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Chapter 4: Problem 32

A skier of mass 65.0 \(\mathrm{kg}\) is pulled up a snow-covered slope at constant speed by a tow rope that is parallel to the ground. The ground slopes upward at a constant angle of \(26.0^{\circ}\) above the horizontal, and you can ignore friction. (a) Draw a clearly labeled free body diagram for the skier. (b) Calculate the tension in the tow rope.?

Short Answer

Expert verified
The tension in the tow rope is approximately 278.7 N.

Step by step solution

01

Drawing the Free Body Diagram

To begin, we should create a free body diagram representing all the forces acting on the skier. The forces include:1. **Gravitational Force (W)**: Acts downward, expressed as the weight of the skier, \( W = mg \), where \( m = 65.0 \, \text{kg} \) and \( g = 9.8 \, \text{m/s}^2 \).2. **Normal Force (N)**: Acts perpendicular to the slope, opposite to the component of gravitational force acting perpendicularly.3. **Tension Force (T)**: This is the force exerted by the tow rope that acts parallel to the ground. We need to calculate this.These forces should be represented on a diagram with the skier on an inclined plane sloped at \( 26.0^{\circ} \).
02

Analyzing Forces Perpendicular to the Slope

The skier is not accelerating up or down the slope (since they are moving at constant speed), hence the forces perpendicular to the slope must be in equilibrium.The component of the gravitational force acting perpendicular to the slope is given by\[ F_{\perp} = mg \cos(\theta), \]where \( \theta = 26.0^{\circ} \).The normal force (N) balances this force, hence \( N = mg \cos(\theta) \).
03

Analyzing Forces Parallel to the Slope

There is a balance between the tension force (T) and the component of the gravitational force that acts parallel to the slope.The component of the gravitational force acting parallel to the slope is given by\[ F_{\parallel} = mg \sin(\theta). \]Since the skier moves at a constant speed without acceleration, this force is equal to the tension in the rope, so:\[ T = F_{\parallel} = mg \sin(\theta). \]
04

Calculate the Tension in the Tow Rope

Now we plug in the values to find the tension.\[ T = (65.0 \, \text{kg})(9.8 \, \text{m/s}^2) \sin(26.0^{\circ}). \]Calculate the sine of \( 26.0^{\circ} \) and multiply the values:\[ T = (65.0)(9.8)(0.4384) = 278.7 \, \text{N}. \]
05

Conclusion

The tension in the tow rope is calculated to be approximately \( 278.7 \, \text{N} \). This tension ensures the skier is pulled up the slope at a constant speed, balancing the component of gravitational force that tries to pull them back down.

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Key Concepts

These are the key concepts you need to understand to accurately answer the question.

Tension
When dealing with motion on an incline, like a skier being pulled up a slope, understanding tension is key. Tension refers to the force transmitted through a rope or cable when it is used to carry a load. In this exercise, the tension is what allows the skier to move uphill at a constant speed without slipping back down.
  • Tension is exerted parallel to the ground, counteracting the gravitational pull down the slope.
  • The magnitude of tension can be calculated by analyzing the forces parallel to the slope.
  • Tension balances the component of gravitational force aimed at pulling the skier down.
Mathematically, you'd find tension by using the formula for the component of gravitational force parallel to the slope, which is then equal to the tension. This is important because the skier鈥檚 speed remains constant. Knowing how to measure and balance tension in real life involves both theoretical knowledge and practical understanding of forces.
Normal Force
Normal force is another crucial concept in motion on inclines. It acts perpendicular to the surface of contact and prevents objects from "falling" into the surface. In the given problem, the normal force keeps the skier from sinking into the inclined slope. This force is balanced by the perpendicular component of gravitational force.
  • The normal force is always perpendicular to the contact surface.
  • It counteracts the component of gravitational force acting perpendicularly.
  • The skier's motion does not affect normal force if there's no acceleration perpendicular to the incline.
The balance of the forces perpendicular to the slope means that the normal force equals the gravitational force component perpendicular to the slope: \[ N = mg \cos(\theta) \]Understanding this helps in predicting how objects behave when placed on inclined planes in a frictionless scenario, as in this exercise.
Gravitational Force
Gravitational force pulls objects towards the center of the Earth, and it is always directed downward, no matter the incline. It is what compels the skier to potentially slide down the slope. The problem simplifies the complexity by stating the skier moves at constant speed, which means forces parallel and perpendicular to the slope must be balanced.
  • Gravitational force acts downward, calculated as the weight of the object: \(W = mg\).
  • It can be divided into two components: parallel and perpendicular to the slope.
  • These components are essential for solving inclined plane problems.
For an incline, the gravitational force component parallel to the slope is given by: \[ F_{\parallel} = mg \sin(\theta) \] While the component perpendicular to the slope is: \[ F_{\perp} = mg \cos(\theta) \]Recognizing these components helps visualize and solve problems involving slopes and inclines.
Motion on Incline
To solve physics problems involving motion on an incline, like a skier being pulled uphill, understanding the dynamics of forces is fundamental. The forces acting on an object on a slope are best analyzed using free body diagrams, where you visually identify all forces at play.
  • Motion on an incline involves forces parallel and perpendicular to the slope.
  • Understanding how to decompose gravitational force is critical.
  • The role of tension, normal, and gravitational forces are central to problem solving.
An object moving at constant speed, as the skier in this problem, suggests a balance of forces. Recognizing how each force acts ensures a complete understanding of the motion and potential energy changes on inclines. Additionally, ignoring friction simplifies calculations but in real-world scenarios, friction often plays a vital role, influencing movement significantly.
Physics Problem Solving
Physics problem solving is both an art and a science; it involves understanding concepts, dissecting them, and applying formulas and logical reasoning. Free body diagrams are especially helpful in visualizing the forces at work.
  • Always identify and label all forces involved in the problem.
  • Break forces into components when dealing with angles or inclines.
  • Use equilibrium concepts for constant speed situations.
Solving problems systematically involves recognizing the balance of forces and using equations that relate these forces. Each component, such as tension, normal force, or gravitational force, plays its own role in keeping objects at rest or in motion. This systematic approach facilitates understanding complex situations, whether involving slopes, friction, or even multi-directional forces.

