A 201-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 0.700 m/s2. Find the tension in the rope.



N

To find the tension in the rope, we need to determine the net force acting on the log.

Let's break down the forces acting on the log:

1. Weight (mg): The weight of the log is given by the mass (m) multiplied by the acceleration due to gravity (g = 9.8 m/s^2). Therefore, the weight is equal to mg = 201 kg × 9.8 m/s^2 = 1974.8 N. The weight acts vertically downward.

2. Normal force (N): The normal force is the force exerted by the ramp perpendicular to its surface. In this case, it acts perpendicular to the inclined surface of the ramp and opposes the weight of the log. Since the log is accelerating up the ramp, the normal force is less than the weight.

3. Friction (f): The friction force acts parallel to the ramp and opposes the motion of the log. The friction force can be calculated using the equation f = µN, where µ is the coefficient of friction and N is the normal force. In this case, f = 0.810 × N.

Now, let's resolve the forces along the ramp:

Since the log is accelerating up the ramp, the net force along the ramp (F_net) can be calculated using the equation F_net = m × a, where m is the mass of the log and a is the acceleration. In this case, F_net = 201 kg × 0.700 m/s^2 = 140.7 N.

The net force along the ramp can be calculated by subtracting the force of friction from the parallel force acting on the log:

F_net = T - f

Substituting the values:

140.7 N = T - 0.810 × N

Now, we have two equations involving the normal force N and the tension T:

N - mg = 0 (Equation 1)
T - 0.810 × N = 140.7 (Equation 2)

Solving these two equations simultaneously will give us the values of N and T.

From Equation 1, we have N = mg:

N = 201 kg × 9.8 m/s^2 = 1974.8 N

Substituting N = 1974.8 N into Equation 2:

T - 0.810 × 1974.8 N = 140.7 N

T - 1598.868 N = 140.7 N

T = 1598.868 N + 140.7 N

T = 1739.568 N

Therefore, the tension in the rope is 1739.568 N.