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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?

Short Answer

Expert verified
The components of the resultant force are \( F_{Rx} = -8.10 \, \mathrm{N} \) and \( F_{Ry} = 2.99 \, \mathrm{N} \). The magnitude of the resultant force is approximately \( 8.64 \, \mathrm{N} \).

Step by step solution

01

Determine Components of Force F1

First, calculate the components of force \( \vec{\boldsymbol{F}}_{1} \) with magnitude \( 9.00 \, \mathrm{N} \) and direction \( 60.0^{\circ} \) above the \( x \)-axis in the second quadrant. In this quadrant, the \( x \)-component will be negative, and the \( y \)-component positive. The formulas to use are \( F_{1x} = -9.00 \, \mathrm{N} \times \cos(60.0^{\circ}) \) and \( F_{1y} = 9.00 \, \mathrm{N} \times \sin(60.0^{\circ}) \).
02

Calculate Components of Force F1

Using the trigonometric identities, calculate \( F_{1x} \) and \( F_{1y} \):\[ F_{1x} = -9.00 \, \mathrm{N} \times \cos(60.0^{\circ}) = -4.50 \, \mathrm{N} \]\[ F_{1y} = 9.00 \, \mathrm{N} \times \sin(60.0^{\circ}) = 7.79 \, \mathrm{N} \]
03

Determine Components of Force F2

Now, calculate the components of force \( \vec{\boldsymbol{F}}_{2} \) with magnitude \( 6.00 \, \mathrm{N} \) and direction \( 53.1^{\circ} \) below the \( x \)-axis in the third quadrant. In this quadrant, both the \( x \)-component and \( y \)-component will be negative. The formulas to use are \( F_{2x} = -6.00 \, \mathrm{N} \times \cos(53.1^{\circ}) \) and \( F_{2y} = -6.00 \, \mathrm{N} \times \sin(53.1^{\circ}) \).
04

Calculate Components of Force F2

Using the trigonometric identities, calculate \( F_{2x} \) and \( F_{2y} \):\[ F_{2x} = -6.00 \, \mathrm{N} \times \cos(53.1^{\circ}) = -3.60 \, \mathrm{N} \]\[ F_{2y} = -6.00 \, \mathrm{N} \times \sin(53.1^{\circ}) = -4.80 \, \mathrm{N} \]
05

Sum the Components to Find Resultant Force

Add the corresponding components of \( \vec{\boldsymbol{F}}_{1} \) and \( \vec{\boldsymbol{F}}_{2} \) to find the components of the resultant force:\[ F_{Rx} = F_{1x} + F_{2x} = -4.50 \, \mathrm{N} + (-3.60 \, \mathrm{N}) = -8.10 \, \mathrm{N} \]\[ F_{Ry} = F_{1y} + F_{2y} = 7.79 \, \mathrm{N} + (-4.80 \, \mathrm{N}) = 2.99 \, \mathrm{N} \]
06

Calculate Magnitude of Resultant Force

The magnitude of the resultant force \( \vec{\boldsymbol{F}}_{R} \) can be found using the Pythagorean theorem:\[ |\vec{\boldsymbol{F}}_{R}| = \sqrt{(F_{Rx})^2 + (F_{Ry})^2} = \sqrt{(-8.10)^2 + (2.99)^2} \approx 8.64 \, \mathrm{N} \]

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Key Concepts

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

Force Components
When working with forces in physics, we often deal with directional forces acting at various angles. To simplify these into manageable calculations, forces are typically broken down into their horizontal and vertical components, namely, the \( x \) and \( y \) components.

For instance, for a force \( \vec{\boldsymbol{F}} \) at an angle \( \theta \), the components can be found using trigonometry:
  • The \( x \)-component, \( F_x \), is determined using the cosine function: \[ F_x = F \times \cos(\theta) \]
  • The \( y \)-component, \( F_y \), is determined using the sine function: \[ F_y = F \times \sin(\theta) \]
Signs of these components depend on the quadrant the vector is in, as we'll discuss further below. Breaking down forces into components makes it easier to analyze their effects along each axis. This way, even when forces act in non-standard directions, they can be simplified into a combination of orthogonal forces.
Resultant Force
Finding the resultant force involves combining multiple vector forces into a single equivalent one. The resultant force (\( \vec{\boldsymbol{F}}_{R} \)) can be envisioned as the single force that has the same effect on an object as all the individual forces applied together.