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Most popular questions from this chapter

A mysterious rocket-propelled object of mass 45.0 \(\mathrm{kg}\) is initially at rest in the middle of the horizontal, frictionless surface of an ice- covered lake. Then a force direted east and with magnitude \(F(t)=(16.8 \mathrm{N} / \mathrm{s}) t\) is applied. How far does the object travel in the first 5.00 \(\mathrm{s}\) after the force is applied?

Insect Dynamics. The froghoper (Philaenus spumarius), the champion leaper of the insect world, has a mass of 12.3\(\mathrm{mg}\) and leaves the ground (in the most energetic jumps) at 4.0 \(\mathrm{m} / \mathrm{s}\) from a vertical start. The jump itself lasts a mere 1.0 \(\mathrm{ms}\) before the insect is clear of the ground. Assuming constant acceleration, (a) draw a free-body diagram of this mighty leaper while the jump is taking place; (b) find the force that the ground exerts on the frog hoper during its jump; and (c) express the force in part (b) in terms of the froghopper's weight.

An astronaut is tethered by a strong cable to a spacecraft. The astronaut and her spacesuit have a total mass of 105 \(\mathrm{kg}\) , while the mass of the cable is negligible. The mass of the spacecraft is \(9.05 \times 10^{4} \mathrm{kg} .\) The spacecraft is far from any large astronomical bodies, so we can ignore the gravitational forces on it and the astronaut. We also assume that both the spacecraft and the astronaut are initially at rest in an inertial reference frame. The astronaut then pulls on the cable with a force of 80.0 \(\mathrm{N}\) (a) What force does the cable exert on the astronaut? (b) since \(\Sigma \vec{\boldsymbol{F}}=m \vec{\boldsymbol{a}}\) , how can a "massless" \((m=0)\) cable exert a force? (c) What is the astronaut's acceleration? (d) What force does the cable exert on the spacecraft? (e) What is the acceleration of the spacecraft?

An oil tanker's engines have broken down, and the wind is blowing the tanker straight toward a reef at a constant speed of 1.5 \(\mathrm{m} / \mathrm{s}(\) Fig. \(\mathrm{P} 4.38) .\) When the tanker is 500 \(\mathrm{m}\) from the reef, the wind dies down just as the engineer gets the engines going again. The rudder is stuck, so the only choice is to try to accelerate straight backward away from the reef. The mass of the tanker and cargo is \(3.6 \times 10^{7} \mathrm{kg}\) , and the engines produce a net horizontal force of \(8.0 \times 10^{4} \mathrm{N}\) on the tanker. Will the ship hit the reef? If it does, will the oil be sate? The hull can withstand an impact at a speed of 0.2 \(\mathrm{m} / \mathrm{s}\) or less. You can ignore the retarding force of the water on the tanker's hull.

A hockey puck with mass 0.160 \(\mathrm{kg}\) is at rest at the origin \((x=0)\) on the horizontal, frictionless surface of the rink. At time \(t=0\) a player applies a force of 0.250 \(\mathrm{N}\) to the puck, parallel to the \(x\) -axis; he continues to apply this force until \(t=2.00 \mathrm{s}\) . (a) What are the position and speed of the puck at \(t=2.00 \mathrm{s}\) ? (b) If the same force is again applied at \(t=5.00 \mathrm{s},\) what are the position and speed of the puck at \(t=7.00 \mathrm{s} ?\)
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