Find the position of the center of mass of two bodies points of masses m1 and m2 joined by a rod of mass negligible in length d. Find the acceleration of the center of mass for the case that m1 = 1 [kg] and m2 = 3 [kg] and the applied forces of the figure with d = 2 [m]

Explanation:

If m₁ is at the origin, then the center of mass is at:

x = (m₁ × 0 m + m₂ × d m) / (m₁ + m₂)

x = m₂ d / (m₁ + m₂)

If m₁ = 1 kg, m₂ = 3 kg, and d = 2 m:

x = (3 kg) (2 m) / (1 kg + 3 kg)

x = 1.5 m

Sum of forces in the x direction:

∑F = ma

16 N = (1 kg + 3 kg) aₓ

aₓ = 4 m/s²

Sum of forces in the y direction:

∑F = ma

20 N = (1 kg + 3 kg) aᵧ

aᵧ = 5 m/s²

pythagoras' theorem

square root (-5^2+9^2)

square root (25+81)

square root (106)

angle is tan^-1(9/5)

neg x, pos y