Question

A tractor pulls a sled of mass \(M=1000\). kg across level ground. The coefficient of kinetic friction between the sled and the ground is \(\mu_{k}=0.600 .\) The tractor pulls the sled by a rope that connects to the sled at an angle of \(\theta=30.0^{\circ}\) above the horizontal. What magnitude of tension in the rope is necessary to move the sled horizontally with an acceleration \(a=2.00 \mathrm{~m} / \mathrm{s}^{2} ?\)

Short answer

Question: A sled with a mass of 1000 kg is being pulled by a rope that makes an angle of 30° with the horizontal. The coefficient of kinetic friction between the sled and the ground is 0.600. If the sled is accelerating horizontally at 2.00 m/s², determine the magnitude of tension in the rope. Answer: The magnitude of tension in the rope necessary to move the sled horizontally with an acceleration of 2.00 m/s² is approximately 2395.36 N.

Step-by-step solution

Analyze the forces acting on the sled.

Draw a free body diagram for the sled, indicating all the forces acting on it. The forces include the gravitational force (M × g), the normal force (Fn), the tension force in the rope (T), and the kinetic friction force (Fk).

Resolve the tension force into horizontal and vertical components.

The tension force can be resolved into a horizontal component (Tx) and a vertical component (Ty) as follows: Tx = T × cos(θ) Ty = T × sin(θ)

Write the equations for the forces in the vertical direction.

In the vertical direction, the sled is in equilibrium. The following equation can be written: Fn - Ty - M × g = 0

Write the equations for the forces in the horizontal direction.

In the horizontal direction, the following equation can be written using Newton's second law of motion: M × a = Tx - Fk

Write the equation for kinetic friction force.

The kinetic friction force can be written as: Fk = μk × Fn

Substitute the given values and solve for the tension T.

Plug in the values for M, g, μk, θ, and a into the previous equations and solve for T. Use the following values: M = 1000 kg, g = 9.81 m/s², μk = 0.600, θ = 30°, and a = 2.00 m/s². First, find Fn using the vertical forces equation: Fn = M × g + Ty Fn = M × g + T × sin(θ) Now, find Fk using the kinetic friction equation: Fk = μk × Fn Substitute Fn and Fk into the horizontal forces equation: M × a = Tx - Fk M × a = T × cos(θ) - μk × Fn Substitute Fn equation in the above equation and solve for T: T = M × a / (cos(θ) - μk × sin(θ)) Finally, substitute the given values and solve for T: T = 1000 × 2 / (cos(30°) - 0.600 × sin(30°)) T ≈ 2395.36 N Thus, the magnitude of tension in the rope necessary to move the sled horizontally with an acceleration of 2.00 m/s² is approximately 2395.36 N.

Key concepts

Kinetic Friction

Kinetic friction is the force that opposes the motion of two surfaces sliding past each other. It is crucial in our problem as it influences how much force the tractor has to exert to move the sled. Kinetic friction can be calculated using the formula:

• \( F_k = \mu_k \times F_n \)

where \( F_k \) is the kinetic friction force, \( \mu_k \) is the coefficient of kinetic friction, and \( F_n \) is the normal force.
In this exercise, the coefficient of kinetic friction is given as 0.600, which means that 60% of the normal force will work against the sled's motion.
The normal force, \( F_n \), is primarily the gravitational force acting on the sled minus any vertical components of forces applied at an angle, such as tension.
Overall, understanding the role of kinetic friction is key since it forms part of the resisting forces that we need to counterbalance to allow for horizontal acceleration.

Newton's Second Law

Newton's Second Law is a fundamental principle of physics that describes how the velocity of an object changes when it is subjected to an external force. It is usually stated as:

• \( F = m \times a \)

This formula indicates that the force applied on an object is equal to the mass of the object times its acceleration.
In our exercise, the sled is being pulled across the ground by a tension force, and kinetic friction is acting against this motion.
We utilize Newton's Second Law in the horizontal direction to equate the net force needed for acceleration to the difference between the tension force's horizontal component and the kinetic friction:

• \( M \times a = T_x - F_k \)

This relationship is critical in determining the necessary tension in the rope to achieve the desired acceleration of the sled.

Free Body Diagram

A Free Body Diagram (FBD) is a graphical representation used to visualize the forces acting on an object. In physics problems such as this, drawing an FBD is a fundamental step to analyze all forces involved.
For the given problem, the FBD of the sled includes:

• Gravitational force \( (M \times g) \)
• Normal force \( (F_n) \)
• Tension, with components either horizontally \( (T_x) \) or vertically \( (T_y) \)
• Kinetic friction force \( (F_k) \)

In the analysis, the vertical direction is balanced, meaning the sum of forces equals zero, while the horizontal direction allows us to apply Newton's Second Law.
Starting with a clear FBD simplifies the understanding and calculation of forces, leading to an easier solution of the entire problem.

Tension Force Analysis

The tension force in the rope has a significant role in this problem as it is responsible for moving the sled. Since the rope is at an angle, the tension force needs to be divided into horizontal \( (T_x) \) and vertical \( (T_y) \) components:

• \( T_x = T \times \cos(\theta) \)
• \( T_y = T \times \sin(\theta) \)

These components are crucial for solving the problem:- The horizontal component, \( T_x \), is what contributes to moving the sled forward.- The vertical component, \( T_y \), affects the normal force and thus, the kinetic friction.
In analyzing a physics problem, decomposing an angled force into its components is essential. This process helps in cleanly applying Newton's laws and understanding how the forces lead to the sled's motion.