A 77 kg sprinter is running the 100 m dash. At one instant, early in the race, his acceleration is 4.7 m/s^2.
a) What total force does the track surface exert on the sprinter? Assume his acceleration is parallel to the ground. Give your answer as a magnitude.
b) What total force does the track surface exert on the sprinter? Assume his acceleration is parallel to the ground, and give your answer as an angle with respect to the horizontal.
c) This force is applied to one foot (the other foot is in the air), which for a fraction of a second is stationary with respect to the track surface. Because the foot is stationary, the net force on it must be zero. Thus the force of the lower leg bone on the foot is equal but opposite to the force of the track on the foot. If the lower leg bone is 60 degrees from horizontal, what is the components of the leg’s force on the foot in the direction parallel to the leg? (Force components perpendicular to the leg can cause dislocation of the ankle joint.)
d) What is the components of the leg’s force on the foot in the direction perpendicular to the leg?
Can someone help me on this problem please? I tried googling it, and found I should use a^2 + b^2 = c^2 for A, but I'm so lost I don't even know what numbers I should be using to plug into that equation with.
Sure, I can help you with this problem. Let's break it down step by step.
a) To find the total force exerted by the track surface on the sprinter, we can apply Newton's second law of motion, which states that force (F) is equal to mass (m) multiplied by acceleration (a). In this case, the mass of the sprinter is given as 77 kg, and the acceleration is given as 4.7 m/s^2. So, we can use the formula F = m * a to find the force.
F = 77 kg * 4.7 m/s^2
F = 361.9 N (rounded to the nearest tenth)
Therefore, the total force exerted by the track surface on the sprinter is approximately 361.9 N.
b) To determine the angle with respect to the horizontal at which the total force is exerted on the sprinter, we need to calculate the angle between the force vector and the horizontal direction. Since we have the acceleration (which is parallel to the ground), the total force will also be parallel to the ground. Thus, the angle with respect to the horizontal would be 0 degrees.
c) In this part, we are considering the force exerted by the lower leg bone on the foot when it is stationary with respect to the track surface. Since the net force on the foot is zero, the force of the lower leg bone on the foot must be equal in magnitude but opposite in direction to the force of the track on the foot.
Given that the lower leg bone is at a 60-degree angle from the horizontal, we can find the component of the leg's force on the foot parallel to the leg. To do so, we'll multiply the force magnitude by the cosine of the angle.
Force parallel to the leg = F * cos(angle)
Force parallel to the leg = 361.9 N * cos(60°)
Force parallel to the leg ≈ 180.9 N (rounded to the nearest tenth)
Therefore, the component of the leg's force on the foot in the direction parallel to the leg is approximately 180.9 N.
d) Similarly, to find the component of the leg's force on the foot perpendicular to the leg, we'll multiply the force magnitude by the sine of the angle.
Force perpendicular to the leg = F * sin(angle)
Force perpendicular to the leg = 361.9 N * sin(60°)
Force perpendicular to the leg ≈ 314.8 N (rounded to the nearest tenth)
Therefore, the component of the leg's force on the foot in the direction perpendicular to the leg is approximately 314.8 N.
I hope this helps! Let me know if you have any further questions.
You can find the solution in
http://www.cabrillo.edu/~cfigueroa/11/11prob_sets/Solutions/11prob_sol_5%20.pdf