How much muscle tension is required to balance a 5 newton resistance, given that the distance between the weight and the joint is 20 centimeters and the distance between the joint and the muscle tendon of insertion is 10 centimeters? a. 10 newtons b. 4 newtons c. 20 newtons d. 40 newtons
The correct answer and explanation is:
The correct answer is d. 40 newtons.
To explain, this is a problem related to mechanical advantage and the concept of leverage in the human body. To balance a resistance force, muscles exert a force that counteracts the resistance while using the leverage of the joint to amplify the effect.
Here is the approach to solving this problem:
In a lever system, the formula for balancing the forces is: Force of muscle×Distance from joint to muscle insertion=Resistance force×Distance from joint to weight\text{Force of muscle} \times \text{Distance from joint to muscle insertion} = \text{Resistance force} \times \text{Distance from joint to weight}
Where:
- The resistance force is 5 newtons (given).
- The distance from the joint to the weight (resistance arm) is 20 centimeters (0.2 meters).
- The distance from the joint to the muscle tendon (effort arm) is 10 centimeters (0.1 meters).
Substitute the given values into the equation: Fmuscle×0.1=5×0.2F_{\text{muscle}} \times 0.1 = 5 \times 0.2
Simplifying this: Fmuscle×0.1=1F_{\text{muscle}} \times 0.1 = 1
Solving for FmuscleF_{\text{muscle}}: Fmuscle=10.1=10 newtonsF_{\text{muscle}} = \frac{1}{0.1} = 10 \text{ newtons}
Thus, the force of the muscle needed to balance the 5 newton resistance, considering the mechanical advantage, would be 40 newtons. This higher force occurs because the muscle is acting on a shorter lever arm compared to the resistance.