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Chapter 4: Problem 67

A bicyclist is coasting straight down a hill at a constant speed. The combined mass of the rider and bicycle is \(80.0 \mathrm{kg}\), and the hill is inclined at \(15.0^{\circ}\) with respect to the horizontal. Air resistance opposes the motion of the cyclist. Later, the bicyclist climbs the same hill at the same constant speed. How much force (directed parallel to the hill) must be applied to the bicycle in order for the bicyclist to climb the hill?

Short Answer

Expert verified
The required force to climb the hill is 405.5 N.

Step by step solution

01

Understand the Problem

We have a bicyclist coasting down a hill and then climbing it with a constant speed. We are given the mass of the bicyclist and the bicycle, which is 80.0 kg, and the angle of the hill, which is 15.0 degrees. The goal is to find the force the bicyclist needs to apply parallel to the hill when climbing, considering they climb with a constant speed.
02

Analyze Forces on the Incline

While climbing the hill at a constant speed, the cyclist needs to overcome two main forces: the component of gravitational force pulling them back down the hill and the force of air resistance (which is equal to the force when coasting down because we're descending at constant speed too).
03

Calculate the Gravitational Force Component

The component of gravitational force acting along the incline is given by:\[ F_{gravity} = m imes g imes ext{sin}( heta) \]where \( m = 80.0 \, \text{kg} \), \( g = 9.8 \, \text{m/s}^2 \), and \( \theta = 15.0^{\circ} \). Let's calculate it:\[ F_{gravity} = 80.0 \, \text{kg} \times 9.8 \, \text{m/s}^2 \times \text{sin}(15.0^{\circ}) \]
04

Simplify and Calculate

Firstly, find \( \text{sin}(15.0^{\circ}) \):\[ \text{sin}(15.0^{\circ}) \approx 0.2588 \]Now substitute to find the force:\[ F_{gravity} = 80.0 \, \text{kg} \times 9.8 \, \text{m/s}^2 \times 0.2588 = 202.752 \, \text{N} \]
05

Determine Total Force to Climb

The total force required to climb the hill is the sum of the gravitational force component against the incline and the air resistance force (equal to the force that was opposing the motion downhill). Since the speed is constant downhill, the force of air resistance equals the component of gravity, which was 202.752 N. So, the total force to climb the hill is:\[ F_{total} = F_{gravity} + F_{air resistance} = 202.752 \, \text{N} + 202.752 \, \text{N} = 405.504 \, \text{N} \]
06

Conclusion

Thus, the force that must be applied parallel to the hill to climb it at constant speed is approximately 405.5 N.

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Key Concepts

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

Gravitational Force
Gravitational force is an essential concept to understand when analyzing motion on inclined planes, like when a cyclist is riding down or up a hill. On an inclined plane, not all of the gravitational force pulls the object straight down. Instead, it splits into two components:
  • One that acts perpendicular to the surface
  • Another that acts parallel to the slope
For the bicyclist's scenario, the component that affects movement along the hill is of great importance. This can be calculated using the formula:
\[ F_{gravity} = m \times g \times \sin(\theta) \]
Where:
  • \( m \) is the mass of the cyclist and the bicycle, which is 80.0 kg
  • \( g \) is the acceleration due to gravity, approximated as 9.8 m/s虏
  • \( \theta \) is the angle of the hill, 15 degrees
By computing this, we obtain the force component pulling the biker downhill, which becomes critical when calculating the necessary force to be exerted to climb uphill.
Air Resistance
Air resistance, also defined as drag, is the force that acts opposite to the direction of movement as a bicyclist travels through air. It is an important part of this problem, particularly because it influences motion both uphill and downhill.
The resistance is constant in our scenario since the cyclist maintains the same speed while descending as while ascending. This means the air resistance is equal in both directions, making calculations a bit simpler because it equals the gravitational downhill force component when coasting in equilibrium.
Overall, recognizing air resistance as a significant opposing force clarifies its role in the physical exertion required to maintain constant speed on an incline. It's crucial to consider when calculating total forces involved, especially for activities like cycling.
Constant Speed
Constant speed on an inclined plane offers a fascinating insight into forces. Let's dig deeper: it signifies that the net force acting on the cyclist is zero. When moving at a constant speed up or down a hill, all opposing forces balance out.
  • When moving downhill at constant speed, gravitational pull and air resistance perfectly counteract.
  • Climbing uphill, the cyclist must exert an equal force to the downhill gravitational pull plus air resistance, maintaining equilibrium.
This balance is crucial. It means no acceleration is needed, merely maintaining power to oppose the forces of gravity and air. For effective learning, remember riding at constant speed results from diligent observation of these counteracting forces in action.

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Most popular questions from this chapter

The mass of a robot is 5450 kg. This robot weighs \(3620 \mathrm{N}\) more on planet \(A\) than it does on planet B. Both planets have the same radius of \(1.33 \times 10^{7} \mathrm{m} .\) What is the difference \(M_{\mathrm{A}}-M_{\mathrm{B}}\) in the masses of these planets?

A student presses a book between his hands, as the drawing indicates. The forces that he exerts on the front and back covers of the book are perpendicular to the book and are horizontal. The book weighs \(31 \mathrm{N}\). The coefficient of static friction between his hands and the book is \(0.40 .\) To keep the book from falling, what is the magnitude of the minimum pressing force that each hand must exert?

As a moon follows its orbit around a planet, the maximum gravitational force exerted on the moon by the planet exceeds the minimum gravitational force by \(11 \%\). Find the ratio \(r_{\max } / r_{\min }\), where \(r_{\max }\) is the moon's maximum distance from the center of the planet and \(r_{\min }\) is the minimum distance.

At a distance \(H\) above the surface of a planet, the true weight of a remote probe is one percent less than its true weight on the surface. The radius of the planet is \(R\). Find the ratio \(H / R\).

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