If a 46-kg person jumps out of a tree from a height of 3.9 m, calculate the force in the knee joints?
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.
882323
The CG is between the two scales. let us say it is at a distance of x1 from scale 1 and x2 from scale 2. (assume the plank remains horizontal)\n\nThen, x1*Fg1 = x2*Fg2 (the moments must be equal)\n\n>> x1/x2 = Fg2/Fg1\nand x1 + x2 = 2\n>>\nx1 = 2 * (310/(450+310)) ~ 0.816 m\n\nNow tell me where the feet of the woman are.
706663
The idea is that all forces at the nodes (connecting points) are balanced. Draw the two free body diagrams.\n\nLeft point:\nFx\n0=-T1cos53+T2cos11\nFy\n0=T1sin53-T2sin11-57\nRight point:\nFx\n0=-T2cos11+T3cos42\nFy\n0=T2sin11+T3sin42-W2\n\nBy inspection we see only the first 2 eqns are required. Solve the system:\n0=-T1cos53+T2cos11\n0=T1sin53-T2sin11-57\n\nT1=83.62 N\n{=}8.5 kg
481398
sum the forces in the direction that they are applied and it will be equal for a static object. \n\nRule of thumb on this one is that you have 100N down ward, then you need 100 N's up ward.\n\nGivens, legs are equal lengths, same material, and no angles to calculate.\n\n4 legs - 4 forces - each at 25 Newtons.  \n\nNow if you want to get fancy, break out the strain gauges and let us see what we really find.  But it helps if you have preformed a few material tests and experimenters first.
896953
The net force is the difference between the two weights.\n\nHowever, there is not enough information in the question to answer without making some assumptions.\n\nYou are not given the mass of the man and/or the bricks. One can calculate the mass if given the force of gravity, however you aren't given that either.\n\nOne method of solving is:\n\nASSUMING that this system is on the surface of the Earth, you can find the two masses, using F=mg with g=9.8N/kg. Once you've got that you can find the acceleration using F=ma since you know the net force and the mass of the system.\n\nMake sure you indicate which direction the acceleration is in (in this case, which object is falling, and which is being raised.
735384
Using the law of conservation of energy determin\n\na) the vertical component of vector V(initial)\n\nIt went up 20 meters, mgh = 1/2mV^2, solve for velocity\n\nsqrt[2(9.8m/sec^2*20meters)] = v = 19.8m/sec\n\nb) the work done by the gravitational force on the particle during its motion from P to the ground.\n\nMaybe this is a trick question, I will take it at face value, when at P it is 60meters above the ground, the work of the gravity bringing it down from P is just mgh = 0.5kg*9.8m/s^2*20m=98J\n\nc) the horizontal and vertical components of the velocity vector when the particle reaches the ground:\n\nHorizontal velocity is constant at 30m/sec, nothing to calculate.\n\nVerticle velocity, keep in mind that the object reached its maximum height at 80 meters high (ie from a verticle perspective is is at zero velocity at that point, and begins to fall, the terminal vertical velocity comes from the potential energy of \nmgh = 0.5kg*9.8m/s^2*80m = 1/2 mV^2, solve for v\n\nvertical velocity is 39.6 m/sec downward\n\nplease check for calculation errors.
120746
If you didn't, the brunt of your body's impact would have to be displaced by all the joints in your body such as your ankles, knees, hips, spine, etc..  The impact could cause distress, and/or damage to one or more parts, depending on the height of the jump.\nBending your knees displaces the force to the muscles in your body as opposed to the joints, which are not as forgiving.
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.
921019
First,\ncompute the components of force perpendicular to the boom due to the weight of the boom and the object:\nBp=cos(48)*588\nOp=cos(48)*1770\n\nthe tension in the support cable can be computed by summing the torques around the pivot point:\n\nTC+TBp+TOp=0\nTC=tension*.66*24\nTBp=-Bp*.5*24\nTOp=-Op*24\nTension=(Bp*.5+Op)/.66\n=2093N\n\nThe vertical component of the tension is\ncos(48)*2093\nThe horizontal component is \nsin(48)*2093\n=1555N\nwhich is equal to the horizontal reaction force in the floor\n\nSince we know the components of force on the floor parallel to the beam as\nBh=sin(48)*588\nOh=sin(48)*1770\n\nthen v^2+h^2=parallel^2\nv=sqrt((sin(48)*588+sin(48)*1770)^2-1555^2)\n=807N\n\n\nj
759333
all object fall 4.9m after 1 sec\n\nX=1/2(9.8m/s^2)(2s)^2\nx=19.6m\n\nx=(1/2)(9.8m/s^2)(3s)^2\nx=44.1m
368233
N is Newtons, a unit that measures force. For every 1kg of something, if it accelerates at 1 meter per second per second (m/s/s or meters per seconds squared), than there is a force of 1N acting on it.\n\nForce is calculated by multiplying the mass (not weight) with the acceleration. For example, if you are 75kg, than there is a force of 75kg X 9.8 m/s/s (acceleration due to gravity: roughly constant on Earth) = 735N pulling you down. BUT, there is a reaction force, in this case the NORMAL, pushing you back up at the same time and force. Therefore, the force pulling you down, 735N, take away the NORMAL, also 735N, and you will have a NET FORCE of 0N because you are stationary. \n\nSo in order to resist something pulling at 14N (ignoring air resistance, friction, etc), it needs to pull back 14N.
458619
46 feet is what comes to mind
820907
Assuming you somehow manage to get the ladder pass the centre and out the other side; could you throw a rope ladder 6,378km towards space?\n\nThen the next hurdle; have you ever tried to walk feet-first up a ladder? Because that's what you'll be doing once you've got past the centre of the Earth.
932992
it is a much easier problem than it seems..\nthe key point is the fact that the body is moving at constant speed. meaning that the net force on it is zero --> no acceleration\n\nthere are two forces. friction and the horizontal component of the force that is exerted at 40 degrees.\n\nso simply ...\n\n\nFx = Friction==>\n976 cos40 = coeff. *N *g\n\n\n now there is equilibrium at the vertical axis as well.\n\nmeaning that \n\nN+Fy = m*g =>\n\nN = (mg- fy)\n\nso we mix them together and hve\n\n976cos40 = 0.667 (mg- 976sin40) =>\nm=50.31kg\n\nin this problem you use that there is equilibrium in both of the x and y axis.
1161505
F = m*a\n\nm = F/a\n\nm = 54/3.4 kg\n\nThis assumes the fisherman is not also on the ice and doesn't move.
935163
Start with the formula, F = ma.  The force, F,  he is able to exert is equal to his weight times the coefficient of friction.  And his weight is his mass times the acceleration of gravity.  So, F = 0.9mg, and 0.9mg = ma.  Mass is on both sides, so it cancels.  0.9g = a, 0.9*9.81 m/s^2 = 8.83 m/s^2.  His final velocity, Vf = SQRT(Vi^2 + 2ad), where Vi is his initial velocity (zero), a is his acceleration, and d is the distance he travels.  So, Vf = SQRT(2*8.83*50) = 29.71 m/s.  Because his acceleration is constant, his average velocity is 29.71/2 = 14.86 m/s.  The time he takes is then d/Vave = 50/14.86 = 3.365 s.
700115
Very simple, in your case you only need to convert from N(Newton) to Kg(kilogram) so just divide by 9.8 (gravity constant)\n\nN/m2---- /9.8 ---> Kg/m2\nN/m2<-- X9.8 ---- Kg/m2