Question

Four masses are at the corners of a square of length ℓ = 20.0 cm and a fifth mass is at the center of the square. The masses are m₁ = 5.00 g, m₂ = 3.00 g, m₃ = 1.00 g, m₄ = 5.00 g, and m₅ = 1.50 g.

a-Draw the free body diagram for fifth mass.

b-Determine the net gravitational force on the fifth mass in unit vector notation.

Answer 1

Its given that,

m1 = 5 g ; m2 = 3 g ; m3 = 1 g ; m4 = 5 g and m5 = 1.5 g (m5 is at center) ; l = 20 cm

We need to find the net gravitational force on the fifth mass. Let it be F(net)

We know that, the gravitational force is given by;

F(gravity) = G M m / r²

the distance of all masses from m5 will be half of the diagonal of the square(as m5 is at center).

r = l x sqrt(2)/2 = 10 x 1.414 cm = 0.1414m

From symmetry we find that, the forces from m1 and m4 on m5 will cancel each other as they are of equal magnitude and opposite direction. So the net force on m5 will be due to m2 and m3.

F(net) = G m5 / r [ m2 - m3] = 6.67 x 10^-11 x 1.5 x 10^-3 x [ 3 - 1 ] x 10^-3 / (0.1414)²

F(net) = 20.01 x 10^-17 / 0.02 = 1 x 10^-14 N

In unit vector notation we will write it as (because the force is directed along the diagonal along m2 as m2 >m1)

Vector F(net) = F(net) [ cos 45 i + sin 45 j] = 1 x 10^-14 [ 0.71 i + 0.71 j] = 7.1 x 10^-15 i + 7.1 x 10^-15 j

Hence, vector F(net) = 7.1 x 10^-15 i + 7.1 x 10^-15 j

[my apologies, i couldnt upload the FBD diagram for part (a)]