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The (Figure 1) shows two thin beams joined at right angles. The vertical beam is 19.0 kg and 1.00 m long and the horizontal beam is 25.0 kg and 2.00 m long. Neglect the lateral dimensions of the beams.
A) Find the center of gravity of the two joined beams. Express your answer in the form x,y, taking the origin at the corner where the beams join.
B) Calculate the gravitational torque on the joined beams about an axis through the corner.

Answer :

A) The center of gravity of the two joined beams is;

(x, y) = (25/44), (19/88)

B) The gravitational torque on the joined beams about an axis through the corner is;

τ = 245 N.m

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 center of gravity on the horizontal beam

(x₂, y₂) is the center of gravity on the vertical beam

(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

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