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\(\mathrm{A}\) man is dragging a trunk up the loading ramp of a mover's truck. The ramp has a a slope angle of \(20.0^{\circ}\) , and the man pulls upward with a force \(\vec{F}\) whose direction makes an angle of \(30.0^{\circ}\) with the ramp (Fig. E4.4). (a) How large a force \(\vec{F}\) is necessary for the component \(F_{x}\) parallel to the ramp to be 60.0 \(\mathrm{N} ?\) (b) How large will the component \(F_{y}\) perpendicular to the ramp then be?

Short Answer

Expert verified
F = 69.3 N; F_y = 34.65 N

Step by step solution

01

Understand the Problem

We are given a force \( \vec{F} \) at a 30° angle relative to the ramp. We need to find the magnitude of \( \vec{F} \) such that its component along the ramp is 60 N, and subsequently find the perpendicular component.
02

Calculate \( F_x \) Component

The parallel component of the force \( F_x \) is given by \( F \cos(30°) \). We set up the equation: \( F \cos(30°) = 60 \).
03

Solve for Force \( F \)

Rearrange the equation to solve for \( F \): \[ F = \frac{60}{\cos(30°)} \]. Substitute \( \cos(30°) = \frac{\sqrt{3}}{2} \) to find \( F \).
04

Calculate

Substitute the value: \[ F = \frac{60}{\sqrt{3}/2} = \frac{60 \cdot 2}{\sqrt{3}} = \frac{120}{\sqrt{3}} \approx 69.3 \text{ N} \].
05

Calculate \( F_y \) Component

The perpendicular component of the force \( F_y \) is given by \( F \sin(30°) \). Use the calculated \( F \) to find \( F_y \).
06

Calculate

Substitute \( F \approx 69.3 \text{ N} \) and \( \sin(30°) = \frac{1}{2} \) into \( F_y = F \sin(30°) \) to find \( F_y \).
07

Substitute and Solve

\( F_y = 69.3 \cdot \frac{1}{2} = 34.65 \text{ N} \). Therefore, the perpendicular component \( F_y \) is \( 34.65 \text{ N} \).

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Key Concepts

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

Force Components
In physics, when we discuss force components, we are talking about the breaking down of a force vector into parts that act along the axes of a coordinate system. This is useful because it allows us to analyze forces separately in each direction.
For instance, consider a man pulling a trunk up a ramp. The force he applies, denoted as \( \vec{F} \), can be split into two key components:
  • The component along the ramp, \( F_x \), which makes the trunk move up the slope
  • The component perpendicular to the ramp, \( F_y \), which affects how hard the trunk presses against the surface of the ramp
By resolving these components, we can focus on how much of the man's effort is being used to actually move the trunk up as opposed to pressing it into the ramp. This way, each component can be individually managed and calculated, making problem-solving simpler and more intuitive.
Trigonometry in Physics
Trigonometry plays a crucial role in physics, especially when dealing with forces in non-standard directions, such as on a slope. When a force acts at an angle, such as \( \vec{F} \) in our problem scenario, we use trigonometric functions to break it down.
Key functions like sine and cosine are used extensively:
  • \( \cos(\theta) \) helps to find the component of force along a given direction, like \( F_x = F \cos(30°) \)
  • \( \sin(\theta) \) helps to find the component perpendicular to the initial direction, like \( F_y = F \sin(30°) \)
In our example, the man pulls the trunk up the ramp, and his pulling force is at an angle to the slope. Trigonometry allows us to predict how much of this force contributes to moving the trunk up the ramp and how much contributes to the friction interaction with the ramp's surface. Through such calculations, it becomes manageable to design better practices for exerting the optimal force necessary.
Slope Angle
The slope angle is a fundamental concept when analyzing forces on inclined planes. It represents the angle of inclination relative to the horizontal plane. In our problem, the ramp has a slope angle of \( 20° \), which sets the context for the forces acting on the trunk.
Understanding slope angles is crucial because:
  • It affects the gravitational force component along the ramp, which is crucial in determining the effort needed to move an object uphill or downhill.
  • It determines the frictional forces, as more incline generally leads to higher friction resistance due to increased normal force.
Essentially, knowing the slope angle helps in calculating how much force is needed to either overcome gravity or work against friction. In practical applications, such as moving loads up ramps, factoring in the slope angle ensures that the exerted forces are both efficient and effective.

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Most popular questions from this chapter

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 \(t=7.00 \mathrm{s} ?\)

A ball is hanging from a long string that is tied to the ceiling of a train car traveling eastward on horizontal tracks. An observer inside the train car sees the ball hang motionless. Draw a clearly labeled free-body diagram for the ball if (a) the train has a uniform velocity, and (b) the train is speeding up uniformly. Is the net force on the ball zero in either case? Explain.

An electron \(\left(\mathrm{mass}=911 \times 10^{-31} \mathrm{kg}\right)\) leaves one end of a TV picture tube with zero initial speed and travels in a straight line to the accelerating grid, which is 1.80 \(\mathrm{cm}\) away. It reaches the grid with a speed of \(3.00 \times 10^{6} \mathrm{m} / \mathrm{s} .\) If the accelerating force is constant, compute (a) the acceleration; (b) the time to reach the grid; (c) the net force, in newtons. (You can ignore the gravitational force on the electron.)

At the surface of Jupiter's moon Io, the acceleration due to gravity is \(g=1.81 \mathrm{m} / \mathrm{s}^{2} .\) A watermelon weighs 44.0 \(\mathrm{N}\) at the surface of the earth. (a) What is the watermelon's mass on the earth's surface? (b) What are its mass and weight on the surface of Io?

Two forces, \(\vec{\boldsymbol{F}}_{1}\) and \(\vec{\boldsymbol{F}}_{2},\) a point. The magnitude of \(\vec{\boldsymbol{F}}_{1}\) is \(9.00 \mathrm{N},\) and its direction is \(60.0^{\circ}\) above the \(x\) -axis in the second quadrant. The magnitude of \(\vec{\boldsymbol{F}}_{2}\) is \(6.00 \mathrm{N},\) and its direction is \(53.1^{\circ}\) below the \(x\) -axis in the third quadrant. (a) What are the \(x\) - and \(y\) -components of the resultant force? (b) What is the magnitude of the resultant force?

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