     F = mg - R  (because the lift is moving downwards,\n                  resultant force is moving downwards too.)\nmg - R = m(a)    (a = g,because the acceleration of the lift\n                 is equals to the acceleration due to the\n                 gravity)\nmg - R = mg\n     R = 0\n\nAs R = 0,the body will not experience any force exerted by the floor of the lift,the body experiences weightlessness.\n\nIn short, weightlessness happens only when a body is in free fall motion, or fall freely under the gravity, or a more accurate explaination, weightlessness happens when a body falls freely with an acceleration which is equals to the acceleration due to gravity.\n\nHope my answer helps you.
858654
the idea is that the fireman is accelerating such that ma=mg-800N, the 800N is exerted by the person holding the fireman and is facing up, the force on fireman that is pointign down is mg (a is his acceleration). where ma is mass of fireman times his acceleration. 1. find a\n2 find v, given that d=v^2/(2a) [from: d=at^2/2, v=at] where d is 10m\n3. knowing v, the ave force of impact is calculated based on impact energy and dissipation of this energy in frictional force acting for 20cm: mv^2/2=F*d, find F..., where d=20cm
744947
First you should draw a free-body diagram to understand the forces acting on the block.  In this case, you have 4 forces:\n\n1) 40 N force moving in the +x direction, let's denote it F_1\n2) frictional force moving in the -x direction, let's denote it F_2\n3) Weight, moving in the -y direction, let's denote it W\n4) Normal force opposing weight, let's denote it N\n\nNow you can write Newton's 2nd law equations:\n\nx-direction: F_1 - F_2 = m*a\ny-direction N - W = 0 (equal to zero because block does not move in this direction)\n\nUsing these equations, you know that N = W = m*g\n\nUsing F_2 = u*N and a = 6 m/s^2m, we can solve for thd desired variables:\n\na)\nF_2 = F_1 - m*a  ---------> 40 - (5)(6) ---------> F_2 = 10 N\n\nb)\nF_2 = u*N = u*m*g, now solve for u:\n\nu = F_2/(m*g) = (10)/(5*9.8) -----------> u = 0.204\n\n--------------\n\nHope this helps
143729
Final veloc. when he reaches the land:\nV=sqr(2*g*h)= 8.743 m/s\nKinetic Energy= Ec= 0.5 * m* V2 = 1872,780 \nThis turns into potential Energy Ep= 1872,780  = mgh\nHe stops in h=0.8m , hence: \nForce = m g = 1872,78 / 0,8 = 2340,795 Newtons\n\nHe has two knees, so F= 2340, 795 / 2 = 1170,488 Newtons in each knee.
1164784
Add up one opposite vector (going up). In physics, they call it NORMAL FORCE.
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
153350
This problem employs the simple concept of conservation of energy (athough there are a number of ways to solve it).\n\nBasically, the person in the wheel chair has an initial kinetic energy at the bottom of thr ramp,\nKE = 1/2 mv^2\nAt the top of the ramp, some of that kinetic energy has been transfered into Gravitational Potential Energy,\nPE = mgh\n\nSince the mass of the person is constant, that drops out of the equation.\n\nSolve for the change in PE and subtract this from the intial KE.  Using the new KE, solve for the final velocity.
257076
dude u need to start doin your own hw. ill do it this time\n5. well you have to check all the forces on the bag. the first force is gravity, which is downwards and its 9.8m where m is the mass (Force = mass x acceleration) so gravity force is 20.5kg * 9.8 = 200.9N. then the 2nd force is the force of your own movement. you accelerate it upwards at 5m/s^2. so you apply a force of 5x20.5 = 102.5N on the bag upwards. but this is the thing. it gets confusing. the force is applied on the bag, not on the items in the bag. the items in the bag  is what  makes the bag weigh so much. so if you pull the bag up with 102.5N force, the items in the bag pull downwards with the same force (another one of newtons laws) which means the total downwards force is 102.5 + 200.9 = 303.4N downwards. which is more than 245N so the bag will rip. \n\n6. well in general F  = ma. but the F is total force. in this case, there is a friction force along with the applied force.\nso Fa + Ff = ma\n   37 + Ff = 5 x 5.8\n   Ff = 29 - 37\n   Ff = -8N     its negative because its opposing the applied force. in this case i can choose which direction i want to be positive, i chose the direction that the object is moving. you could put negatives in front of the applied force, friction force and acceleration   and youll get the same answer.  so friction force is 8N.\nb. the coefficient of friction is calculated by Ff = u x Fn\nwhere Fn is the normal force and u is the coefficient. you already know Ff. the normal force is easy its basically  the force tha topposes gravity. if the object isnt moving up or down (if its on a surface). so then gravity is 5 kg x 9.8 = 49N \nso then Ff = u x 49\n    8 = u x 49\n u = 8/49\n u = 0.1632..... coefficient of friction is 0.1632\n\n7.\na.  the force is 107N and the mass is 5.1kg. so acceleration is 20.98m/s^2\nb. so acceleration = Vf-Vi/t\n  20.98 = Vf - 0 / 10s\n  Vf = 10 x 20.98 = 209.8m/s\nfinal velocity is 209.8m/s\nc. the net force that acts on the object is only the downward gravity force which is 9.8x 5.1 = 49.98.\nd. the direction of their velocity goes down when the velocity becomes 0. so you can do this by saying acceleration is gravitational and it is -9.8. initial velocity is 209.8 and final velocity is 0. and now u gotta find time. so  a = Vf-Vi/t\n-9.8 = -209.8 / t\nt = 209.8/9.8\nt = 21.4s   so it starts going down   21.4s after 10s after its been released (the first 10 s  its accelerating upwards)   so it starts going down 31.4s after its been realeased.