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Chapter 3: Problem 46

Having hauled it to the top of a tilted driveway, a child is holding a wagon from rolling back down. The driveway is inclined at \(20^{\circ}\) to the horizontal. If the wagon weighs \(150 \mathrm{~N}\), with what force must the child pull on the handle if the handle is parallel to and pointing up the incline?

Short Answer

Expert verified
The child must exert a force of 51.3 N along the incline.

Step by step solution

01

Analyze the Force Components

The force exerted by the child must counteract the component of gravitational force pulling the wagon down the inclined plane. First, identify that the weight of the wagon (150 N) acts vertically downward. To find the force component along the incline, we use the formula for the component of the weight along the plane. This force is given by the equation: \( F_{gravity} = mg \sin{\theta} \), where \( m \) is the mass, \( g \) is the gravitational acceleration (9.8 m/s虏), and \( \theta = 20^{\circ} \).
02

Calculate the Force Component Down the Incline

We need the component of the weight acting along the incline. Using \( F_{gravity} = 150 \sin{20^{\circ}} \), compute this component. First, find \( \sin{20^{\circ}} \), which is approximately 0.342. Then the force component is \( 150 \times 0.342 \approx 51.3 \mathrm{~N} \).
03

Determine the Required Force

The child must exert an equal and opposite force to keep the wagon from rolling down the incline. Therefore, the required pulling force, \( F_{child} \), is equal to the gravitational force component along the incline: \( F_{child} = 51.3 \mathrm{~N} \) to counteract the pull downward.

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Key Concepts

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

Force Components
When dealing with inclined plane problems, one of the first steps is to break forces into components. This helps us analyze the direction and magnitude of the forces involved. For our scenario, the weight of the wagon is the main force to consider.

The weight of an object acts vertically downward, but on an inclined surface, it can be split into two components:
  • A component parallel to the incline that tries to pull the wagon down.
  • A component perpendicular to the incline that pushes into the surface.
Understanding these force components is crucial to figuring out how much force is needed to hold or move the wagon along the incline.
Gravitational Force
Gravitational force is the force exerted by the Earth on objects, pulling them towards its center. In our problem, gravitational force influences how much force the child must use to stop the wagon from rolling back down the incline.

The gravitational force on the wagon is calculated as its weight, given by:\[ F_{gravity} = mg \]where \( m \) is mass and \( g \) is the acceleration due to gravity (approximately 9.8 m/s虏). Here, the weight is given as 150 N.

We need to find the component of this gravitational force that acts down the incline, determined by:\[ F_{gravity,\ parallel} = mg \sin{\theta} \]This component is the one pulling the wagon down, which equals 51.3 N in this scenario.
Incline Analysis
Analyzing an inclined plane involves understanding how angles affect force components. We want to identify which forces act along and perpendicular to the surface. Here's what typically happens on an incline:

  • The angle of incline, \( \theta \), affects how much of the gravitational force is pulling the object down the slope.
  • Lower angles mean less force exerted down the slope, while higher angles increase this force.
  • We use trigonometric functions to determine the force components: \( \sin{\theta} \) gives the down-slope component, and \( \cos{\theta} \) the perpendicular component.
In our exercise, using \( \sin{20^{\circ}} \), we calculated the parallel force component since the handle aligns precisely with this angle. This careful analysis helps determine the counteracting force the child needs to apply.
Newton's Laws of Motion
To understand why the child needs to pull on the handle with a specific force, we refer to Newton's laws of motion. These foundational principles describe how objects move and react to forces. For this problem:

  • Newton's First Law (Law of Inertia) implies the wagon will remain at rest on the incline only if the forces are balanced.
  • Newton's Second Law states that the net force acting on an object is equal to its mass times its acceleration \( (F = ma) \). Here, since we want no acceleration, the net force should be zero, meaning the pulling force equals the gravitational pull down the incline.
  • The child's force of 51.3 N directly balances the gravitational downslope force, keeping the wagon still.
Applying Newton's laws helps to rationalize why specific force amounts are necessary to maintain equilibrium on an incline.

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Most popular questions from this chapter

A 700-N man stands on a scale on the floor of an elevator. The scale records the force it exerts on whatever is on it. What is the scale reading if the elevator has an acceleration of ( \(a\) ) \(1.8 \mathrm{~m} / \mathrm{s}^{2}\) up? (b) \(1.8 \mathrm{~m} / \mathrm{s}^{2}\) down? (c) \(9.8 \mathrm{~m} / \mathrm{s}^{2}\) down?

A car whose weight is \(F_{W}\) is on a ramp which makes an angle \(\theta\) to the horizontal. How large a perpendicular force must the ramp withstand if it is not to break under the car's weight? As rendered in Fig. \(3-6\), the car's weight is a force \(\overrightarrow{\mathbf{F}}_{W}\) that pulls straight down on the car. We take components of \(\overrightarrow{\mathbf{F}}\) along the incline and perpendicular to it. The ramp must balance the force component \(F_{W} \cos \theta\) if the car is not to crash through the ramp. In other words, the force exerted on the car by the ramp, upwardly perpendicular to the ramp, is \(F_{N}\) and \(F_{N}=F_{W} \cos \theta\).

Floating in space far from anything else are two spherical asteroids, one having a mass of \(20 \times 10^{10} \mathrm{~kg}\) and the other a mass of \(40 \times 10^{10} \mathrm{~kg}\). Compute the force of attraction on each one due to gravity when their center-to-center separation is \(10 \times 10^{6} \mathrm{~m}\).

A horizontal force of \(200 \mathrm{~N}\) is required to cause a \(15-\mathrm{kg}\) block to slide up a \(20^{\circ}\) incline with an acceleration of \(25 \mathrm{~cm} / \mathrm{s}^{2}\). Find \((a)\) the friction force on the block and \((b)\) the coefficient of friction.

A 200-N wagon is to be pulled up a \(30^{\circ}\) incline at constant speed. How large a force parallel to the incline is needed if friction effects are negligible? The situation is shown in Fig. \(3-16(a)\). Because the wagon moves at a constant speed along a straight line, its velocity vector is constant. Therefore, the wagon is in translational equilibrium, and the first condition for equilibrium applies to it. We isolate the wagon as the object. Three non-negligible forces act on it: (1) the pull of gravity \(F_{W}\) (its weight), directed straight down; (2) the applied force \(F\) exerted on the wagon parallel to the incline to pull it up the incline; (3) the push \(F_{N}\) of the incline that supports the wagon. These three forces are shown in the free-body diagram in Fig. \(3-16\). For situations involving inclines, it is convenient to take the \(x\) -axis parallel to the incline and the \(y\) -axis perpendicular to it. After taking components along these axes, we can write the first condition for equilibrium: $$ \begin{array}{lll} +\sum F_{x}=0 & \text { becomes } & F-0.50 F_{W}=0 \\ +\sum F_{y}=0 & \text { becomes } & F_{N}-0.866 F_{W}=0 \end{array} $$ Solving the first equation and recalling that \(F_{W}=200 \mathrm{~N}\), we find that \(F=0.50 F_{W}\). The required pulling force to two significant figures is \(0.10 \mathrm{kN}\).
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