To determine the resultant force:
  • Add together all the \( x \)-components of each force to find \( F_{Rx} \).
  • Add together all the \( y \)-components of each force to find \( F_{Ry} \).
Once the components \( F_{Rx} \) and \( F_{Ry} \) are known, the magnitude of the resultant force can be calculated using the Pythagorean theorem:\[|\vec{\boldsymbol{F}}_{R}| = \sqrt{(F_{Rx})^2 + (F_{Ry})^2}\]This formula consolidates the orthogonal components into a single force magnitude, giving us a direct understanding of the net effect of all applied forces.
Quadrants in Physics
A clear understanding of the quadrants in a two-dimensional coordinate system is crucial when working with vectors. In physics, the plane is divided into four quadrants, each of which affects the sign of vector components differently.

Here's a breakdown of the quadrant system:
  • First Quadrant: Both \( x \)- and \( y \)-components are positive.
  • Second Quadrant: \( x \)-component is negative while \( y \)-component is positive.
  • Third Quadrant: Both \( x \)- and \( y \)-components are negative.
  • Fourth Quadrant: \( x \)-component is positive while \( y \)-component is negative.
For example, a force in the second quadrant, like \( \vec{\boldsymbol{F}}_1 \) in the original problem, implies using negative cosine for the \( x \)-component and positive sine for the \( y \)-component. Understanding the quadrant location helps in correctly applying signs to the calculated components, ensuring accurate vector addition and analysis.

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Most popular questions from this chapter

A large box containing your new computer sits on the bed of your pickup truck. You are stopped at a red light. The light turns green and you stomp on the gas and the truck accelerates. To your horror, the box starts to slide toward the back of the truck. Draw clearly labeled free-body diagrams for the truck and for the box. Indicate pairs of forces, if any, that are third-law action- reaction pairs. (The bed of the truck is not friction less.)

Insect Dynamics. The froghoper (Philaenus spumarius), the champion leaper of the insect world, has a mass of 12.3\(\mathrm{mg}\) and leaves the ground (in the most energetic jumps) at 4.0 \(\mathrm{m} / \mathrm{s}\) from a vertical start. The jump itself lasts a mere 1.0 \(\mathrm{ms}\) before the insect is clear of the ground. Assuming constant acceleration, (a) draw a free-body diagram of this mighty leaper while the jump is taking place; (b) find the force that the ground exerts on the frog hoper during its jump; and (c) express the force in part (b) in terms of the froghopper's weight.

A 4.9 -N hammer head is stopped from an initial downward velocity of 3.2 \(\mathrm{m} / \mathrm{s}\) in a distance of 0.45 \(\mathrm{cm}\) by a nail in a pine board. In addition to its weight, there is a 15 downward force on the hammer head applied by the person using the hammer. Assume that the acceleration of the hammer head is constant while it is in contact with the nail and moving downward. (a) Draw a free-body diagram for the hammer head. Identify the reaction force to each action force in the diagram. (b) Calculate the downward force \(\vec{\boldsymbol{F}}\) exerted by the hammer head on the nail while the hammer head is in contact with the nail and moving downward. (c) Suppose the nail is in hardwood and the distance the hammer head travels in coming to rest is only 0.12 \(\mathrm{cm}\) . The downward forces on the hammer head are the same as in part (b). What then is the force \(\vec{\boldsymbol{F}}\) exerted by the hammer head on the nail while the hammer head is in contact with the nail and moving downward?

Two horses pull horizontally on ropes attached to a stump. The two forces \(\vec{F}_{1}\) and \(\vec{F}_{2}\) that they apply to the stump are such that the net (resultant) force \(\vec{R}\) has a magnitude equal to that of \(\vec{\boldsymbol{F}}_{1}\) and makes an angle of \(90^{\circ}\) with \(\vec{\boldsymbol{F}}_{1 .}\) Let \(F_{1}=1300 \mathrm{N}\) and \(R=1300 \mathrm{N}\) also. Find the magnitude of \(\vec{\boldsymbol{F}}_{2}\) and its direction \((\mathrm{rela}-\)tive to \(\vec{\boldsymbol{F}}_{1} ).\)

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.

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