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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What scientific concept do you need to know in order to solve this problem?

Our tutors have indicated that to solve this problem you will need to apply the Torque Due to Weight concept. You can view video lessons to learn Torque Due to Weight. Or if you need more Torque Due to Weight practice, you can also practice Torque Due to Weight practice problems.

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