A 588 N uniform boom is supported by a cable. The boom is pivoted at the bottom and a 1770 N object hangs?
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
110772
0.58 (with a horizontal bar right above the 58)
923264
i wonder if my @$$ is going to show from under this mini.........\nand a tube top?  i hope that it dosnt fall down.
1137609
there is insulation on the wire...........
592688
assuming the board is also of uniform dimensions the CM is at 2.325m from left end
760545
If a weight is hanging from a V formed by two ropes, the angle of the V cannot be 180 deg, unless the weight is 0.\n\nThis is because the Tensions in the two ropes must compose to a force equal and opposite to the weight, which is acting vertically down.\n\nIf the angle is 0, then each tension is half the weight = 500/2 = 250 N\n\nFor any other angle, we need to know the relative lengths of the ropes.  If they are equal, the tensions are equal and\n2Tcos(a/2) = weight.\nwhere a is the "V" angle.
769928
T = 2pisqrt(l/g)\nl = g(T/(2pi))^2 = 9.80(1.0/(2pi))^2 = 0.25 m\n\n(The object's weight is irrelevant for small oscillations.)
895549
suspension and floating
527998
Twice half the height.
458035
my jigsaw has a framulator. Do your own homework.\nMIKE
923169
this is to scale: http://en.wikipedia.org/wiki/Image:NewSolarSystem2.jpg
759031
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.)
321671
Ah, this brings me back!\nThis is conservation of momentum problem. That means that the initial momentum must equal the final. You must break up the values to x and y directions.\n\nThe 900kg mass is traveling 18mps * cos(30) = 15.6mps in +x\n                            18mps * sin(30) = 8.98mps in -y\nThe 750kg mass is traveling 0mps in x\n                            23mps in +y\n\nWith these values take the initial momentum as\n\ninit mom in x = 900kg(15.6mps) + 0 = final mom = 1650kg*(v)\nwhere v is final speed in x = 8.5mps\n\ninit mom in y = 900kg(-8.98mps) + 750kg(23mps) \n\nthat equals the final mom in y = 1650kg*(v)\n\nso v=+5.6\n\nIf you know the speed in the x and y directions, convert to vector notation by using s^2=vx^6+vy^2 = 10.2mps\nand the angle is atn(vy/vx)\n\nHope this helps
880505
assume piece 1 = x and piece 2 = y\ny = 4x\nx + y = 140\nx + 4x = 140\n5x = 140\nx = 28\n\nso x = 28 m and y = 4*28 = 112 m
367999
It weighs the amount that the air inside it weighs.
917650
It seems to me that the net force in the X direction would need to be zero while the net force in the Y direction would need to be 5.\n\nAssuming the angle the wires make at the point where the microphone is affixed is 90 degrees, the acute angle T makes with the wall  on the right would be 60 degrees.  That is a huge assumption, but I see no other way to solve the problem without making that assumption.\n\nC exerts a force of C cos 30 in the Y direction and C sin 30 in the X direction.  T exerts a force of T cos 60 in the Y direction and T sin 60 in the X direction.\n\nsin 30 = cos 60 = 1/2\ncos 30 = sin 60 = sqrt(2)/2\n\nY = 5 = C sqrt(2)/2 + T/2\nX = 0 = T sqrt(2)/2 - C/2\n\nor C = T sqrt(2)\n\nplugging that into the Y equation, \n\n5 = T * sqrt(2) * sqrt(2) / 2 + T/2\n5 = T + T/2 = 3T/2\n10 = 3T\nT = 10/3 = about 3.3333 N\nC = 10 sqrt(2) / 3 = about 4.7140452079103168293389624140323 N\n\nCheck it by inserting the C and T calculated into the X and Y equations above:\n\n0 = T sqrt(2)/2 - C/2 = (10/3) (sqrt(2)/2) - (10sqrt(2)/3)2 =\n(10sqrt(2))/6 - (10 sqrt(2))/6  = 0\n\n5 = C sqrt(2)/2 + T/2 = (10sqrt(2)/3)*(sqrt(2)/2) + (10/3)/2 =\n(10*2)/(2*3) + 10/6 = 10/3 + 10/6 = 20/6 + 10/6  =30/6 =5\n\nSo it checks, C = 10sqrt(2)/3 and T = 10/3 .
643953
depends on the direction of the force