Question

to determine the location of her center of mass, a physics student lies on a lightweight plank supported by two scales 2.50 m apart, as indicated in the figure below. if the left scale reads 290 n, and the right scale reads 122 n, find the distance from the students head to her center of mass.
axis of rotation
0
2.50 m

a. 0.15 m
b. 0.28 m
c. 0.34 m
d. 0.74 m
e. 0.96 m

Explanation:

Step1: Calculate the total weight of the student

The total weight \(W\) of the student is the sum of the forces read by the two scales. Let \(F_1 = 290\ N\) be the force on the left - hand scale and \(F_2=122\ N\) be the force on the right - hand scale. So \(W=F_1 + F_2=290+122 = 412\ N\).

Step2: Set up the torque equation

Take the left - hand scale as the axis of rotation. Let the distance from the student's head (left - hand side) to the center of mass be \(x\). The distance from the left - hand scale to the right - hand scale is \(L = 2.50\ m\). The torque \(\tau\) about the left - hand scale is zero since the plank is in rotational equilibrium. The weight of the student acts at the center of mass and creates a clock - wise torque, and the force on the right - hand scale creates a counter - clockwise torque. The torque due to the weight of the student is \(\tau_{W}=W\times x\) (clock - wise) and the torque due to the right - hand scale is \(\tau_{F_2}=F_2\times L\) (counter - clockwise). In equilibrium, \(\tau_{W}=\tau_{F_2}\), so \(W\times x=F_2\times L\).

Step3: Solve for \(x\)

We know \(W = 412\ N\), \(F_2 = 122\ N\) and \(L = 2.50\ m\). Rearranging the equation \(W\times x=F_2\times L\) for \(x\), we get \(x=\frac{F_2\times L}{W}\). Substitute the values: \(x=\frac{122\times2.50}{412}\).
\[x=\frac{305}{412}\approx0.74\ m\]

Answer:

d. \(0.74\ m\)