A 287-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 0.810, and the log has an acceleration of 0.900 m/s2. Find the tension in the rope.
Tension= friction+ma-gravity down the ramp
tension= mu*mg*cos30+ma-mgsinTheta
To find the tension in the rope, we need to consider the forces acting on the log. There are three main forces at play: the force of gravity acting vertically downward, the normal force acting perpendicular to the ramp, and the force of friction acting parallel to the ramp.
Let's break down each force:
1. Force of gravity: This force is given by the formula F_gravity = m * g, where m is the mass of the log and g is the acceleration due to gravity (approximated as 9.8 m/s^2). So, the force of gravity acting on the log is F_gravity = (287 kg) * (9.8 m/s^2) = 2813.6 N.
2. Normal force: The normal force is the force exerted by a surface to support the weight of an object resting on it. In this case, the normal force is equal in magnitude and opposite in direction to the force of gravity. So, the normal force is N = 2813.6 N.
3. Force of friction: The force of friction can be calculated using the formula F_friction = μ * N, where μ is the coefficient of kinetic friction and N is the normal force. Plugging in the given values, we have F_friction = (0.810) * (2813.6 N) = 2278.04 N. Since the log is accelerating, the force of friction acts in the opposite direction of the log's motion.
Now, let's determine the net force acting on the log. The net force is given by the formula F_net = m * a, where m is the mass of the log and a is the acceleration. Plugging in the values, we have F_net = (287 kg) * (0.900 m/s^2) = 258.3 N.
The tension in the rope is equal to the sum of the force of friction and the net force: Tension = F_friction + F_net = 2278.04 N + 258.3 N = 2536.34 N.