Four equal masses of 4.0 kg are placed at the vertices of a square of side 8.0 m. Find the force on the mass at the top left vertex.

Fx = +G m m /8^2 + G m m /(8sqrt2)^2 = G m^2 [1/64 + 1/128]

Fy =-G m^2[ 1/64 +1/128]

Doesn't make sense. Why would putting mass on the corners of a square cause a force on one of the masses?

Force of gravity?

(a) Given that the energy released per fission of 235U nucleus is 200MeV, calculate (i) the number of fissions and (ii) the amount of energy (in joules) available if 50g of 235U is completely fissioned.

To find the force on the mass at the top left vertex, we can use the concept of gravitational force.

The force between two masses due to gravity can be calculated using Newton's Law of Universal Gravitation:

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

Where:
F is the force between the two masses,
G is the gravitational constant (approximately 6.674 × 10^-11 N m^2 / kg^2),
m1 and m2 are the masses of the two objects, and
r is the distance between the centers of the two masses.

In this case, we want to find the force on the top left vertex of the square. The force is being exerted by the other three masses at the remaining vertices.

Since all the masses are equal, let's assume each mass is "m".

The distance between the two masses is the diagonal of the square. The square has a side length of 8.0 m, which means the diagonal can be found using the Pythagorean theorem:

d = sqrt(8.0^2 + 8.0^2)

Now we can calculate the force:

F = G * (m * m) / d^2

Substituting the values:

F = 6.674 × 10^-11 N m^2 / kg^2 * (4.0 kg * 4.0 kg) / (sqrt(8.0^2 + 8.0^2))^2

Simplifying this equation will give you the force on the mass at the top left vertex.