Three uniform spheres of masses m1 = 3.50 kg, m2 = 4.00 kg, and m3 = 5.50 kg are placed at the corners of a right triangle (see figure below). Calculate the resultant gravitational force on the object of mass m2, assuming the spheres are isolated from the rest of the Universe.

Maybe you tries to copy and paste the figure but that does not work here.

You need the trig to get the vector components.

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To calculate the resultant gravitational force on the object of mass m2, we can use the formula for gravitational force:

F = G * (m1 * m2) / r^2 + G * (m3 * m2) / r^2

where F is the gravitational force, G is the gravitational constant (approximately 6.67430 × 10^-11 m^3 kg^-1 s^-2), m1, m2, and m3 are the masses of the spheres, and r is the distance between the spheres.

First, we need to calculate the distance between the spheres. Since they are arranged at the corners of a right triangle, we can use the Pythagorean theorem to find r. Let's assume the triangle sides are labeled as follows: a, b, and c.

In this case, we can assign m3 as the hypotenuse of the right triangle and m1 and m2 as the other two sides.

Using the Pythagorean theorem:
c^2 = a^2 + b^2

Replacing c by m3 and a by m1, and b by m2:
m3^2 = m1^2 + m2^2

Simplifying and solving for m3:
m3 = sqrt(m1^2 + m2^2)
m3 = sqrt((3.50 kg)^2 + (4.00 kg)^2)
m3 ≈ 5.13 kg

Now we have the values of the masses and the distance between the spheres. Plugging these values into the formula for gravitational force, we get:

F = G * (m1 * m2) / r^2 + G * (m3 * m2) / r^2
F = (6.67430 × 10^-11 m^3 kg^-1 s^-2) * ((3.50 kg * 4.00 kg) / r^2) + (6.67430 × 10^-11 m^3 kg^-1 s^-2) * ((5.13 kg * 4.00 kg) / r^2)

Calculating the sum of these two terms will give us the resultant gravitational force on the object of mass m2.