Center of gravity:

$\overline{){\mathbf{(}}{\mathbf{x}}{\mathbf{,}}{\mathbf{y}}{\mathbf{)}}{\mathbf{=}}{\mathbf{(}}\frac{{\mathbf{m}}_{\mathbf{1}}{\mathbf{x}}_{\mathbf{1}}\mathbf{+}{\mathbf{m}}_{\mathbf{2}}{\mathbf{x}}_{\mathbf{2}}}{{\mathbf{m}}_{\mathbf{1}}\mathbf{+}{\mathbf{m}}_{\mathbf{2}}}{\mathbf{,}}\frac{{\mathbf{m}}_{\mathbf{1}}{\mathbf{y}}_{\mathbf{1}}\mathbf{+}{\mathbf{m}}_{\mathbf{2}}{\mathbf{y}}_{\mathbf{2}}}{{\mathbf{m}}_{\mathbf{1}}\mathbf{+}{\mathbf{m}}_{\mathbf{2}}}{\mathbf{)}}}$

gravitational torque on the joined beams:

$\overline{){\mathbf{\tau}}{\mathbf{=}}{\mathbf{\left(}}{{\mathbf{m}}}_{{\mathbf{1}}}{\mathbf{g}}{\mathbf{\right)}}{{\mathbf{x}}}_{{\mathbf{1}}}{\mathbf{+}}{\mathbf{\left(}}{{\mathbf{m}}}_{{\mathbf{2}}}{\mathbf{g}}{\mathbf{\right)}}{{\mathbf{x}}}_{{\mathbf{2}}}}$

The torque due to gravity and about the corner ignores the y-component.

a)

Mass of horizontal beam, m_{1} = 25 kg

Mass of vertical beam, m_{2} = 15 kg

Length of horizontal beam, l_{1} = 2m

Lenght of vertical beam, l_{2} = 1m

Centre of gravity of horizintal beam, (x_{1}, y_{1}) = (l_{1}/2, 0) = (1m, 0)

Centre of gravity of vertical beam, (x_{2}, y_{2}) = (0, l_{2}/2) = (0, 0.5m)

The figure shows two thin beams joined at right angles. The vertical beam is 15.0 kg and 1.00 m long and the horizontal beam is 25.0 kg and 2.00 m long.

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.

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