Question

The (Figure 1) shows two thin beams joined at right angles. The vertical beam is 16.0 kg and 1.00 m long and the horizontal beam is 26.0 kg and 2.00 m long. Neglect the lateral dimensions of the beams.

Part 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 Part B Calculate the gravitational torque on the joined beams about an axis through the corner. 1.00 m 2.00 m

Answer 1

Part A centre of gravity T_cm = r_1 m_1 + r_2 m_2/m_1 + m_2 Therefore r_1 = (0, 0.5 m) m_1 = 16 kg r_2 = (1.00 m, 0.00 m) m_2 = 26 kg Therefore x_cm = m_1 x_1 + m_2 x_2/m_1 + m_2 = 0 + 1 times 26/16 + 26 m x_cm = 0.619 m y_cm = m_1 y_1 + m_2 y_2/m_1 + m_2 = (0.5) times 16/16 + 26 m = 0.19 m Therefore x_cm = 0.619 m y_cm = 0.19 m r_cm = (0.619 m, 0.19 m) Part B Torque tau_0 (x_1 i) times m_2 g (-j) tau_0 = (1 i) times (26 times 9.8 (-j)) tau_0 = -254.8 k N-m or, tau_0 = -255 k N-m rightarrow torque is into the plane of paper