Question

Very large forces are produced in joints when a person jumps from some height to the ground.

(a) Calculate the magnitude of the force produced if an 80.0-kg person jumps from a 0.600–m-high ledge and lands stiffly, compressing joint material 1.50 cm as a result. (Be certain to include the weight of the person.)

(b) In practice the knees bend almost involuntarily to help extend the distance over which you stop. Calculate the magnitude of the force produced if the stopping distance is 0.300 m.

(c) Compare both forces with the weight of the person.

Short answer

(a) The magnitude of the total force produced is\(3.21 \times {10^4}{\rm{ N}}\).

(b) The magnitude of the force produced if the stopping distance is 0.300 m is\(2.35 \times {10^3}{\rm{ N}}\).

(c) The magnitude of the force produced in jump 41 times of the weight and the force produced if the stopping distance is 0.300 m is 3 times of the weight.

Step-by-step solution

Weight

Weight is a force which acts on a mass due to acceleration due to gravity.

Mathematically,

\({W_g} = mg\)

Here, m is the mass of the body, and g is the acceleration due to gravity.

Total magnitude of the force produced

(a)

The weight of the person is,

\({W_g} = mg\)

Here, m is the mass of the person\(\left( {m = 80.0{\rm{ kg}}} \right)\)and g is the acceleration due to gravity\(\left( {g = 9.8{\rm{ m}}/{{\rm{s}}^2}} \right)\).

Putting all known values,

\(\begin{aligned}{W_g} &= \left( {80.0{\rm{ kg}}} \right) \times \left( {9.8{\rm{ m}}/{{\rm{s}}^2}} \right)\\ &= 784{\rm{ N}}\end{aligned}\)

The work done by conservative force is,

\(\begin{aligned}W& = mgh\\Fd &= mgh\end{aligned}\)

Here, F is the force on the joint, d is the compression in joint\(\left( {d = 1.50{\rm{ cm}}} \right)\), m is the mass of the person\(\left( {m = 80.0{\rm{ kg}}} \right)\)and g is the acceleration due to gravity\(\left( {g = 9.8{\rm{ m}}/{{\rm{s}}^2}} \right)\), and h is the height of ledge\(\left( {h = 0.600{\rm{ m}}} \right)\).

Putting all known values,

The expression for the force on the joint is,

\(F = \frac{{mgh}}{d}\)

Putting all known values,

\(\begin{aligned}F& = \frac{{\left( {80.0{\rm{ kg}}} \right) \times \left( {9.8{\rm{ m}}/{{\rm{s}}^2}} \right) \times \left( {0.600{\rm{ m}}} \right)}}{{\left( {1.50{\rm{ cm}}} \right)}}\\ &= \frac{{\left( {80.0{\rm{ kg}}} \right) \times \left( {9.8{\rm{ m}}/{{\rm{s}}^2}} \right) \times \left( {0.600{\rm{ m}}} \right)}}{{\left( {1.50{\rm{ cm}}} \right) \times \left( {\frac{{1{\rm{ m}}}}{{100{\rm{ cm}}}}} \right)}}\\ &= 3.136 \times {10^4}{\rm{ cm}}\end{aligned}\)

The total force on the joint is,

\({F_{total}} = F + {W_g}\)

Putting all known values,

\(\begin{aligned}{F_{total}}& = \left( {3.136 \times {{10}^4}{\rm{ N}}} \right) + \left( {784{\rm{ N}}} \right)\\ &= 3.21 \times {10^4}{\rm{ N}}\end{aligned}\)

Therefore, the required magnitude of the total force produced is \(3.21 \times {10^4}{\rm{ N}}\).

Magnitude of the force produced

(b)

The expression for the force on the joint is,

\({F_s} = \frac{{mgh}}{d}\)

Here, F is the force on the joint, d is the distance of stopping\(\left( {d = 0.300{\rm{ m}}} \right)\), m is the mass of the person\(\left( {m = 80.0{\rm{ kg}}} \right)\)and g is the acceleration due to gravity\(\left( {g = 9.8{\rm{ m}}/{{\rm{s}}^2}} \right)\), and h is the height of ledge\(\left( {h = 0.600{\rm{ m}}} \right)\).

Putting all known values,

\(\begin{aligned}F& = \frac{{\left( {80.0{\rm{ kg}}} \right) \times \left( {9.8{\rm{ m}}/{{\rm{s}}^2}} \right) \times \left( {0.600{\rm{ m}}} \right)}}{{\left( {{\rm{0}}{\rm{.300 m}}} \right)}}\\ &= 1568{\rm{ cm}}\end{aligned}\)

The total force on the joint is,

\({F_{total}} = {F_s} + {W_g}\)

Putting all known values,

\(\begin{aligned}{F_{total}} &= \left( {{\rm{1568 N}}} \right) + \left( {784{\rm{ N}}} \right)\\ &= 2.35 \times {10^3}{\rm{ N}}\end{aligned}\)

Therefore, the magnitude of the force produced if the stopping distance is \({\rm{0}}{\rm{.300 m}}\) is \(2.35 \times {10^3}{\rm{ N}}\).

Compare both forces with the weight of the person

(c)

The ratio of force in part (a) to the weight is,

\(\begin{aligned}\frac{F}{{{W_g}}} &= \frac{{3.21 \times {{10}^4}{\rm{ N}}}}{{784{\rm{ N}}}}\\ \approx 41\end{aligned}\)

The ratio of force in part (b) to the weight is,

\(\begin{aligned}\frac{{{F_s}}}{{{W_g}}} &= \frac{{2.35 \times {{10}^3}{\rm{ N}}}}{{784{\rm{ N}}}}\\ \approx 3\end{aligned}\)

Therefore, the force in part (a) is 41 and in part (b) is 3 times of the weight.