a) Find the magnitude and direction of the net force acting on the 2.0 kg mass i
ID: 1694067 • Letter: A
Question
a) Find the magnitude and direction of the net force acting on the 2.0 kg mass if x = 0.36 m and y = 0.15 m. ---------N -------- degree(b) How do your answers to part (a) change (if at all) if all sides of the rectangle are doubled in length? The magnitude of the force will be unchanged.
The magnitude of the force will be reduced by a factor of two.
The magnitude of the force will be reduced by a factor of four.
The direction will remain unchanged.
The direction will shift clockwise.
The direction will shift counterclockwise.
Explanation / Answer
The masses are m1 = 1 kgm2 = 2 kg
m3 = 3 kg
m4 = 4 kg
The lengths of the rectangle x = 0.36 m
y = 0.15 m
The length of the diagonal d = sqrt(x^2+y^2)
= 0.39 m
The force acting on m2 due to m1 is In X direction: F1-2 = Gm1m2/x^2 = (6.67*10^-11 Nm^2/kg^2)(1 kg)(2 kg)/(0.36 m)^2 = 1.029*10^-9 N The force acting on m2 due to m4 is In X direction: F4x-2 = Gm4m2cos45/d^2 = (6.67*10^-11 Nm^2/kg^2)(4 kg)(2 kg)cos45/(0.39 m)^2 = 2.480*10^-9 N In X direction: F4x-2 = Gm4m2cos45/d^2 = (6.67*10^-11 Nm^2/kg^2)(4 kg)(2 kg)cos45/(0.39 m)^2 = 2.480*10^-9 N In Y direction: F4y-2 = Gm4m2sin45/d^2 = (6.67*10^-11 Nm^2/kg^2)(4 kg)(2 kg)sin45/(0.39 m)^2 = 2.480*10^-9 N The force acting on m2 due to m3 is In Y direction: F3-2 = Gm3m2/x^2 = (6.67*10^-11 Nm^2/kg^2)(3 kg)(2 kg)/(0.36 m)^2 = 3.087*10^-9 N The net force acting on m2 in X direction is Fx = F1-2 + F4x-2 = 1.029*10^-9 N + 2.480*10^-9 N = 3.509*10^-9 N The net force acting on m2 in Y direction is Fy = F4y-2 + F3-2 = 2.480*10^-9 N+ 3.087*10^-9 N = 5.567*10^-9 N The net force on m2 is F = 6.580 N The direction of the force is theta = tan^-1(Fy/Fx) = 57.77 degree with respect to X axis If the distances are doubled, the net force will be decreased because the gravitational force varies inversly with the square of the distance.(1/4 th of the initial value) The direction of the force is unaltered F4y-2 = Gm4m2sin45/d^2 = (6.67*10^-11 Nm^2/kg^2)(4 kg)(2 kg)sin45/(0.39 m)^2 = 2.480*10^-9 N The force acting on m2 due to m3 is In Y direction: F3-2 = Gm3m2/x^2 = (6.67*10^-11 Nm^2/kg^2)(3 kg)(2 kg)/(0.36 m)^2 = 3.087*10^-9 N The net force acting on m2 in X direction is Fx = F1-2 + F4x-2 = 1.029*10^-9 N + 2.480*10^-9 N = 3.509*10^-9 N The net force acting on m2 in Y direction is Fy = F4y-2 + F3-2 = 2.480*10^-9 N+ 3.087*10^-9 N = 5.567*10^-9 N The net force on m2 is F = 6.580 N The direction of the force is theta = tan^-1(Fy/Fx) = 57.77 degree with respect to X axis If the distances are doubled, the net force will be decreased because the gravitational force varies inversly with the square of the distance.(1/4 th of the initial value) The direction of the force is unaltered In Y direction: F3-2 = Gm3m2/x^2 = (6.67*10^-11 Nm^2/kg^2)(3 kg)(2 kg)/(0.36 m)^2 = 3.087*10^-9 N The net force acting on m2 in X direction is Fx = F1-2 + F4x-2 = 1.029*10^-9 N + 2.480*10^-9 N = 3.509*10^-9 N The net force acting on m2 in Y direction is Fy = F4y-2 + F3-2 = 2.480*10^-9 N+ 3.087*10^-9 N = 5.567*10^-9 N The net force on m2 is F = 6.580 N The direction of the force is theta = tan^-1(Fy/Fx) = 57.77 degree with respect to X axis If the distances are doubled, the net force will be decreased because the gravitational force varies inversly with the square of the distance.(1/4 th of the initial value) The direction of the force is unaltered The net force acting on m2 in Y direction is Fy = F4y-2 + F3-2 = 2.480*10^-9 N+ 3.087*10^-9 N = 5.567*10^-9 N The net force on m2 is F = 6.580 N The direction of the force is theta = tan^-1(Fy/Fx) = 57.77 degree with respect to X axis If the distances are doubled, the net force will be decreased because the gravitational force varies inversly with the square of the distance.(1/4 th of the initial value) The direction of the force is unaltered
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