Chapter 7, Problem 55P

To determine

The position of the center of mass of an arm bent at a right angle.

Answer to Problem 55P

The center of mass of a bent arm is $x_{cm}=21.7 \text{cm}, y_{cm}=7.6 \text{cm}$

Explanation of Solution

Given:

From table of mass proportions, mass of the upper arm is 6.6 kg, the lower arm is 4.2 kg, and the hand is 1.7 kg. Assuming the height of the person is 155 cm.

Formula used:

The center of mass is calculated as,

  $x_{cm}=\frac{{\displaystyle \sum mx}}{{\displaystyle \sum m}}$

Calculation:

Considering origin at the shoulder and the upper arm is parallel to the x-axis and lower arm and hard in the y-axis.

  $x_{upper arm}=81.2-71.7=9.5$ ; $x_{lower arm}=81.2-62.2=19.0$ ; $x_{hand}=81.2-62.2=19.0$

For calculating the relative mass of a single-arm, no need to consider the factor of two. Since in the denominator, the total mass of an arm is considered.

To calculate the center of mass, relative mass percentage can be used. Thus, the center of mass of a bent arm is

  $\begin{array}{c}{x}_{cm}=\frac{x_{upper arm}{m}_{upper arm}+x_{lower arm}{m}_{lower arm}+x_{hand}{m}_{hand}}{m_{upper arm}+m_{lower arm}+m_{hand}}\\ =\frac{(9.5)(6.6)+(19)(5.9)}{6.6+4.2+1.7}\\ =14\end{array}$

  $\begin{array}{c}{y}_{cm}=\frac{y_{upper arm}{m}_{upper arm}+y_{lower arm}{m}_{lower arm}+y_{hand}{m}_{hand}}{m_{upper arm}+m_{lower arm}+m_{hand}}\\ =\frac{(0)(6.6)+(62.2-55.3)(4.2)+(62.2-43.1)(1.7)}{6.6+4.2+1.7}\\ =4.92\end{array}$

Now, calculating center of mass for the person’s actual height that is for 155 cm by converting, distance percentage with the actual distance.

  $x_{cm}=(14\%)(155 \text{cm})=21.7 \text{cm}$

  $y_{cm}=(4.92\%)(155 \text{cm})=7.6 \text{cm}$

Conclusion:

The center of mass of a bent arm is $x_{cm}=21.7 \text{cm}, y_{cm}=7.6 \text{cm}$