A 255 kg log is pulled up a ramp by means of a rope that is parallel to the surface of the ramp. The ramp is inclined at 30.0° with respect to the horizontal. The coefficient of kinetic friction between the log and the ramp is 0.870, and the log has an acceleration of 0.900 m/s2. Find the tension in the rope.
To find the tension in the rope, 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.
First, let's find the net force acting on the log. There are two forces acting on the log: the force of gravity and the force of friction.
1. Force of Gravity: The force of gravity is given by the equation F_gravity = m * g, where m is the mass of the log and g is the acceleration due to gravity (approximately 9.8 m/s^2).
F_gravity = (255 kg) * (9.8 m/s^2) = 2499 N (rounded to three significant figures).
2. Force of Friction: The force of friction is given by the equation F_friction = coefficient_of_friction * normal_force. In this case, the normal force is equal to the component of gravity acting perpendicular to the ramp, given by the equation normal_force = m * g * cos(theta), where theta is the angle of the ramp (30 degrees).
normal_force = (255 kg) * (9.8 m/s^2) * cos(30 degrees) = 2211 N (rounded to three significant figures).
F_friction = (0.870) * (2211 N) = 1921 N (rounded to three significant figures).
Now, let's calculate the net force by subtracting the force of friction from the force of gravity:
Net Force = F_gravity - F_friction = 2499 N - 1921 N = 578 N (rounded to three significant figures).
Finally, we can use Newton's second law to find the tension in the rope:
Net Force = m * a, where m is the mass of the log and a is the acceleration.
Tension = Net Force + Force of friction = (255 kg) * (0.900 m/s^2) + 1921 N = 679 N (rounded to three significant figures).
Therefore, the tension in the rope is approximately 679 N.