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gravitation (urgent)
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13. Gravitational force Fo on m before hollowing Fo = G(0.431).(2.95)/0.09^2 N Mass of the lead sphere taken out from the big sphere = 2.95 x (2/4)^3 kg = 2.95/8 kg Gravitational force F' due to this sphere F' = G.(0.431).(2.95/8)/(0.09 - 0.02)^2 N Hence, resultant gravitational foce = F0 - F' = [G(0.431).(2.95)/0.09^2 - G.(0.431).(2.95/8)/(0.09 - 0.02)^2 ] N = 8.31 x 10^-9 N 15. Let the coordinate of mass D be (x, y, z) Force of attraction of mass D on mass A Fd = G(4m)(m)/(x^2+y^2+z^2)^(3/2) The x-component of force on mass A, Fx, is Fx = G(4m)(m)(x)/(x^2+y^2+z^2)^(3/2) The y-component of force on mass A, Fy, is Fy = G(4m)(m)(y)/(x^2+y^2+z^2)^(3/2) The z-component of force on mass A, Fz, is Fz = G(4m)(m)(z)/(x^2+y^2+z^2)^(3/2) Similarly, we can calculate the resultant x-component of graviational force on mass A from masses B and C respectively, Fb = G(2m)(m)(2d)/d^3.(2^2+1^2+2^2)^(3/2) = 0.148/d^2 Fc = G(3m)(m).(-d)/d^3.(1^2+2^2+3^2)^(3/2) = -0.0573/d^2 Hence, net force on A from B and C = Gm^2(0.0907)/d^2 Therefore, for the net force on A to be zero Fx = -Gm^2(0.0907)/d^2 i.e. G(4m)(m)(x)/(x^2+y^2+z^2)^(3/2) = -Gm^2(0.0907)/d^2 4x/(x^2+y^2+z^2)^(3/2) = -0.0907/d^2 ---------------- (1) By similar way, we can find the y and z components, 4y/(x^2+y^2+z^2)^(3/2) = -0.1886/d^2 ------------ (2) and 4z/(x^2+y^2+z^2)^(3/2) = 0.0238/d^2 ------------ (3) solve the 3 equations for x, y and z 35. The two graphs give the following equations: T = 600(y/L) Fh = 300 - 300(y/L) Take moment about the hinge of the beam [T.cos(theta)].L = (Fa)y i.e. 600(y/L).cos(theta).L = (Fa)y 600.cos(theta) = Fa ----------------------- (1) Take moment about the top of the beam (Fh).L = Fa(L-y) [300-300(y/L)].L = Fa(L-y) 300(L-y) = Fa(L-y) hence, Fa = 300 N substitute into (1) cos(theta) = Fa/600 = 300/600 = 0.5 hence, (theta) = 60 degrees
其他解答:E2A5F59BAA12C031
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圖片參考：http://farm7.static.flickr.com/6224/6283088672_e91caeaa57_b.jpg ans: 13: 8.31 x 10^-9 N 15a: -1.88d, b: -3.9d, c: 0.489d 35a: 60(degree), 300N thanks 更新: for 35, why Fh is 300-300y/L but not -300y/L ?最佳解答:
13. Gravitational force Fo on m before hollowing Fo = G(0.431).(2.95)/0.09^2 N Mass of the lead sphere taken out from the big sphere = 2.95 x (2/4)^3 kg = 2.95/8 kg Gravitational force F' due to this sphere F' = G.(0.431).(2.95/8)/(0.09 - 0.02)^2 N Hence, resultant gravitational foce = F0 - F' = [G(0.431).(2.95)/0.09^2 - G.(0.431).(2.95/8)/(0.09 - 0.02)^2 ] N = 8.31 x 10^-9 N 15. Let the coordinate of mass D be (x, y, z) Force of attraction of mass D on mass A Fd = G(4m)(m)/(x^2+y^2+z^2)^(3/2) The x-component of force on mass A, Fx, is Fx = G(4m)(m)(x)/(x^2+y^2+z^2)^(3/2) The y-component of force on mass A, Fy, is Fy = G(4m)(m)(y)/(x^2+y^2+z^2)^(3/2) The z-component of force on mass A, Fz, is Fz = G(4m)(m)(z)/(x^2+y^2+z^2)^(3/2) Similarly, we can calculate the resultant x-component of graviational force on mass A from masses B and C respectively, Fb = G(2m)(m)(2d)/d^3.(2^2+1^2+2^2)^(3/2) = 0.148/d^2 Fc = G(3m)(m).(-d)/d^3.(1^2+2^2+3^2)^(3/2) = -0.0573/d^2 Hence, net force on A from B and C = Gm^2(0.0907)/d^2 Therefore, for the net force on A to be zero Fx = -Gm^2(0.0907)/d^2 i.e. G(4m)(m)(x)/(x^2+y^2+z^2)^(3/2) = -Gm^2(0.0907)/d^2 4x/(x^2+y^2+z^2)^(3/2) = -0.0907/d^2 ---------------- (1) By similar way, we can find the y and z components, 4y/(x^2+y^2+z^2)^(3/2) = -0.1886/d^2 ------------ (2) and 4z/(x^2+y^2+z^2)^(3/2) = 0.0238/d^2 ------------ (3) solve the 3 equations for x, y and z 35. The two graphs give the following equations: T = 600(y/L) Fh = 300 - 300(y/L) Take moment about the hinge of the beam [T.cos(theta)].L = (Fa)y i.e. 600(y/L).cos(theta).L = (Fa)y 600.cos(theta) = Fa ----------------------- (1) Take moment about the top of the beam (Fh).L = Fa(L-y) [300-300(y/L)].L = Fa(L-y) 300(L-y) = Fa(L-y) hence, Fa = 300 N substitute into (1) cos(theta) = Fa/600 = 300/600 = 0.5 hence, (theta) = 60 degrees
其他解答:E2A5F59BAA12C031
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