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Chapter 7: Q59P (page 170)

When a high jumper is in a position such that his arms and lower legs are hanging vertically, and his thighs, trunk, and head are horizontal just above the bar, estimate how far below the torso鈥檚 median line the CM will be. Will this CM be outside the body? Use Table 7鈥�1.

Short Answer

Expert verified

The center of mass is 4.3% of the full body height, below the torso鈥檚 medium line. It is not outside the body.

Step by step solution

01

Define the center of mass (CM)

The CM of a two-particle system lies on the line joining the particles. It is closer to the heavier object. It is the point at which the whole mass of a body is concentrated.

02

Given information

From table 7-1,

The percent mass of the upper arm is \({m_{{\rm{UA}}}} = 6.6\;{\rm{units}}\).

The percent mass of the lower arm is \({m_{{\rm{LA}}}} = 4.2\;{\rm{units}}\).

The percent mass of the hand is \({m_{\rm{H}}} = 1.7\;{\rm{units}}\).

The percent mass of the lower leg is \({m_{{\rm{LL}}}} = 9.6\;{\rm{units}}\).

The percent mass of feet is \({m_{\rm{F}}} = 3.4\;{\rm{units}}\).

The percent mass of the full body is \({m_{{\rm{Full}}\;{\rm{body}}}} = 100\;{\rm{units}}\).

The distance of the upper arm from the center of mass is \({y_{{\rm{UA}}}} = \left( {81.2 - 71.7} \right)\;{\rm{units}}\).

The distance of the lower arm from the center of mass is \({y_{{\rm{LA}}}} = \left( {81.2 - 55.3} \right)\;{\rm{units}}\).

The distance of the hand from the center of mass is \({y_{\rm{H}}} = \left( {81.2 - 43.1} \right)\;{\rm{units}}\).

The distance of the lower leg from the center of mass is \({y_{{\rm{LL}}}} = \left( {28.5 - 18.2} \right)\;{\rm{units}}\).

The distance of the foot from the center of mass is \({y_{\rm{F}}} = \left( {28.5 - 1.8} \right)\;{\rm{units}}\).

03

Calculate the center of mass

The center of mass can be calculated as shown below:

\(\begin{array}{l}{y_{{\rm{cm}}}} = \frac{{{m_{{\rm{UA}}}}{y_{{\rm{UA}}}} + {m_{{\rm{LA}}}}{y_{{\rm{LA}}}} + {m_{\rm{H}}}{y_{\rm{H}}} + {m_{{\rm{LL}}}}{y_{{\rm{LL}}}} + {m_{\rm{F}}}{y_{\rm{F}}}}}{{{m_{{\rm{Full}}\;{\rm{body}}}}}}\\{y_{{\rm{cm}}}} = \frac{{\left[ \begin{array}{l}\left( {6.6} \right)\left( {81.2 - 71.7} \right) + \left( {4.2} \right)\left( {81.2 - 55.3} \right) + \left( {1.7} \right)\left( {81.2 - 43.1} \right)\\ + \left( {9.6} \right)\left( {28.5 - 18.2} \right) + \left( {3.4} \right)\left( {28.5 - 1.8} \right)\end{array} \right]}}{{\left( {100} \right)}}\\{y_{{\rm{cm}}}} = 4.25 \approx 4.3\end{array}\)

The center of mass is 4.3% of the full body height, below the torso鈥檚 medium line.

For a person of height 1.7 m, this will be

\(\begin{array}{l}{y_{{\rm{cm}}}} = \left( {4.3\% } \right)\left( {1.7\;{\rm{m}}} \right)\left( {\frac{{100\;{\rm{cm}}}}{{1\;{\rm{m}}}}} \right)\\{y_{{\rm{cm}}}} = 7.24\;{\rm{cm}}{\rm{.}}\end{array}\)

The thickness of the torso from the median line is about 10 cm. Therefore, the CM is well inside the body and not outside it.

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Most popular questions from this chapter

Astronomers estimate that a 2.0-km-diameter asteroid collides with the Earth once every million years. The collision could pose a threat to life on Earth. (a) Assume a spherical asteroid has a mass of 3200 kg for each cubic meter of volume and moves toward the Earth at\(15\;{\rm{km/s}}\). How much destructive energy could be released when it embeds itself in the Earth? (b) For comparison, a nuclear bomb could release about\(4.0 \times {10^{16}}\;{\rm{J}}\). How many such bombs would have to explode simultaneously to release the destructive energy of the asteroid collision with the Earth?

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Three cubes, of side \({l_ \circ }\), \(2{l_ \circ }\), and \(3{l_ \circ }\), are placed next to one another (in contact) with their centers along a straight line as shown in Fig. 7鈥�38. What is the position, along this line, of the CM of this system? Assume the cubes are made of the same uniform material.

FIGURE 7-38

Problem 52.
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