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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, \(46 \mathrm{~N}\). (b) The lifting force is the vertical component, \(39 \mathrm{~N}\).

Short Answer

Expert verified
(a) 46 N; (b) 39 N

Step by step solution

01

Identify the Problem Components

The problem provides that a force of 60 N is applied at an angle of 40 degrees to the ground. We need to resolve this force into its horizontal and vertical components.
02

Calculate the Horizontal Component (along the ground)

The horizontal component of the force can be calculated using the formula \( F_{x} = F \cdot \cos(\theta) \), where \( F \) is the total force (60 N) and \( \theta \) is the angle (40 degrees).
03

Compute the Horizontal Force

Substituting the given values, we calculate: \( F_{x} = 60 \cdot \cos(40^{\circ}) \). Calculate \( \cos(40^{\circ}) \approx 0.766 \). Thus, \( F_{x} \approx 60 \times 0.766 \approx 45.96 \mathrm{~N} \).
04

Calculate the Vertical Component (lifting force)

The vertical component of the force can be calculated using the formula \( F_{y} = F \cdot \sin(\theta) \), where \( F \) is the total force (60 N) and \( \theta \) is the angle (40 degrees).
05

Compute the Vertical Force

Substituting the given values, we calculate: \( F_{y} = 60 \cdot \sin(40^{\circ}) \). Calculate \( \sin(40^{\circ}) \approx 0.643 \). Thus, \( F_{y} \approx 60 \times 0.643 \approx 38.58 \mathrm{~N} \).

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Key Concepts

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

Horizontal Component
When dealing with forces, it is essential to understand how the horizontal component plays a role. The horizontal component is the "part" of the force that acts along a surface, such as the ground. In the context of our exercise, this represents how much of the child's effort is making the sled move along the path. To find the horizontal component of a force that is applied at an angle, we use the cosine function from trigonometry.

Here's the formula you need:
  • Formula: \( F_{x} = F \cdot \cos(\theta) \)
  • Where: \( F_{x} \) is the horizontal component, \( F \) is the total force, and \( \theta \) is the angle of force.
By identifying these values and performing the calculation, we can understand how effectively the force can move things horizontally, just like calculating that the sled moves with approximately 45.96 N of force along the ground.
Vertical Component
The vertical component represents the part of the force that works against gravity, such as lifting or pushing something upward. In many situations, this is crucial for determining how much force is available to fight the weight of an object. The vertical component is calculated by using the sine function from trigonometry.

For our specific question, the formula plugs in as follows:
  • Formula: \( F_{y} = F \cdot \sin(\theta) \)
  • Where: \( F_{y} \) is the vertical component, \( F \) is the total force, and \( \theta \) is the angle of force.
Using these calculations, Fig. 3-5 showed that the lifting force exerted is about 38.58 N. This outcome implies how much of the child's pull is trying to lift the sled upward, although not all of it will be effective due to gravity.
Angle of Force
The angle at which a force is applied significantly influences how the force is distributed into horizontal and vertical components. The direction of the angle determines how much of the force acts in each direction.

For example, in our case, an angle of 40 degrees implies a specific alignment of the force relative to the ground. This affects how much of the force moves the sled forward versus upward.

Understanding the angle is crucial for correctly setting up equations to solve for components:
  • Angles closer to 0 degrees will predominantly affect the horizontal component, making it larger.
  • As angles increase towards 90 degrees, more force contributes to the vertical component.
By accurately understanding the angle of force, we ensure precise calculations of the components, enhancing our grasp of the problem's dynamics.
Trigonometric Functions
Trigonometric functions, particularly sine and cosine, are indispensable in force decomposition problems like the one involving the sled. These functions help us break down a force into its horizontal and vertical components based on the angle.
  • Cosine function: Used to find the horizontal component. It helps determine how much force moves along the direction of an object鈥檚 surface.
  • Sine function: Used for the vertical component. It indicates the force working perpendicular to the surface, showing lifting potential.
Understanding these functions and how they relate to angles is key for problems involving angled forces. Simply put, they let us decompose a single, often complex, force into manageable parts so that we can make more precise calculations and predictions about how objects move.

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Most popular questions from this chapter

A man who weighs \(1000 \mathrm{~N}\) on Earth stands on a scale on the surface of the mythical nonspinning planet Mongo. That body has a mass which is \(4.80\) times Earth's mass and a diameter which is \(0.500\) times Earth's diameter. Neglecting the effect of the Earth's spin, how much does the scale read?

How large a horizontal force in addition to \(F_{T}\) must pull on block- \(A\) in Fig. \(3-21\) to give it an acceleration of \(0.75 \mathrm{~m} / \mathrm{s}^{2}\) toward the left? Assume, as in Problem \(3.31\), that \(\mu_{k}=0.20, m_{A}=25 \mathrm{~kg}\), and \(m_{B}=15 \mathrm{~kg}\). Redraw Fig 3-21 for this case, including a force \(F\) pulling toward the left on \(A\). In addition, the retarding friction force \(F_{\mathrm{f}}\) must be reversed in direction. As in Problem 3.31, \(F_{\mathrm{f}}=49.1 \mathrm{~N}\). Write \(F=m a\) for each block in turn, taking the direction of motion (to the left and up) to be positive. We have $$ \begin{aligned} \pm \sum F_{x A}=F-F_{T}-49.1 \mathrm{~N} &=(25 \mathrm{~kg})\left(0.75 \mathrm{~m} / \mathrm{s}^{2}\right) \text { and }+\uparrow \sum F_{y B}=F_{T}-(15)(9.81) \mathrm{N} \\ &=(15 \mathrm{~kg})\left(0.75 \mathrm{~m} / \mathrm{s}^{2}\right) \end{aligned} $$ Solve the last equation for \(F_{T}\) and substitute in the previous equation. Then solve for the single unknown \(F\), and find it to be \(226 \mathrm{~N}\) or \(0.23 \mathrm{kN}\).

A tow rope will break if the tension in it exceeds \(1500 \mathrm{~N}\). It is used to tow a \(700-\mathrm{kg}\) car along level ground. What is the largest acceleration the rope can give to the car? (Remember that 1500 has four significant figures; see Appendix A.) The forces acting on the car are shown in Fig. \(3-12\). Only the \(x\) -directed force is of importance, because the \(y\) -directed forces balance each other. Indicating the positive direction with a \(+\) sign and a little arrow, we write, $$ \pm \sum F_{x}=m a_{x} \quad \text { becomes } \quad 1500 \mathrm{~N}=(700 \mathrm{~kg})(a) $$ from which \(a=2.14 \mathrm{~m} / \mathrm{s}^{2}\).

A \(12.0-\mathrm{g}\) bullet is accelerated from rest to a speed of \(700 \mathrm{~m} / \mathrm{s}\) as it travels \(20.0 \mathrm{~cm}\) in a gun barrel. Assuming the acceleration to be constant, how large was the accelerating force? [Hint: Be careful of units.]

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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