A beam 4m long is supproted at both ends. A weight of 5N is attached 1m from end A; a weight of 80N is attached 1.5m from end B and a weight of 30N is attached 1.25m from end A.calculate the force exerted by the supports at each end of the beam.
131996
So, this looks like a homework problem, but this is how you would set it up.  And, it looks like the beam is massless...\n\nFirst, you know that the beam's not going anywhere, so the sum of the forces on it must be equal to zero.  That is, the sum of the masses must be equal to the sum of the forces exerted by the two supports-- 5N+80N+30N=s1+s2, where s1, s2 are the two forces.\n\nNext, you know that the beam's not rotating, so the sum of the torques on it  around any point must be zero.  To make things simpler, I'll set up the equation with the torques about end A--  5N*1m+80N*2.5m+30N*1.25m-s2*4m=0.  Note the difference in sign of the torque produced by s2 relative to the other forces.\n\nNow you see that the second equation has only one unknown-- s2.  Solve for s2, then plug that value into the first equation.  Then solve for s1.
476423
People above neglected the vertical component of 460 N.  This must be added to the weight of the box.  Also, not sure why they took 350 N and multiplied it by acceleration.\n\nso F_vert=460sin35.6=267.78 N (downward)\n     F_hor=460cos35.6=374.0 N\n\ncalculating F_frict=mu x (F_normal)\n\nF_normal=weight of box + F_vert = 267.78+350=617.78\n\nso F_frict=0.57x617.78=352.13 N\n\nnow the net force on the box = 374.0 - 352.13 = 21.87 N\n\nand we need mass of the box, m=350/9.81=35.7 kg\n\nso a=21.87/35.7=0.6126 m/s2\n\nfinally,  d=V_initial x t + 0.5 x a x t^2, V_initial=0\n\nso t=((4.2)/(0.5x0.6126))^0.5=3.7 sec\n\n**********************************************************\n\nHowever, should you wish to ignore the vert component\n\nF_frict=mu x (F_normal)=0.57 x 350 = 199.5 N\n\nso net horizontal force is:\n\nF= 374.0 - 199.5 = 174.5 N, a=174.5/35.7= 4.89 m/s2\n\nplug a into:\n\nd=V_initial x t + 0.5 x a x t^2, and solve for t\n\nt=((4.2)/(0.5x4.89))^0.5 = 1.31 sec
416477
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
486540
Force acting down at the center:\ng*m = 9.81*60 = 589 N\n\nTwo tension forces, one to the left and one to the right at angle 4.8 degrees\n\nForce acting up at the center equals force acting down\n2*T*sin(4.8) = 589\n\nSolve for T\nT = 589/(2*sin(4.8)) = 3520 N
487305
My best guess would be a simple ratio calulation:\nLet x = the unknown\nLet d = the distancefrom weight\nLet p = total length of the pole\nLet w = total weight\n\nThe formula would look something like this:\nf(x)=(d/p)w\n\nor in english: (3.5/6) x 216 and (2.5/6) x 216 or 125.28 and 90.72
291457
answer is 350newton\nsince the system is in equilibrium, we have,\n400*1.4=W(3-1.4)\nthat is W=400*1.4/1.6=350\nand the force on the pivot would be 400-350=50newton
500313
a=F/m\n\nF=30N\nm=60kg\n\na=30/60\na=.5m/s^2
243117
If the forces are both acting in the same direction, add them together. = 65 Newtons
318618
I would think you would use the force created by compressing the spring.\n\nThis force is given by the equation:\n\nF = k(Li - Lc)\n\nWhere:\nF = force\nk = spring constant\nLi = uncompressed spring length\nLc = compressed spring length\n\nThe below link is for a calculator for this force.\n\nhttp://www.efunda.com/DesignStandards/springs/calc_comp_Fstress.cfm
41563
Draw a force diagram to help you figure this out.\n\nRemember, perpendicular vectors are independent.\n\nF=ma
413477
Depends on the supports, whether they are simple (pivot possible), or completely fixed.\n\nAlso depends on whether it is a point load, or a distributed load.\n\nThere is a technique that is taught in engineering such that you can work it out for almost any configuration, but it is a bit too involved to detail. Do a web search on 'Bending Moment Diagrams' and you will find lots on the techniques.\n\nFor specific calculation, I found a neat online calculator, go to http://www.mechengcalculations.com/index.html, and look for your beam under section 'Beam Flexure'
585369
letter b is the answer.. because the weight of the object is approx. 627.20kg multiplied by 9.8(acceleration due to gravity) so the product is the force..
48071
cmon...dont u know basic physics...just calculate the resultant of these 2 forces\nanswer= [(132)^2 + (181)^2]^1/2\nif this makes sense!!
455385
I can't help you with the first question, but to figure out the second one you just need to use Hooke's law (I'm assuming the spring to be almost perfectly elastic.) In equation form the law looks like: F=k*x\nWhere F is the force applied, k is the spring constant, and x is the deformation.\nThe force applied will be the objects weight, so 9.8*.5=4.9N, and the deformation is .04m.\nSo 4.9N=k*.04m\nk=4.9N/.04m\nk=122.5N/m
365500
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.!!!
386032
I agree with the first answerer...where is the figure? But let's see if we can solve it just the same. We have to guess a little regarding the location of the supports.\n\nGiven: length of board=5.0m\n           mass of board=54kg\n    Hence,  weight of board =54g N.\n    Place this at the middle of the board because it is a uniform board as given in the problem.\n\nFirst support is 3.4m from the end, and the second is 4.6m form the end. \n\nDiver stands at end of the board. He weighs 65g N.\n\nFind: F1 at first support and F2 at second support.\n\nTake torque about F1:\n\n54g*(2.5-0.4)+65g*4.6=\nF2*(1.6-.4)\n\n54g*2.1+65g*4.6=1.2F2      Call this equation 1.\n113.4g+299g=1.2F2\nF2=(1111.3+2930.2)/1.2\n    =3368N upward.\n\nTake torque about F2:\n\n54g*(3.4-2.5)+65g*3.4=\nF1*(1.6-.4)\n\n54g*0.9+65g*3.4=1.2F1       Call this equation 2.\n\n48.6g+221g=1.2F1\nF1=2642/1.2\n    =2201.7N downward\n\nCheck your answer:\n\nF1+F2=65*9.8+54*9.7\n-2201.7+3368=637+529.2\n\n3368=637+529.2+2201.7\n       =3367.9\n0.1 difference is acceptable. so our solution is correct. (Assuming my assumptions as to locations of supports are same as in your figure.)
561727
Use Newton's universal gravitation equation, then sum the forces.\n\nF = G*m1*m2/r^2\n\nF = force (newtons)\nG = gravitational constant (6.6742 x 10^-11)\nm1 = mass of object 1 (kg)\nm2 = mass of object 2 (kg)\nr = distance between objects (meters) ==> don't forget to convert the units\n\nIn this case, you do it twice:\nF1 = G * 390 * 390 / 0.11^2 = 0.84mN to the left\nF2 = G * 390 * 390 / 0.39^2 = 0.067mN to the right \n\nNet Force = 0.84 - 0.067 = 0.772mN to the left