To find the horizontal acceleration of the tire, we need to consider the forces acting on the tire and 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 scenario, the forces acting on the tire are:
Applied force (F_applied): The force applied by the athlete, which is 255 N at an angle of 30° upward.
Force due to friction (F_friction): The force opposing the motion of the tire, given by the coefficient of kinetic friction (μ_k) multiplied by the normal force (N). The normal force is the force exerted by the surface on the tire perpendicular to the surface.
Weight (F_weight): The force due to gravity acting on the tire, given by the mass of the tire (m) multiplied by the acceleration due to gravity (g ≈ 9.8 m/s^2).
Since we are interested in the horizontal acceleration, we need to consider only the horizontal components of the forces.
The horizontal component of the applied force (F_applied) can be calculated using trigonometry: F_applied_horizontal = F_applied * cos(30°)
The force due to friction (F_friction) is given by: F_friction = μ_k * N
The normal force (N) is equal in magnitude and opposite in direction to the weight of the tire: N = F_weight = m * g
Now, we can write the equation for the net force acting on the tire: Net force = F_applied_horizontal - F_friction
Since the net force is equal to the mass of the tire (m) multiplied by its acceleration (a), we have: m * a = F_applied_horizontal - F_friction
Substituting the known values into the equation and solving for the horizontal acceleration (a) will give us the answer.
Let's calculate it step by step:
Given: m (mass of the tire) = 36 kg F_applied = 255 N μ_k (coefficient of kinetic friction) = 0.73 g (acceleration due to gravity) ≈ 9.8 m/s^2
Calculations: F_applied_horizontal = F_applied * cos(30°) F_applied_horizontal = 255 N * cos(30°) ≈ 220.971 N
F_weight = m * g F_weight = 36 kg * 9.8 m/s^2 ≈ 352.8 N
N = F_weight = 352.8 N
F_friction = μ_k * N F_friction = 0.73 * 352.8 N ≈ 257.424 N
Net force = F_applied_horizontal - F_friction Net force = 220.971 N - 257.424 N ≈ -36.453 N (negative because the friction force opposes the applied force)
m * a = Net force 36 kg * a = -36.453 N
Solving for a: a = (-36.453 N) / (36 kg) a ≈ -1.013 m/s^2
The horizontal acceleration of the tire is approximately -1.013 m/s^2 (negative sign indicates that it is in the opposite direction of the applied force).