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Chapter 3: Problem 5

A child pulls on a rope attached to a sled with a force of \(60 \mathrm{~N}\). The rope makes an angle of \(40^{\circ}\) to the ground. (a) Compute the effective value of the pull tending to move the sled along the ground. (b) Compute the force tending to lift the sled vertically. As depicted in Fig. \(3-5\), the components of the \(60 \mathrm{~N}\) force are \(39 \mathrm{~N}\) and \(46 \mathrm{~N}\). ( \(a\) ) The pull along the ground is the horizontal component, 46 N. (b) The lifting force is the vertical component, \(39 \mathrm{~N}\).

Short Answer

Expert verified
Horizontal pull is 46 N; vertical lift is 39 N.

Step by step solution

01

Understanding the Problem

We're given a force of 60 N acting at an angle of 40掳 to the ground. We need to find the horizontal component (force along the ground) and the vertical component (force lifting the sled).
02

Setting Up the Equations

To find the components, we'll use trigonometric functions. The horizontal component can be found using \( F_x = F \cdot \cos(\theta) \), and the vertical component can be calculated with \( F_y = F \cdot \sin(\theta) \).
03

Computing Horizontal Component

The horizontal component \( F_x = 60 \cdot \cos(40^{\circ}) \). Calculate \( \cos(40^{\circ}) \) and multiply by 60 N to get the effective pull along the ground.
04

Computing Vertical Component

The vertical component \( F_y = 60 \cdot \sin(40^{\circ}) \). Calculate \( \sin(40^{\circ}) \) and multiply by 60 N to get the force tending to lift the sled.
05

Understanding Given Components

The problem states the components are 39 N for vertical and 46 N for horizontal, which aligns with the calculations: \( F_x = 46 \text{ N} \) and \( F_y = 39 \text{ N} \).

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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 forces in physics, understanding force components is crucial. Force components refer to the division of a force into its horizontal and vertical parts. This is particularly useful in scenarios where a force is applied at an angle. Instead of considering the force as a whole, breaking it down into these components provides clarity for calculations.

In our problem, the gravity of the situation can be distilled into two main components:
  • Horizontal Component: This component is parallel to the surface (along the ground) and responsible for moving the sled forward.
  • Vertical Component: Perpendicular to the ground, this component attempts to lift the sled.
Understanding these components allows us to predict and calculate the effects of the force more accurately. It simplifies complex problems by focusing on each direction separately, making it easier to solve.
Trigonometric functions
Trigonometry is the study of relationships between angles and sides of triangles, and it's especially handy in physics when dealing with angles. Trigonometric functions like sine and cosine allow us to break down forces into their components.

Here are two main functions involved:
  • Cosine (\(\cos(\theta)\)): Used when calculating the horizontal component of the force. It's the adjacent side ratio over the hypotenuse in a right triangle, useful for forces acting parallel to a surface.
  • Sine (\(\sin(\theta)\)):
Used for determining the vertical component. This function represents the opposite side ratio over the hypotenuse, suitable for lifting or counteracting weight.

In our example, these trigonometric functions allow us to reshape a force into useful parts, confirming the calculated components of 46 N horizontally and 39 N vertically.
Vector decomposition
Vector decomposition is a technical term for separating a vector into two or more simpler vectors. In physics, vectors are quantities that have both magnitude and direction, like force. When a force is applied at an angle, it behaves like a vector. We can separate this force into components with vector decomposition.

The process involves breaking the vector down using a coordinate system. Typically, this means using horizontal and vertical axes, aligning with our force components. The original force becomes the hypotenuse of a right triangle, while the components form the triangle鈥檚 perpendicular legs.

Vector decomposition simplifies complex physics problems, making it easier to understand the effects of forces acting at angles. In our sled scenario, it explains why we can discuss a force of 60 N as two smaller forces, effectively moving or lifting the sled.
Angle of force application
The angle at which a force is applied crucially influences how it impacts an object. An angle determines how much of the force contributes to horizontal movement versus vertical lifting. This angle effectively divides a single applied force into multiple effects, governed by trigonometric relationships.

In our problem, the 40掳 angle means that part of the initial force of 60 N pulls horizontally, while another part provides a lifting force. By understanding how angles affect force application, we can understand why only part of the total force moves the sled along the ground, resulting in the horizontal component of 46 N. Similarly, another portion lifts it, resulting in a vertical force of 39 N.

This concept is vital for many real-world applications like determining how to angle a pull when dragging an object, or even when designing equipment to maximize efficiency based on angle and force distribution.

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Most popular questions from this chapter

Two forces, \(80 \mathrm{~N}\) and \(100 \mathrm{~N}\), acting at an angle of \(60^{\circ}\) with each other, pull on an object. (a) What single force would replace the two forces? (b) What single force (called the equilibrant) would balance the two forces? Solve algebraically.

An unknown force acting on a \(50.0\) -g body floating in space produces a constant acceleration of \(20.0 \mathrm{~cm} / \mathrm{s}^{2}\). If the same force is now made to act on a different body, also in space, producing a constant acceleration of \(40.0 \mathrm{~cm} / \mathrm{s}^{2}\), what is the mass of that body?

A 600 -kg car is coasting along a level road at \(30 \mathrm{~m} / \mathrm{s}\). (a) How large a retarding force (assumed constant) is required to stop it in a distance of \(70 \mathrm{~m} ?(b)\) What is the minimum coefficient of friction between tires and roadway if this is to be possible? Assume the wheels are not locked, in which case we are dealing with static friction- there's no sliding. ( \(a\) ) First find the car's acceleration from a constant- \(a\) equation. It is known that \(\mathrm{v}_{i x}=30 \mathrm{v}_{f x},=0\), and \(x=70 \mathrm{~m}\). Use \(v_{k}^{2}=v_{u}^{2}+2 a x\) to find $$ a=\frac{v_{f x}^{2}-v_{i r}^{2}}{2 x}=\frac{0-900 \mathrm{~m}^{2} / \mathrm{s}^{2}}{140 \mathrm{~m}}=-6.43 \mathrm{~m} / \mathrm{s}^{2} $$ Now write $$ F=m a=(600 \mathrm{~kg})\left(-6.43 \mathrm{~m} / \mathrm{s}^{2}\right)-3858 \mathrm{~N} \text { or }-3.9 \mathrm{kN} $$ (b) Assume the force found in \((a)\) is supplied as the friction force between the tires and roadway. Therefore, the magnitude of the friction force on the tires is \(F_{\mathrm{f}}=3858 \mathrm{~N}\). The coefficient of friction is given by \(u=F_{f} / F_{N}\), where \(F_{N}\) is the normal force. In the present case, the roadway pushes up on the car with a force equal to the car's weight. Therefore, that $$ \begin{array}{l} F_{N}=F_{W}=m g=(600 \mathrm{~kg})\left(9.81 \mathrm{~m} / \mathrm{s}^{2}\right)=5886 \mathrm{~N} \\ \mu_{s}=\frac{F_{\mathrm{f}}}{F_{N}}=\frac{3858}{5886}=0.66 \end{array} $$ The coefficient of friction must be at least \(0.66\) if the car is to stop within \(70 \mathrm{~m}\).

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?

A 300 -g mass hangs at the end of a string. A second string hangs from the bottom of that mass and supports a 900 -g mass. \((a)\) Find the tension in each string when the masses are accelerating upward at \(0.700 \mathrm{~m} / \mathrm{s}^{2}\). Don't forget gravity. ( \(b\) ) Find the tension in each string when the acceleration is \(0.700 \mathrm{~m} / \mathrm{s}^{2}\) downward.
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