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Forces \(\vec{F_1}\) and \(\vec{F_2}\)act at a point. The magnitude of \(\vec{F_1}\) is 9.00 N, and its direction is 60.0\(^\circ\) above the \(x\)-axis in the second quadrant. The magnitude of \(\vec{F_2}\) is 6.00 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 resultant force has components \(-8.11\, N\) (x-axis) and \(3.00\, N\) (y-axis), with magnitude 8.65 N.

Step by step solution

01

Understanding the Forces

We have two forces, \(\vec{F_1}\) and \(\vec{F_2}\). The force \(\vec{F_1}\) has a magnitude of 9.00 N and is directed 60.0\(^\circ\) above the \(x\)-axis in the second quadrant. \(\vec{F_2}\) has a magnitude of 6.00 N and is directed 53.1\(^\circ\) below the \(x\)-axis in the third quadrant.
02

Breaking Down \(\vec{F_1}\) into Components

To find the \(x\)- and \(y\)-components of \(\vec{F_1}\), use trigonometry:\[F_{1x} = F_1 \cdot \cos(180^\circ - 60^\circ) = 9.00 \cdot \cos(120^\circ) = 9.00 \cdot (-0.5) = -4.50 \text{ N}\]\[F_{1y} = F_1 \cdot \sin(180^\circ - 60^\circ) = 9.00 \cdot \sin(120^\circ) = 9.00 \cdot 0.866 = 7.79 \text{ N}\]
03

Breaking Down \(\vec{F_2}\) into Components

For \(\vec{F_2}\), also use trigonometry:\[F_{2x} = F_2 \cdot \cos(180^\circ + 53.1^\circ) = 6.00 \cdot \cos(233.1^\circ) = 6.00 \cdot (-0.6018) = -3.61 \text{ N}\]\[F_{2y} = F_2 \cdot \sin(180^\circ + 53.1^\circ) = 6.00 \cdot \sin(233.1^\circ) = 6.00 \cdot (-0.7986) = -4.79 \text{ N}\]
04

Calculating the Resultant Force Components

Add the \(x\)-components and \(y\)-components of \(\vec{F_1}\) and \(\vec{F_2}\):\[R_x = F_{1x} + F_{2x} = -4.50 + (-3.61) = -8.11 \text{ N}\]\[R_y = F_{1y} + F_{2y} = 7.79 + (-4.79) = 3.00 \text{ N}\]
05

Finding the Magnitude of the Resultant Force

The magnitude of the resultant force \(R\) can be found using the Pythagorean theorem:\[R = \sqrt{R_x^2 + R_y^2} = \sqrt{(-8.11)^2 + (3.00)^2} = \sqrt{65.84 + 9.00} = \sqrt{74.84} = 8.65 \text{ N}\]

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Key Concepts

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

Resultant Force
In physics, a resultant force is the single force that represents the vector sum of two or more individual forces acting on a point. To understand the resultant force, imagine this: if you have multiple forces pulling or pushing an object in different directions, the resultant force is what's left when you combine all those forces together.
The resultant force not only has a specific magnitude but also a unique direction. It determines the acceleration and motion of the object according to Newton's second law of motion. In the provided exercise, we were dealing with two forces, \(\vec{F_1}\) and \(\vec{F_2}\), acting at a point in different directions. By adding these forces vectorially, we found the resultant force that acts as if it were a single force exerted on the point.
To determine the magnitude and direction of the resultant force, both the individual components of the forces are considered. This involves breaking down each force into its horizontal (x-axis) and vertical (y-axis) components. Once we have these components, they are summed up separately to get the resultant components (i.e., \(R_x\) and \(R_y\)), which we then use to calculate the overall magnitude and direction of the resultant force.
Force Components
The concept of force components is crucial for solving problems involving forces acting at angles. Any force can be resolved into two perpendicular components: one along the x-axis and the other along the y-axis. These are known as the horizontal and vertical components of the force, respectively.
For instance, in the exercise, \(\vec{F_1}\) was resolved into \(F_{1x}\) and \(F_{1y}\), and \(\vec{F_2}\) was resolved into \(F_{2x}\) and \(F_{2y}\).
Here's how it works:
  • The x-component of a force is found by multiplying the force's magnitude by the cosine of its angle relative to the positive x-axis.
  • The y-component is similarly found by multiplying the force's magnitude by the sine of the angle.
These components are vector quantities themselves and can be positive or negative depending on the quadrant in which the force is acting. As seen in the exercise, correctly resolving these components is key to determining the overall effect resulting forces have on an object.
Trigonometric Functions
Trigonometric functions like sine and cosine play an indispensable role in physics, especially when dealing with vector quantities like forces. These functions help translate angled forces into their respective x and y components, which is essential when adding vectors.
In the exercise, each force was represented in terms of its magnitude and direction. The angles given were utilized with cosine and sine functions:
  • Cosine function (cos) deals with the adjacent side (x-component) in a right triangle formed by the force and axes.
  • Sine function (sin) deals with the opposite side (y-component).
This process is grounded in the fundamental principles of trigonometry, which state that for a given angle in a right triangle, the cosine of the angle is the ratio of the length of the adjacent side to the hypotenuse, while the sine is the ratio of the opposite side to the hypotenuse. Therefore, trigonometric functions bridge the angle forces make with coordinate axes, making it possible to solve for component forces operating perpendicularly on the axes.

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Most popular questions from this chapter

To study damage to aircraft that collide with large birds, you design a test gun that will accelerate chicken-sized objects so that their displacement along the gun barrel is given by \(x =\) (9.0 \(\times\) 10\(^3\) m/s\(^2)t^2\) - (8.0 \(\times\) 10\(^4\) m/s\(^3)t^3\). The object leaves the end of the barrel at \(t =\) 0.025 s. (a) How long must the gun barrel be? (b) What will be the speed of the objects as they leave the end of the barrel? (c) What net force must be exerted on a 1.50-kg object at (i) \(t =\) 0 and (ii) \(t =\) 0.025 s?

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

A dockworker applies a constant horizontal force of 80.0 N to a block of ice on a smooth horizontal floor. The frictional force is negligible. The block starts from rest and moves 11.0 m in 5.00 s. (a) What is the mass of the block of ice? (b) If the worker stops pushing at the end of 5.00 s, how far does the block move in the next 5.00 s?

You have landed on an unknown planet, Newtonia, and want to know what objects weigh there. When you push a certain tool, starting from rest, on a frictionless horizontal surface with a 12.0-N force, the tool moves 16.0 m in the first 2.00 s. You next observe that if you release this tool from rest at 10.0 m above the ground, it takes 2.58 s to reach the ground. What does the tool weigh on Newtonia, and what does it weigh on Earth?

Crates \(A\) and \(B\) sit at rest side by side on a frictionless horizontal surface. They have masses \(m_A\) and \(m_B\), respectively. When a horizontal force \(\vec{F}\) is applied to crate \(A\), the two crates move off to the right. (a) Draw clearly labeled free-body diagrams for crate \(A\) and for crate \(B\). Indicate which pairs of forces, if any, are third-law action-reaction pairs. (b) If the magnitude of \(\vec{F}\)is less than the total weight of the two crates, will it cause the crates to move? Explain.

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