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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.

Short Answer

Expert verified
(a) 156.2 N at 33.69°; (b) 156.2 N, opposite direction.

Step by step solution

01

Understand the Situation

We have two forces, 80 N and 100 N, acting at an angle of 60° with each other. We are to find the resultant force and the equilibrant force.
02

Set up the Problem using Vectors

The 80 N force may be represented as a vector \( \mathbf{F_1} = 80 \mathbf{i} \). The 100 N force can be divided into components: \( F_{2x} = 100 \cos(60^{\circ}) \) and \( F_{2y} = 100 \sin(60^{\circ}) \). The forces form a triangle with the angle between them, so we use trigonometry to resolve them.
03

Calculate Horizontal and Vertical Components

Calculate the components of the 100 N force.\[ F_{2x} = 100 \cdot \cos(60^{\circ}) = 100 \cdot 0.5 = 50 \text{ N} \]\[ F_{2y} = 100 \cdot \sin(60^{\circ}) = 100 \cdot \frac{\sqrt{3}}{2} = 100 \cdot 0.866 = 86.6 \text{ N} \].
04

Find the Resultant Force Components

Add the horizontal components of the forces: \[ F_{Rx} = 80 + 50 = 130 \text{ N} \]Add the vertical components of the forces: \[ F_{Ry} = 0 + 86.6 = 86.6 \text{ N} \].
05

Calculate the Magnitude of the Resultant Force

Use Pythagorean theorem to find the magnitude of the resultant force:\[ R = \sqrt{F_{Rx}^2 + F_{Ry}^2} \]Substitute the components:\[ R = \sqrt{130^2 + 86.6^2} \]\[ R = \sqrt{16900 + 7494.76} \]\[ R = \sqrt{24394.76} \]\[ R \approx 156.2 \text{ N} \].
06

Determine the Direction of the Resultant

The angle \( \theta \) of the resultant force from the horizontal is given by:\[ \theta = \tan^{-1}\left(\frac{F_{Ry}}{F_{Rx}}\right) \]\[ \theta = \tan^{-1}\left(\frac{86.6}{130}\right) \]\[ \theta \approx 33.69^{\circ} \].
07

Find the Equilibrant Force

The equilibrant force is equal in magnitude and opposite in direction to the resultant force. Hence, its magnitude is 156.2 N, and its direction is opposite. If the resultant force makes an angle of 33.69° with the horizontal axis, the equilibrant will make an angle of 33.69° with the negative axis.

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Key Concepts

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

Equilibrant Force
In physics, to find an equilibrant force, think about achieving a balance. When multiple forces are acting on an object, such as in our exercise, the task is to find a single force that can counteract all the others to bring the system into equilibrium.
This force, known as the "equilibrant force," is crucial for stability. In this exercise, after calculating the resultant force which combines all forces acting on the object, the equilibrant is simply the same magnitude but in the opposite direction.
This opposite direction ensures that the net force becomes zero, effectively balancing the object. This means that if the resultant force points northeast, the equilibrant will point southwest.
  • Equilibrant is equal in magnitude to the resultant force.
  • It is exactly opposite in direction.
  • Its purpose is to balance all applied forces.
Vector Addition
Vector addition is a method used to combine two or more forces acting on a point. It helps us understand how these forces interact.
In simple terms, you perform vector addition by placing vectors head to tail and then drawing a vector from the start of the first vector to the end of the last. This new vector is the resultant.
In our exercise, the forces of 80 N and 100 N are combined. By breaking vectors into components and summing these components, vector addition allows precise calculation of the resultant magnitude and direction.
  • Combining forces acting at angles involves breaking them into components.
  • Add horizontal (x) components and vertical (y) components separately.
  • The vector sum gives a single force representing all applied forces.
Trigonometric Components
Breaking down forces into trigonometric components allows for simplified calculations. This process involves using the sine and cosine functions to resolve forces into horizontal and vertical components.
For example, the 100 N force in our exercise is split using trigonometry, with its x-component calculated using cosine of the angle, and its y-component using sine. These components help in straightforward summation using vector addition.
This method capitalizes on right-angle trigonometry to separate forces making analysis easier and applicable in various physics problems.
  • Use cosine for horizontal (x) components: \( F_x = F \cos(\theta) \).
  • Use sine for vertical (y) components: \( F_y = F \sin(\theta) \).
  • Trigonometry simplifies the interaction of forces at angles.
Pythagorean Theorem
The Pythagorean Theorem is a mathematical principle often used in physics to find the magnitude of resultant forces. It fits perfectly within vector analysis when forces form right triangles.
In our exercise, after determining the x and y components of the forces, the theorem helps us calculate the resultant force. By creating a right triangle with components as legs, you apply the theorem: \(c^2 = a^2 + b^2\).
This relation allows solving for the hypotenuse—or the resultant force in this context—by taking the square root of the sum of the squares of the components.
  • Utilize when vectors form a right triangle.
  • Results in the magnitude of the resultant from component forces.
  • Involves simple squaring, adding, and taking the square root.

