a long non uniform board of length 8 m and mass m = 12 kg is suspended by two ropes if the tensions in the ropes are mg/3 (on left) and 2mg/3 (on right) what is the location of the board's center of mass
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To find the location of the board's center of mass, we need to consider the distribution of its mass along its length.
Since the board is non-uniform, we can assume that its mass is distributed continuously along its length. Let's denote the position along the length of the board as x, where x = 0 is the left end and x = 8 m is the right end.
We can find the position of the center of mass (denoted as x_cm) using the concept of torque. The torque acting on the board about any point must be balanced for it to be in rotational equilibrium.
Let's choose the left end of the board as the reference point. The torque exerted by the left rope (T1) is given by T1 * 0, as its distance from the reference point is 0. The torque exerted by the right rope (T2) is given by T2 * (8 m), as its distance from the reference point is 8 m.
For rotational equilibrium, the sum of the torques must be equal to zero:
T2 * (8 m) - T1 * 0 = 0
Given T1 = mg/3 and T2 = 2mg/3:
(2mg/3) * (8 m) - (mg/3) * 0 = 0
Simplifying the equation:
16mg/3 = 0
This implies that the left and right torques are equal, suggesting that the board has a symmetrical distribution of mass. In this case, the center of mass would be located exactly at the midpoint of the board, which is x_cm = 8 m / 2 = 4 m from the left end.
Therefore, the center of mass of the board is located 4 m from the left end.