A winch is used to drag a 375 N crate up a ramp at a constant speed of 75 cm/s by means of a rope that pulls parallel to the surface of the ramp. The rope slopes upward at 33 degreesabove the horizontal, and the coefficient of kinetic friction between the ramp and the crate is 0.25.

Question

A winch is used to drag a 375 N  crate up a ramp at a constant speed of 75 cm/s  by means of a rope that pulls parallel to the surface of the ramp. The rope slopes upward at 33 degreesabove the horizontal, and the coefficient of kinetic friction between the ramp and the crate is 0.25.

Bot · Answer

To solve this problem, we can break it down into a few steps:

1. Find the gravitational force acting on the crate.
2. Find the normal force between the crate and the ramp.
3. Calculate the frictional force acting on the crate.
4. Determine the net force acting on the crate.
5. Calculate the tension in the rope.

Step 1: Gravitational force
The gravitational force acting on the crate is given by F_gravity = m * g, where m is the mass of the crate and g is the acceleration due to gravity (approximately 9.81 m/s²). We can find the mass of the crate by dividing the weight by the acceleration due to gravity.

mass = (375 N) / (9.81 m/s²)
mass ≈ 38.2 kg

F_gravity = (38.2 kg) * (9.81 m/s²)
F_gravity = 375 N

Step 2: Normal force
The normal force is the component of the gravitational force that is perpendicular to the ramp. To find it, we can use the angle between the ramp and the horizontal (33°).

F_normal = F_gravity * cos(33°)

F_normal = 375 N * cos(33°)
F_normal ≈ 315 N

Step 3: Frictional force
The frictional force is given by F_friction = μ * F_normal, where μ is the coefficient of kinetic friction between the crate and the ramp.

F_friction = 0.25 * 315 N
F_friction = 78.75 N

Step 4: Net force
Since the crate is moving up the ramp at a constant speed, the net force acting on it along the ramp must be zero. Therefore, the force exerted by the rope pulling the crate must balance the sum of the frictional force and the component of the gravitational force acting parallel to the ramp. To find this component, we can use the sine of the angle between the ramp and the horizontal.

F_gravity_parallel = F_gravity * sin(33°)

F_gravity_parallel = 375 N * sin(33°)
F_gravity_parallel ≈ 205.5 N

Now, we can find the net force:

F_net = F_rope - F_gravity_parallel - F_friction

Since F_net = 0, the force exerted by the rope (F_rope) is:

F_rope = F_gravity_parallel + F_friction

F_rope = 205.5 N + 78.75 N
F_rope ≈ 284.25 N

So, the tension in the rope is approximately 284.25 N.