Three spheres are arranged in the xy plane as in the figure below. The first sphere, of mass

m1 = 13.4 kg,
is located at the origin; the second sphere, of mass
m2 = 4.56 kg
is located at (−6.00, 0.00) m; and the third sphere, of mass
m3 = 4.00 kg
is located at (0.00, 5.00) m. Assuming an isolated system, what is the net gravitational force on the sphere located at the origin?

I keep getting 1.82502801*10^(-10). Am I wrong?

no idea.

where's your work?

6.67 x 10⁻¹¹(13.4*4.56)/-6^2 = 1.13280026*10^(-10)

6.67 x 10⁻¹¹(13.4*4)/5^2 = 1.4309056*10^(-10)

\sqrt((1.4309056*10^(-10))^(2)+(-1.13280026*10^(-10))^(2)) = 1.82502801*10^(-10).

Fx = - G m2 m1 / 36 = - G m1 (m2/36)

Fy = + G m3 m1 / 25 = G m1 (m3/25)
|F| = sqrt (Fx^2 + Fy^2)
G = 6.67*10^-11
G m1 = 6.67*10^-11 * 13.4 = 8.94 *10^-10
Fx= -G m1 (m2/36) = - 8.94*10^-10 * 0.127 = - 1.14*10^-10
Fy =+G m1(m3/25) = +8.94 *10^-10 * 0.16 = + 1.43 * 10^-10
so 10^-10 sqrt (1.14^2 + 1.43^2)\\\
10^-10 sqrt (1.3 + 2.05) = 1.83 * 10^-10
I agree with you.

To find the net gravitational force on the sphere located at the origin, we need to calculate the gravitational force between each pair of spheres and then sum them up.

The formula for the gravitational force between two spheres is given by Newton's Law of Universal Gravitation:

F = G * (m1 * m2) / r^2

Where:
F is the gravitational force between the two spheres,
G is the gravitational constant (approximately 6.67430 × 10^(-11) N m^2/kg^2),
m1 and m2 are the masses of the two spheres, and
r is the distance between the centers of the two spheres.

First, let's calculate the gravitational force between the sphere at the origin (m1 = 13.4 kg) and the sphere located at (-6.00, 0.00) m (m2 = 4.56 kg). The distance between their centers is the magnitude of the position vector difference:

r12 = √[(x2 - x1)^2 + (y2 - y1)^2]
r12 = √[(-6.00 - 0.00)^2 + (0.00 - 0.00)^2]
r12 = √[36 + 0]
r12 = √36
r12 = 6.00 m

Using Newton's Law of Universal Gravitation:

F12 = G * (m1 * m2) / r12^2
F12 = (6.67430 × 10^(-11) N m^2/kg^2) * (13.4 kg * 4.56 kg) / (6.00 m)^2
F12 ≈ 7.24615116 × 10^(-10) N

Now, let's calculate the gravitational force between the sphere at the origin (m1 = 13.4 kg) and the sphere located at (0.00, 5.00) m (m3 = 4.00 kg). The distance between their centers is:

r13 = √[(x3 - x1)^2 + (y3 - y1)^2]
r13 = √[(0.00 - 0.00)^2 + (5.00 - 0.00)^2]
r13 = √[0 + 25]
r13 = √25
r13 = 5.00 m

Using Newton's Law of Universal Gravitation:

F13 = G * (m1 * m3) / r13^2
F13 = (6.67430 × 10^(-11) N m^2/kg^2) * (13.4 kg * 4.00 kg) / (5.00 m)^2
F13 ≈ 6.77979968 × 10^(-10) N

So the net gravitational force on the sphere located at the origin is the vector sum of F12 and F13:

Fnet = F12 + F13
Fnet ≈ 7.24615116 × 10^(-10) N + 6.77979968 × 10^(-10) N
Fnet ≈ 1.80269528 × 10^(-9) N

So the net gravitational force on the sphere located at the origin is approximately 1.80269528 × 10^(-9) N.

Therefore, the value you obtained, 1.82502801 × 10^(-10) N, is not correct.