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Most popular questions from this chapter

A 700 -N man stands on a scale on the floor of an elevator. The scale records the force it exerts on whatever is on it. What is the scale reading if the elevator has an acceleration of (a) \(1.8 \mathrm{~m} / \mathrm{s}^{2}\) up? (b) \(1.8 \mathrm{~m} / \mathrm{s}^{2}\) down? (c) \(9.8 \mathrm{~m} / \mathrm{s}^{2}\) down?

Suppose, as depicted in Fig. 3-14, that a 70 -kg box is pulled by a 400-N force at an angle of \(30^{\circ}\) to the horizontal. The coefficient of kinetic friction is \(0.50\). Find the acceleration of the box. Fig. \(3-14\) Because the box does not move up or down, we have \(\sum F_{y}=m a_{y}=\) 0\. From Fig. \(3-14\), this equation is $$ +\uparrow \sum_{y}=F_{N}+200 \mathrm{~N}-m g=0 $$ But \(m g=(70 \mathrm{~kg})\left(9.81 \mathrm{~m} / \mathrm{s}^{2}\right)=686.7 \mathrm{~N}\), and it follows that \(F_{N}=\) \(486.7 \mathrm{~N}\) Next find the friction force acting on the box: $$ F_{\mathrm{f}}=\mu_{k} F_{N}=(0.50)(486.7 \mathrm{~N})=243.4 \mathrm{~N} $$ Now write \(\sum F_{x}=m a_{x}\) for the box. It is $$ (346-243.4) \mathrm{N}=(70 \mathrm{~kg})\left(a_{X}\right) $$ from which \(a_{x}=1.466 \mathrm{~m} / \mathrm{s}^{2}\) or \(1.5 \mathrm{~m} / \mathrm{s}^{2}\)

The coefficient of static friction between a box and the flat bed of a truck is 0.60. What is the maximum acceleration the truck can have along level ground if the box is not to slide? The box experiences only one \(x\) -directed force, the friction force. When the box is on the verge of slipping, \(F_{f}=\mu \mathrm{s} F_{W}\), where \(F_{W}\) is the weight of the box. As the truck accelerates, the friction force must cause the box to have the same acceleration as the truck; otherwise, the box will slip. When the box is not slipping, \(\sum F_{x}=m a_{x}\) applied to the box gives \(F_{f}=m a_{x}\). However, if the box is on the verge of slipping, \(F_{f}\) \(=\mu_{s} F_{W}\) so that \(\mu_{s} F_{W}=m a_{x} .\) Because \(F_{W}=m g\), $$ a_{x}=\frac{\mu_{s} m g}{m}=\mu_{s} g=(0.60)\left(9.81 \mathrm{~m} / \mathrm{s}^{2}\right)=5.9 \mathrm{~m} / \mathrm{s}^{2} $$ as the maximum acceleration without slipping.

A 20 -kg wagon is pulled along the level ground by a rope inclined at \(30^{\circ}\) above the horizontal. A friction force of \(30 \mathrm{~N}\) opposes the motion. How large is the pulling force if the wagon is moving with (a) constant speed and \((b)\) an acceleration of \(0.40 \mathrm{~m} / \mathrm{s}^{2}\) ?

A force of \(100 \mathrm{lb}\) acting on a body weighing \(500 \mathrm{lb}\) causes the body to accelerate uniformly. What would happen to the acceleration if the force is increased to \(200 \mathrm{lb}\) ? [Hint: Units are not important here as long as you are consistent.] Assume no friction.

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