A person with a body mass of 65 kg was running over level ground with a strong tailwind, while
wearing a 10 kg weighted vest. The tailwind applied a force of 125 N to the runner’s body in the horizontal
direction. Early in the stance phase, the runner generated a vertical ground reaction force of 1400 N and an
anterior-posterior ground reaction force of -350 N. What was the resultant acceleration of the runner’s center
of mass (magnitude and direction)?
To find the resultant acceleration of the runner's center of mass, we need to analyze the forces acting on the runner.
1. Calculate the net horizontal force:
The tailwind force and anterior-posterior ground reaction force are the only horizontal forces acting on the runner. We need to subtract the force exerted by the tailwind from the anterior-posterior ground reaction force to find the net force in the horizontal direction.
Net horizontal force = Anterior-posterior ground reaction force - Tailwind force
Net horizontal force = -350 N - (-125 N)
Net horizontal force = -350 N + 125 N
Net horizontal force = -225 N
2. Calculate the net vertical force:
The vertical ground reaction force and the force exerted by gravity on the runner are the only vertical forces acting on the runner. We need to subtract the force exerted by gravity from the vertical ground reaction force to find the net force in the vertical direction.
Net vertical force = Vertical ground reaction force - Weight
Net vertical force = 1400 N - (65 kg + 10 kg) * 9.8 m/s^2
Net vertical force = 1400 N - 75 kg * 9.8 m/s^2
Net vertical force = 1400 N - 735 N
Net vertical force = 665 N
3. Calculate the acceleration:
Now that we have the net horizontal force and net vertical force, we can use Newton's second law (F = ma) to calculate the acceleration.
Resultant acceleration = √((Net horizontal force)^2 + (Net vertical force)^2) / mass
Resultant acceleration = √((-225 N)^2 + (665 N)^2) / (65 kg + 10 kg)
Resultant acceleration = √(50625 N^2 + 442225 N^2) / 75 kg
Resultant acceleration = √(492850 N^2) / 75 kg
Resultant acceleration = √(492850 N^2) / √(75 kg^2)
Resultant acceleration = 701.27 N / 8.66 kg (approximate)
Resultant acceleration = 80.97 m/s^2 (approximate)
The resultant acceleration of the runner's center of mass is approximately 80.97 m/s^2.
To find the resultant acceleration of the runner's center of mass, we can use Newton's second law of motion, which states that the net force acting on an object is equal to the mass of the object multiplied by its acceleration.
In this case, we have the following forces acting on the runner:
1. The tailwind force in the horizontal direction: 125 N
2. The vertical ground reaction force: 1400 N
3. The anterior-posterior ground reaction force: -350 N
To find the resultant acceleration, we need to find the net force acting on the runner in both the horizontal and vertical directions.
In the horizontal direction, the net force is given by:
Net horizontal force = Tailwind force + Anterior-posterior ground reaction force
Net horizontal force = 125 N + (-350 N)
Net horizontal force = -225 N
In the vertical direction, the net force is given by:
Net vertical force = Vertical ground reaction force
Net vertical force = 1400 N
Now that we have the net forces in both directions, we can find the resultant acceleration using Newton's second law.
Net horizontal force = mass × horizontal acceleration
-225 N = (65 kg + 10 kg) × horizontal acceleration
-225 N = 75 kg × horizontal acceleration
horizontal acceleration = -225 N / 75 kg
horizontal acceleration = -3 m/s² (negative sign indicates that the runner is decelerating)
Net vertical force = mass × vertical acceleration
1400 N = (65 kg + 10 kg) × vertical acceleration
1400 N = 75 kg × vertical acceleration
vertical acceleration = 1400 N / 75 kg
vertical acceleration = 18.67 m/s²
The resultant acceleration of the runner's center of mass is approximately 3 m/s² in the horizontal direction (deceleration) and 18.67 m/s² in the vertical direction.