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Chapter 12: Problem 87

A particle is acted on by forces given, in newtons, by \(\vec{F}_{1}=\) \(8.40 \hat{\mathrm{i}}-5.70 \hat{\mathrm{j}}\) and \(\vec{F}_{2}=16.0 \hat{\mathrm{i}}+4.10 \hat{\mathrm{j}} .\) (a) What are the \(x\) component and (b) \(y\) component of the force \(\vec{F}_{3}\) that balances the sum of these forces? (c) What angle does \(\vec{F}_{3}\) have relative to the \(+x\) axis?

Short Answer

Expert verified
\(F_{3x} = -24.4, F_{3y} = 1.6; \alpha = -3.75^\circ \).

Step by step solution

01

Calculate the sum of forces

To find the net force, add the components of \( \vec{F}_1 \) and \( \vec{F}_2 \). The \( x \)-components are added together: \[ F_{1x} = 8.40, \quad F_{2x} = 16.0 \] \[ F_x = F_{1x} + F_{2x} = 8.40 + 16.0 = 24.4 \, \hat{i} \] and the \( y \)-components are added: \[ F_{1y} = -5.70, \quad F_{2y} = 4.10 \] \[ F_y = F_{1y} + F_{2y} = -5.70 + 4.10 = -1.60 \, \hat{j} \] So, the total force is \( \vec{F} = 24.4 \, \hat{i} - 1.60 \, \hat{j} \).
02

Determine components of \( \vec{F}_3 \)

The force \( \vec{F}_3 \) acts to balance the sum of the forces, meaning it must exactly cancel out \( \vec{F} \). Hence, \[ F_{3x} = -F_x = -24.4 \] \[ F_{3y} = -F_y = 1.60 \] Thus, the components of \( \vec{F}_3 \) are \( -24.4 \, \hat{i} \) and \( 1.60 \, \hat{j} \).
03

Calculate angle \( \alpha \) with the \( +x \) axis

The angle \( \alpha \) \( \vec{F}_3 \) makes with the \( +x \)-axis is given by the inverse tangent function: \[ \tan \alpha = \frac{F_{3y}}{F_{3x}} = \frac{1.60}{-24.4} \] Calculating this gives:\[ \alpha = \tan^{-1}\left(\frac{1.60}{-24.4}\right) \approx -3.75^\circ \] This indicates that \( \vec{F}_3 \) is \( 3.75^\circ \) above the \( -x \) axis.

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Key Concepts

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

Component Form of Vectors
Vectors can be represented in what's called component form. This involves breaking down a vector into two perpendicular parts: the horizontal component and the vertical component. These components are expressed using unit vectors, typically referred to as \( \hat{i} \) for the x-axis (horizontal) and \( \hat{j} \) for the y-axis (vertical).
  • The x-component represents how far the vector extends along the horizontal axis.
  • The y-component shows the vector's reach along the vertical axis.
To find these components, especially when given vectors like \( \vec{F}_1 = 8.40 \hat{i} - 5.70 \hat{j} \), focus on the coefficients in front of the unit vectors.
  • The x-component of \( \vec{F}_1 \) is \( 8.40 \).
  • The y-component of \( \vec{F}_1 \) is \( -5.70 \).
By knowing the component form, you can easily add vectors by simply adding respective components, leading to a clearer understanding of the vector's overall influence in a given direction.
Equilibrium of Forces
When dealing with forces, equilibrium occurs when all the forces acting on a particle balance out to produce no net force. In other words, for equilibrium, the sum of all forces should equal zero. For example, if we have a resulting force \( \vec{F}= 24.4 \hat{i} - 1.60 \hat{j} \), to achieve equilibrium, a balancing force \( \vec{F}_{3} \) must act such that:
  • The x-component of \( \vec{F}_{3} \) is opposite to \( \vec{F} \)'s x-component: \( F_{3x} = -24.4 \).
  • The y-component should also cancel out: \( F_{3y} = 1.60 \).
Thus, \( \vec{F}_{3} \)'s components ensure that the particle remains stationary, maintaining equilibrium by counteracting the effects of other forces. This concept is crucial in physics to understand systems where stability and balance are essential.
Angle Calculation with Trigonometry
Trigonometry offers handy tools for finding angles between vectors and the coordinate axes. When a vector's direction concerning an axis is needed, we calculate the angle using trigonometric functions, such as tangent.For the vector \( \vec{F}_{3} \), once its components are known, the angle \( \alpha \) it makes with the positive x-axis can be found using:\[ \tan \alpha = \frac{F_{3y}}{F_{3x}} \]In practical terms:
  • Use the inverse tangent function, helping calculate \( \alpha = \tan^{-1}\left(\frac{1.60}{-24.4}\right) \).
  • This works out to an angle of approximately \( -3.75^\circ \), showing the direction is slightly above the negative x-axis.
Calculating angles allows us to comprehend how vectors are oriented in space, giving a complete view beyond just their magnitude.

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Most popular questions from this chapter

The leaning Tower of Pisa is \(59.1 \mathrm{~m}\) high and \(7.44 \mathrm{~m}\) in diameter. The top of the tower is displaced \(4.01 \mathrm{~m}\) from the vertical. Treat the tower as a uniform, circular cylinder. (a) What additional displacement, measured at the top, would bring the tower to the verge of toppling? (b) What angle would the tower then make with the vertical?

A uniform cube of side length \(8.0 \mathrm{~cm}\) rests on a horizontal floor. The coefficient of static friction between cube and floor is \(\mu .\) A horizontal pull \(\vec{P}\) is applied perpendicular to one of the vertical faces of the cube, at a distance \(7.0 \mathrm{~cm}\) above the floor on the vertical midline of the cube face. The magnitude of \(\vec{P}\) is gradually increased. During that increase, for what values of \(\mu\) will the cube eventually (a) begin to slide and (b) begin to tip? (Hint: At the onset of tipping, where is the normal force located?)

A horizontal aluminum rod \(4.8 \mathrm{~cm}\) in diameter projects \(5.3 \mathrm{~cm}\) from a wall. A \(1200 \mathrm{~kg}\) object is suspended from the end of the rod. The shear modulus of aluminum is \(3.0 \times 10^{10} \mathrm{~N} / \mathrm{m}^{2}\). Neglecting the rod's mass, find (a) the shear stress on the rod and (b) the vertical deflection of the end of the rod.

A crate, in the form of a cube with edge lengths of \(1.2 \mathrm{~m}\), contains a piece of machinery; the center of mass of the crate and its contents is located \(0.30 \mathrm{~m}\) above the crate's geometrical center. The crate rests on a ramp that makes an angle \(\theta\) with the horizontal. As \(\theta\) is increased from zero, an angle will be reached at which the crate will either tip over or start to slide down the ramp. If the coefficient of static friction \(\mu_{s}\) between ramp and crate is \(0.60,(\) a) does the crate tip or slide and (b) at what angle \(\theta\) does this occur? If \(\mu_{s}=0.70\), (c) does the crate tip or slide and (d) at what angle \(\theta\) does this occur? (Hint: At the onset of tipping, where is the normal force located?)

A trap door in a ceiling is \(0.91 \mathrm{~m}\) square, has a mass of \(11 \mathrm{~kg}\), and is hinged along one side, with a catch at the opposite side. If the center of gravity of the door is \(10 \mathrm{~cm}\) toward the hinged side from the door's center, what are the magnitudes of the forces exerted by the door on (a) the catch and (b) the hinge?
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