If you apply a force of 100 N down into the middle of a circular stool with 4 legs,what is force in each leg?
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.
160197
The relationship between force(F) and pressure(P) is:\n\nP = F/a\n\nWhere: a is the area over which the force acts.\n\nSo, if you apply 50 N of force on an area that is 0.5 m^2, you will end up with 100 N/m^2 --- or ---- 100 Pa\n\n[A Pascal is a typical unit of pressure. 1 Pa = 1 N/m^2]\n\nIf you increase the area to say 1 m^2, you get:\n\nP = 50 N/1 m^2 = 50 Pa\n\nThe pressure dropped by one half. The force is spread across a larger area tehrefore it's effect is less.\n\nHope this help out.
657765
1. the object is pushed up by the weight of its equivalent volume of water, 4-3.2=0.8l\nSo the object's density is equal to its mass divided by its volume:\n4kg/0.8l=4kg/0.0008m3=40,000kg/8m3=5000kg/m3\n\n2. the force decreases with the square of distance, so it will be 3^2=9 times smaller, which is 6N.
839427
Assuming you mean minimum force required to move the mass up the slope, the answer is 100N.\n\nI also assumed you meant distance parallel to the slope was 200, meaning the angle between the slope and the horizontal would have sin=20/200 = 0.1. I will call this angle "&".\n\nDraw a free body diagram with a force (parallel to the slope) acting on a block, and a reaction force (perpendicular to the slope) under the block. The weight of the block acts straight downwards with a force of 100*10= 1000N\n\nJust as the block is about to move upwards (i.e not going up or down), all horizontal forces are equal 0\n\n=> Fcos&=Rsin&\n\nAll vertical forces are also equal 0\n\n=> Fsin&+Rcos&=1000\n\nRearranging first equation:\nR=Fcos&/sin&\nsub R into second equation:\n=>Fsin&+Fcos^2&/sin&=1000\nMultiply all terms by sin&\nFsin^2& +Fcos^2&=1000sin&\nTake out F\nF(sin^2&+cos^2&)=1000sin&\nLaw: sin^2+cos^2=1\n=>F=1000sin&\n=>F=100N
800064
For these problems, it would be helpful to draw a free-body diagram to represent all the forces on this problem.  Here you have 4:\n\n1) Weight in the vertical direction going down, let's denote it W\n2) Force, moving in the x-direction, denote it F_1\n3) frictional force, moving opposite of Force, denote it F_2\n4) Normal force, perpendicular place of contact (opposite direction of W), denote it N\n\nNow use Newton's 2nd law to write equations in each direction:\n\ny-direction: \nN - W =0 (because the block does not move in the y-direction).  Therefore:\n\nN = W, where W = m*g\n\nx-direction:\nF_1 - F_2 = 0 [because the block is just about to move when F_1 is applied].  Therefore:\n\nF_1 = F_2, where F_2 = u*N\n\nMaking the necessary substitutions for F_1 and N, you get:\n\n30 = u*(4.4)(9.8).  Solving for u, you get --------->\n\nu = 0.7\n\n--------\n\nHope this helps
693390
Gravity, centrifugal forces and centripetal forces
769351
this question was on this yr's midyear exam, ill give u a clue.\nyou need to seperate the forces acting on the wire into two types \n1) acting horizontally (the force to the left equals the force to the right)\n2)acting vertically (the force to the bottom {or the normal force = 20 *9.8} equals the force to the top)\n\nequate them both and you should get the answer
621321
If the density is 'd' kg/m^3 then the weight is '9.81d' Newton. [remember weight is force]
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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.
832042
The vertical, horizontal components of all forces should balance.\n       Let the tensions in the ropes be T1 and T2 newtons.\n       T1*cos(90-71) + T2*cos(90-28) = 254.8\n       i.e., T1*sin71 + T2*sin28 = 254.8\n\n       For horizontal componets (which are opp. & equal)\n       T1*cos71 + T2*cos28 = 0\n\nSolve for T1, T2 please.
152239
AIR RESISTANCE CANCELS OUT THE FORCE DUE TO GRAVITY , THERE IS NO ACCELERATION BECAUSE THE OPPOISTE FORCE IN THE SAME.\nSO YOU CAN SAY 0
210676
Use free body diagram to balance the forces and u can derive the equation of motion \nIf u want i can give u full derivation.\n"::
757274
READ THIS CAREFULLY\nAs the block just slides on the table the net force just equals the limiting friction acting on the body\n\nhence,\nF=coeff. of friction * Nor,mal rxn\n\nnormal rxn= weight of body on a plane\n\nhence F= 20*9.8*0.4\n              =78.4N\n\nACC \nit is just begg to slide and the given data is just unsufficient 2 find the acc. of the box
989214
F = -G(m1*m2)/r^2\nF = force\nG = gravitational constant = (6.6742 +/- 0.001)*10^(-11) N * m^2 / kg^2\nm1 = mass of object 1 = 47 kg\nm2 = mass of object 2 = 6.00×10^24 kg\nr = distance between objects\n\nF = -(6.6742 +/- 0.001)*10^(-11) * (47 kg * 6.00*10^24 kg) / r^2\nF = -4.225 * 10^14 N*m^2 / r^2\nSince the distanct between the Earth and the person weren't given this is all the farther the force can be calculated to.  The force will be equal on both the objects (person and Earth) only that the acceleration of the Earth will be almost nothing due to its size and the force will be in the opposite direction.\n\nBut based on the acceleration of the person, you can calculate a force, but it will only be an estimate\nF = m*a = 47 kg * 9.8*10 m / s^2 = 460.6 N\nForce on Earth is 460.6 N\n\nNow the acceleration of the Earth is\n460.6 = 6.00*10^24 * a\na = 7.68 * 10^(-23) m/s^2
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
481598
There are three objects; 5kg, 9kg, and a cable of negligible mass.\n\nWhen the three move, they must have the SAME acceleration.\n\nTherefore, consider the three as a single object of mass 5 + 9 = 14 kg, with acceleration ‘a’.\n\nThe only force that is pulling the three is the force given by the 9kg object. (The gravitational pull is only on this 9kg.)\n\nThis force is 9 x 9.8 N = F.\n\nSince F = M a,   9 x 9.8 = 14 x a.\n\nFrom this a = 6.3 m/s^2.\n\nAll the three objects are moving with an acceleration 6.3 m/s^2.\n\nIf the body 5kg is moving with an acceleration of 6.3 m/s^2, then what is the force acting on it?\n\nWe know that f = ma; hence, f = 5 x 6.3 = 31.5 Newton.\n\nSince the tension of the rope is the pulling force, tension =31.5 Newton.\n\nThe tension can be calculated using the forces acting on the other end of the cable also.\n\nThe downward force on 9kg is 9 x 9.8 N; the tension is the upward force T. \n\nThe net force acting on 9kg is (9 x 9.8 - 9 x 6.3) = 9 x 3.5 = 31.5 N = T.
859844
The previous answerer has the right idea but the wrong details.\n\nThe four parts should weigh 1, 3, 9 and 27 pounds.  The logic is that the desired weight is written in "balanced ternary" notation, where the values of the digit positions are 1, 3, 9, 27, 81, . . . (that's "ternary"), but the allowed digit values instead of being 0, 1, 2 (that's regular ternary) are -1, 0, 1 (that's balanced ternary).  Then the digit values for any desired total weight written down in balanced ternary tell you whether the weight for that column is to go in one pan, go in the other pan, or stay on the floor.