✔ 最佳答案
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
