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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

Two crates, one with mass 4.00 kg and the other with mass 6.00 kg, sit on the frictionless surface of a frozen pond, connected by a light rope (\(\textbf{Fig. P4.39}\)). A woman wearing golf shoes (for traction) pulls horizontally on the 6.00-kg crate with a force \(F\) that gives the crate an acceleration of 2.50 m/s\(^2\). (a) What is the acceleration of the 4.00-kg crate? (b) Draw a free-body diagram for the 4.00-kg crate. Use that diagram and Newton's second law to find the tension \(T\) in the rope that connects the two crates. (c) Draw a free-body diagram for the 6.00-kg crate. What is the direction of the net force on the 6.00-kg crate? Which is larger in magnitude, \(T\) or \(F\)? (d) Use part (c) and Newton's second law to calculate the magnitude of \(F\).

Basketball player Darrell Griffith is on record as attaining a standing vertical jump of 1.2 m (4 ft). (This means that he moved upward by 1.2 m after his feet left the floor.) Griffith weighed 890 N (200 lb). (a) What was his speed as he left the floor? (b) If the time of the part of the jump before his feet left the floor was 0.300 s, what was his average acceleration (magnitude and direction) while he pushed against the floor? (c) Draw his free-body diagram. In terms of the forces on the diagram, what was the net force on him? Use Newton's laws and the results of part (b) to calculate the average force he applied to the ground.

(a) An ordinary flea has a mass of 210 \(\mu\)g. How many newtons does it weigh? (b) The mass of a typical froghopper is 12.3 mg. How many newtons does it weigh? (c) A house cat typically weighs 45 N. How many pounds does it weigh, and what is its mass in kilograms?

The fastest served tennis ball, served by "Big Bill" Tilden in 1931, was measured at 73.14 m/s. The mass of a tennis ball is 57 g, and the ball, which starts from rest, is typically in contact with the tennis racquet for 30.0 ms. Assuming constant acceleration, (a) what force did Big Bill's tennis racquet exert on the ball if he hit it essentially horizontally? (b) Draw free-body diagrams of the ball during the serve and just after it moved free of the racquet.

A hockey puck with mass 0.160 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 N to the puck, parallel to the \(x\)-axis; she continues to apply this force until \(t =\) 2.00 s. (a) What are the position and speed of the puck at \(t =\) 2.00 s? (b) If the same force is again applied at \(t =\) 5.00 s, what are the position and speed of the puck at \(t =\) 7.00 s?

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