when measuring force like pulling test it should resist a force of 14n?
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.
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
198626
You could measure the strength of an electromagnet by hanging small iron \nweights from the magnet.  When the applied weight (force of gravity acting \non the weights) exceeds the force of the magnet, the weights will fall \noff.  Steel Flat washers of the same diameter as the coil would be \ninexpensive and simple to use.  You can weigh several washers on a food \nscale and calculate the average weight of each washer.\n\nThe equation for the magnetic field strength, H0, in Ampere-turns per \nmeter for an empty (air) coil is:\nH0 = the applied voltage / (the number of turns of wire in the coil \nwinding X Area within the coil winding X the permeability of air).\nThe area is pie X D squared/4, where D is the diameter of the coil winding \nin meters.\nThe permeability of air,  , is 4 X pie X 10 to the - 7  henry/meter\n\nThe Equation is for a direct current.  An alternating current applied to \nthe coil cannot be used for lifting magnetic objects.  \n\nBy substituting different numbers for the applied voltage, the number of \nturns of wire and the diameter of the coil, you can calculate how each \nparameter will effect the strength of the magnet.\n\nElectromagnets can be fun, however,  keep them away from recorded tapes or \nthey will erase the contents.
853309
N is the normal force.\n\nN = mg\nN = 200*9.8\nN = 1960\n\nf = uN\nf = 0.30*1960\nf = 588 Newtons
784797
Suppose there is load and two men are pulling it in one direction each one giving a force of 5N.\n\nIn the language of vector, two vectors of magnitude 5 are acting in the same direction,\n\nThe resultant is a vector of magnitude 10N in the same direction. \n\nIf they are pulling it in opposite direction the load will not move.\n\nIn this case the two vectors of equal magnitude are acting in opposite direction.\n\nThe resultant of these two vectors is zero.
826246
smaller tension = 246 N\nlarger tension = 654 N\n\narrived at by summing the moments around one end of the scaffold (equation #1) and then doing a force balance on the scaffold (equation #2)
349619
One end of the galvanometer is connected between P and Q; the other end between R and S.\n\nYou have one sided deflection even after checking for any wrong connections.\n\nIt means proper ratio of resistances is not chosen.\n\nThe resistances cannot be taken at random.  P/ Q must be nearly equal to R/S.\n\nAdjust the resistances so that P/ Q must be nearly equal to R/S.\n\nWhen these ratios differ much you will get one sided deflection.
731589
Well To solve this problem, you need to break up forces in x and y components. Once you find both x and y components, you can find magnitude of force and direction ken must pull. \n\nTo keep the bear from moving, some of the forces (in both x and y direction) equal to zero. \n\nSo, sum of the forces in X-direction = 350sin(60) - 400sin(45) - F(in x direction) = 0\nSum of the forces in y-direction = 400cos(45) + 350cos(60) - F(in y direction) = 0\n\nSolve for F(in x direction) and F(in y direction), you will get these: F(in x direction) = 20.266 lb east and F(in y-direction) = 457.84 lb south\nTo find magnitude = ||F||= sqrt((F(x))^2+(F(y))^2) = 458.29 lb\n\nTo find angle= tan(theta) =(F(y))/(F(x)) where theta is the angle\ntheta = arctan (F(y)/F(x))= 87.5 degree east south\n\nSo Ken must pull the rop with force 458.29 lb in the direction East 87.5 degree South to keep the bear from moving. \nI think this is right. Good Luck.!!!
266808
Create an electromagnet by coiling the wire around the steel rod and attatching the wire to the power supply (put the ammeter in the circuit somewhere (in series). Use the retort stands to hold the weighing scales above the electromagnet. Place the 6 magnets in the weighing scale whilie the power supply is off. See what the weight of the magnets is. switch the power on and see whether there is any change in the weight. try repeating this a few times using different currents (measured on the ammeter). 100grams=approx 1Newton of force
119697
It depends if it is acting as a Force or a Pressure (a distributed Force). If it is a force, it has units of N (Newtons, or kg m/s²) and if it is a pressure it has units of Pa (Pascals, or N/m² which is the same as kg/(m s²))
150308
There is no net force acting on it. When there is a net, or resultant force acting on an object it causes an acceleration. As in this case a terminal velocity has been reached, that is no further acceleration is occuring, the resultant force is zero.\n\nThis is represented in the equation known as Newton's Second Law, or in Applied Mathematics, Newton Two. The equation is:\n\nF = ma and it means Force = Mass x Acceleration where the units are:\n\nNewtons (N) = Kilograms (Kg) x Meters per Second (ms^-1).
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.
871196
Don't you just need to balance forces in orthogonal directions and then use these components to get the forces along the cable?\n\nSince the sign is stationary, vector sum of forces must be zero.  Sketch a free body diagram and label known forces.
1164784
Add up one opposite vector (going up). In physics, they call it NORMAL FORCE.
136523
44.89Nm which is 44.9 Nm\n\nFormula is   torque=force*distance
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.
250025
The net force is zero.  Remember the equation F=MA:  force equals mass times acceleration.  If something is at rest, A is zero, so F must be zero.\n\nThe gravitational force on the brick is exactly balanced by the force of the table holding the brick up.  Net force is zero.