A) The center of gravity of the two joined beams is;
<u><em>(x, y) = (25/44), (19/88)</em></u>
B) The gravitational torque on the joined beams about an axis through the corner is;
<u><em>τ = 245 N.m</em></u>
We are given;
Mass of vertical beam; m₂ = 19 kg
Mass of horizontal beam, m₁ = 25 kg
Length of horizontal beam; L₁ = 2 m
Length of vertical beam; L₂ = 1 m
A) Formula for center of gravity of the two joined beams is;
(x, y) = [(m₁x₁ + m₂x₂)/(m₁ + m₂)], [(m₁y₁ + m₂y₂)/(m₁ + m₂)]
Where;
(x₁, y₁) is the <em>center of gravity</em> on the <em>horizontal beam</em>
(x₂, y₂) is the <em>center of gravity</em> on the <em>vertical beam</em>
(x₁, y₁) = (L₁/2, 0)
Plugging in the relevant values gives;
(x₁, y₁) = (2/2, 0)
(x₁, y₁) = (1, 0)
(x₂, y₂) = (0, L₂/2)
Plugging in the relevant values gives;
(x₂, y₂) = (0, 1/2)
(x₂, y₂) = (0, 0.5)
Thus, center of gravity of the 2 joined beams is;
(x, y) = [((25 × 1) + (19 × 0))/(25 + 19)], [((25 × 0) + (19 × 0.5)))/(25 + 19)]
(x, y) = (25/44), (19/88)
B) Formula for the gravitational torque on the joined beams about an axis through the corner is given as;
τ = (m₁g)x₁ + (m₂g)x₂
Plugging in the relevant values;
τ = (25 × 9.8)1 + (19 × 9.8)0
τ = 245 N.m
Read more at; brainly.com/question/14825257
