Four masses are positioned at the corners of a rectangle, as indicated in the fi

Question

Four masses are positioned at the corners of a rectangle, as indicated in the figure below (not to scale).
mass 1 = 1.0kg
mass 2= 2.0kg
mass 3= 3.0kg
mass4= 4.0kg


(a) Find the magnitude and direction of the net force acting on the 2.0 kg mass if x = 0.30 m and y = 0.13 m.
N (magnitude)
° counterclockwise from the x-axis, which points to the right
(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 kg
m2 = 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 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

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