Question

The center of mass of a person may be determined by an arrangement A light plank rests on two scales separated by a distance of d = 1.80 m and reading fg1 = 495 N and fg2 = 365 N. Determine the distance of the girl's center of mass from her feet?

Answer 1

To find the forces F1 and F2 required to hold the plank in equilibrium, we calculate the torques and use the static equilibrium conditions. The calculated force magnitudes are F1 = 65.33 N and F2 = 130.67 N.

Step-by-step explanation:

To determine the magnitudes of the forces F1 and F2 when a person carries a plank of wood, we will use the concept of torque and the equilibrium condition. The plank is 2.00 m long and has a mass of 20.0 kg, with its center of gravity at the midpoint. The hand holding the plank is 0.50 m from one end.

First, we convert the mass of the plank to weight (W = mg) using the acceleration due to gravity (g = 9.8 m/s2):

W = 20.0 kg × 9.8 m/s2 = 196 N

The plank is in static equilibrium, so the sum of forces and sum of torques must be zero. Considering torques about one end of the plank (where F1 is applied), and taking counter-clockwise torque as positive, the torque due to the weight of the plank is:

Torque due to weight = 196 N × 1.00 m = 196 Nm

For equilibrium, the torque due to F2 should be equal and opposite:

Torque due to F2 = F2 × 1.50 m = -196 Nm

From this, F2 can be calculated:

F2 = 196 Nm / 1.50 m = 130.67 N

Using the equilibrium condition for forces:

F1 + F2 = W

We find F1:

F1 = 196 N - 130.67 N = 65.33 N

Thus, the magnitudes of the forces are F1 = 65.33 N and F2 = 130.67 N.