Question

Three 3 kg masses are located at points in the
xy plane.
x=40 cm
y=53 cm
What is the magnitude of the resultant
force (caused by the other two masses) on
the mass at the origin? The universal gravitational
constant is 6.6726 × 10−11 N · m2
/kg2
.

At what angle from the positive x-axis will the
resultant force point? Let counterclockwise
be positive, within the limits −180◦
to 180◦
.
Answer in units of ◦
.

Answer 1

Heya!

Check out the attachment:))

Hope it's right and helpful.

CORRECTION: I converted 53cm to 0.53m and 40cm to 0.4m but in the diagram I made the mistake of writing 0.5 cm and 0.4cm.

Answer 2

Answer:

The magnitude of the resultant force on an object B is $4.3\times 10^{-9}N$ and the angle of the resultant force from the positive x-axis is -150.40°.

Explanation:

Given:

$m_{1}=3kg\\ m_{2}=3kg\\ m_{3}=3kg$

$r_{AB}=0.53m\\r_{CB}=0.40m$

$G=6.6726\times 10^{-11} Nm^{2}/kg^{2}$

Here,

As we all know, the gravitational constant is denoted as G.

The mass of object A is denoted by $m_{1}$.

The mass of object B is denoted by $m_{2}$.

The mass of object C is denoted by $m_{3}$.

The force on object B due to object A is denoted by $F_{2}$.

The force on object B due to object C is denoted by $F_{1}$.

The distance between objects A and B is denoted by $r_{AB}$.

The distance between objects A and C is denoted by $r_{AC}$.

The resultant force of forces $F_{1}$ and $F_{2}$ is denoted by $F_{r}$.

The angle of the resultant force from the positive x-axis is denoted by $\theta$.

Now,

By the equation,

$F_{2}= \frac{Gm_{1} m_{2} }{r_{AB} ^{2} } \\F_{2}= \frac{6.6726\times 10^{-11} \times 3\times 3 }{0.53^{2} }\\ F_{2}=2.13\times 10^{-9}N$

Then,

By the equation,

$F_{1}= \frac{Gm_{2} m_{3} }{r_{CB} ^{2} } \\F_{1}= \frac{6.6726\times 10^{-11} \times 3\times 3 }{0.40^{2} }\\ F_{1}=3.75\times 10^{-9}N$

Now,

By the equation,

$F_{r} =\sqrt{F_{1} ^{2}+ F_{2} ^{2} } \\F_{r} =\sqrt{(3.75\times 10^{-9} ) ^{2}+ (2.13\times 10^{-9}) ^{2} } \\F_{r}= 4.3\times 10^{-9}N$

Then,

$\theta= -180+tan^{-1} \frac{F_{2}}{F_{1}}\\\theta= -180+tan^{-1} \frac{2.13\times 10^{-9}}{3.75\times 10^{-9}}$

$\theta=$ -150.40°

So, the magnitude of the resultant force on an object B is $4.3\times 10^{-9}N$ and the angle of the resultant force from the positive x-axis is -150.40°.