A fun classroom activity

The torques on Maurice's body are in equilibrium. There is one clockwise torque from the force of gravity at his center of mass; and two counterclockwise torques from the supporting scales.

For the torque analysis, we can take any point as the reference point; we might as well take the origin $x = 0$ (top of head) as reference point. Thus we find $mgx_{CM} = m_1gx_1 + m_2gx_2,$ where $m_1,m_2$ are the scale readings and $x_1,x_2$ the positions of the scales. Cancel $g$ . Note also that $m = m_1 + m_2$ : together, the two scales support Maurice's weight. $x_{CM} = \frac{m_1x_1 + m_2x_2}{m} = \frac{40\cdot 30 + 35\cdot 160}{75} \approx \boxed{\SI{91}{cm}};$ or also $x_{CM} = x_2 - \frac{m_1}m(x_2 - x_1) = 160 - \frac{40}{75}\cdot 130 \approx \boxed{\SI{91}{cm}}.$

Ans. 35/75*130+30=90.6667 or 91

Let $m_1=30kg,\space m=75kg,\space d_1=0.3m,\space d_2=1.6m$ . We ought to find $x$ , the distance from Maurice's head to his center of mass. There are exactly 3 forces involved in this process: The reaction force, $N_1$ , which has its origin at $d_1$ from Maurice's head, $N_2$ , the other reaction force and $F_W$ , the weight of the man. Translational equilibrium tells us that: $\vec{N_1}+\vec{N_2}+\vec{F_W}=0\implies N_2=F_W-N_1$ Applying rotational equilibrium relative to the origin of the graph, Maurice's head: $\large \tau_{N_1}+\tau_{N_2}+\tau_{F_W}=0$ And therefore: $N_1d_1+N_2d_2=F_Wx\implies m_1gd_1+(m-m_1)gd_2=mgx$ $x=\frac{m_1d_1+(m-m_1)d_2}{m}$

Numerically speaking, this is equivalent to $x=\frac{68}{75}m=90,(6)cm\approx \boxed{91 cm}$ .

Using the definition that an objects center of mass is the location where all the mass is averaged out we can take the weighted average of the scale's results. It reads 40 kilograms 30 centimeters from the top of his head so it must weigh 35 kilograms 160 centimeters from the top of his head. $\frac{40}{75}$ represents the fraction of his mass 30 centimeters from his head $\frac{35}{75}$ represents the fraction of his mass 160 centimeters from his head We can simply find the sum of $\frac{40}{75}$ times 30 and $\frac{35}{75}$ times 160 to find the weighted average of his mass. This results in a value of 90 $\frac{2}{3}$ centimeters which is approximately 91 centimeters.

The sum of the moments about the origin must be = 0. all we need is a single CCW moment arm with 75 kg to balance. Sum of CW moments are 40 x 30 + 35 x 160 = 6800 kg-cm.... 6800 / 75 = 90.666.. or 91 cm

Treating Maurice as a homogeneous bar, this is torquewise equivalent to balancing a bar of 130 cm on a fulcrum, with 40 kg and 35 kg on the ends. The torques on the ends must be equal, with the center of mas at a distance $d$ from where the head is supported:

$40\cdot d=35(130-d)$

This solves to $d=60.67$ , so rounding and adding the 30 cm from the top of the head yields 91 cm.

Obs.: There is an implicit a hypothesis that you must treat Maurice as a homogeneous bar.

As a simpleton, I mean pilot I'll show you we use weight & balance daily to solve this exact problem.

WxA=M. (Weight, Arm is the fixed distance from a reference point, Moment is are working variable). Add you moments together and divide by your total weight to get the Arm again, this becomes your center of gravity or center of mass in this case.

40kg*30cm=1200

35kg*160cm=5600

6800÷75=

                           90.666666

40/75 = 53.33 53.33 * 160 = approx 91. I feel like I lucked out on this one.

Hideous! The answer is 90+2/3. Why should we be rounding without warning before selecting from the multiple choice? Is the poser unable to find numbers thst yield an integer answer? Or does he/she delight in trying to convince the responder that their carefully thought-out logic must be wrong (when it isn’t)?

We can solve this problem as if it was a static structure.

Because it is static, we can say that the sum of the moments around any point equals 0.

When we take the sum of the moment around the right scale, we get:

40.g.1,3 = 75.g.X

X = $\frac{40.1,3}{75}$ = 0,69 cm to the left of the right scale, wich is +/- 91 cm from the top of his head.

Using the definition that an objects center of mass is the location where all the mass is averaged out we can take the weighted average of the scale's results. It reads 40 kilograms 30 centimeters from the top of his head so it must weigh 35 kilograms 160 centimeters from the top of his head. represents the fraction of his mass 30 centimeters from his head represents the fraction of his mass 160 centimeters from his head We can simply find the sum of times 30 and times 160 to find the weighted average of his mass. This results in a value of 90 centimeters which is approximately 91 centimeters.