Q: A force $$F = 260\,\text{N}$$ acts at an angle of $$30^\circ$$ above the horizontal. What are the $$x$$ and $$y$$ components of this force?

$$F_x = 260 \cos 30^\circ = 225\,\text{N},\ F_y = 260 \sin 30^\circ = 130\,\text{N}$$

Q: Given two forces, $$F_1 = 150$$\,\text{N}$$ at 30$$^\circ$$ above the horizontal and F_2$ = 210\,\text{N} at$$ $$115^\circ$$ from the horizontal, what is the magnitude of their resultant using the law of cosines?

$$F_R = \sqrt{150^2$ + 210^2$ + 2 \times 150 \times 210 \cos 85^\circ} = 213\,\text{N}$$

Q: If a force $$F = 200\,\text{lb} is $$resolved into two components $$F_x$$ and $$F$$ such that $$F_x$$ acts at $$60^\circ$$ and $$F at$$ $$45^\circ$$ from the resultant, what is the value of $$F$$ using the law of sines?

$$F = \frac{200}{\sin 60^\circ} \sin 45^\circ = 245\,\text{lb}$$

Q: A link is subjected to two forces: $$F_1 = 600$$\,\text{N}$$ at 30$$^\circ$$ above the x$-axis and F_2$ = 400\,\text{N} at$$ $$45^\circ$$ below the $$x-$$axis. What is the $$y-$$component of the resultant force?

$$F_{Ry} = 600 \sin 30^\circ - 400 \sin 45^\circ = -82.8\,\text{N}$$

Q: Given $$F_1 = 600$$\,\text{N}$$ at 30$$^\circ$$ and F_2$ = 400\,\text{N} at$$ $$-45^\circ$$, what is the magnitude of the resultant force?

$$F_R = \sqrt{(518.8)^2$ + (-82.8)^2$} = 526.4\,\text{N}$$

Question 1

A force $$F = 260\,	ext{N}$$ acts at an angle of $$30^\circ$$ above the horizontal. What are the $$x$$ and $$y$$ components of this force?

Accepted Answer

$$F_x = 260 \cos 30^\circ = 225\,	ext{N},\ F_y = 260 \sin 30^\circ = 130\,	ext{N}$$

Question 2

Given two forces, $$F_1 = 150$$\,	ext{N}$$ at 30$$^\circ$$ above the horizontal and F_2$ = 210\,	ext{N} at$$ $$115^\circ$$ from the horizontal, what is the magnitude of their resultant using the law of cosines?

Accepted Answer

$$F_R = \sqrt{150^2$ + 210^2$ + 2 	imes 150 	imes 210 \cos 85^\circ} = 213\,	ext{N}$$

Question 3

A $$600\,	ext{lb}$$ force acts horizontally. It is resolved into two components along axes $$a$$ and $$n$$ that are $$30^\circ$$ and $$120^\circ$$ from the force direction, respectively. Using the law of sines, what is the magnitude of the $$a-$$component?

Accepted Answer

Question 4

If a force $$F = 200\,	ext{lb} is $$resolved into two components $$F_x$$ and $$F$$ such that $$F_x$$ acts at $$60^\circ$$ and $$F at$$ $$45^\circ$$ from the resultant, what is the value of $$F$$ using the law of sines?

Accepted Answer

$$F = \frac{200}{\sin 60^\circ} \sin 45^\circ = 245\,	ext{lb}$$

Question 5

A link is subjected to two forces: $$F_1 = 600$$\,	ext{N}$$ at 30$$^\circ$$ above the x$-axis and F_2$ = 400\,	ext{N} at$$ $$45^\circ$$ below the $$x-$$axis. What is the $$y-$$component of the resultant force?

Accepted Answer

$$F_{Ry} = 600 \sin 30^\circ - 400 \sin 45^\circ = -82.8\,	ext{N}$$

Question 6

Which of the following best describes the advantage of using Cartesian vector notation for force addition?

Accepted Answer

It allows for direct addition of vector components, which is especially useful in three-dimensional problems.

Question 7

Three concurrent coplanar forces act at a point: $$F_1 = 250$$\,	ext{N}$$ at 45$$^\circ$$, F_2$ = 200\,	ext{N} at$$ $$-30^\circ$$, and $$F_3 = 300$$\,	ext{N}$$ at 180$$^\circ$$. What is the x$-component of the resultant force?

Accepted Answer

$$F_{Rx} = 250 \cos 45^\circ + 200 \cos (-30^\circ) + 300 \cos 180^\circ = 298.8\,	ext{N}$$

Question 8

A force $$F = 240\,	ext{N}$$ acts at $$30^\circ$$ below the $$x-$$axis. What is its Cartesian vector representation?

Accepted Answer

$$F = 240 \cos 30^\circ \mathbf{i} - 240 \sin 30^\circ \mathbf{j}$$

Question 9

If the resultant of two forces $$F_1$$ and $$F_2 is to be $$found using the parallelogram law, which geometric property is essential for constructing the solution?

Accepted Answer

The angle between the two forces at their point of application.

Question 10

A $$600\,	ext{lb}$$ force is resolved into two components along axes that are not perpendicular. What is a possible outcome regarding the magnitudes of the components compared to the original force?

Accepted Answer

A component can have a magnitude greater than the original force.

Question 11

Given $$F_1 = 600$$\,	ext{N}$$ at 30$$^\circ$$ and F_2$ = 400\,	ext{N} at$$ $$-45^\circ$$, what is the magnitude of the resultant force?

Accepted Answer

$$F_R = \sqrt{(518.8)^2$ + (-82.8)^2$} = 526.4\,	ext{N}$$

Question 12

When resolving a force into components along two axes that are not perpendicular, which law is most appropriate to use?

Accepted Answer

Law of sines

Question 13

A force $$F is $$resolved into $$F_x$$ and $$F_y$$ such that $$F_x = 200\,	ext{N}$$ and $$F_y = 100\,	ext{N}. $$What is the magnitude of $$F$$?

Accepted Answer

$$F = \sqrt{200^2$ + 100^2$} = 223.6\,	ext{N}$$

Question 14

If a force $$F is $$resolved into two perpendicular components $$F_x$$ and $$F_y$$, and $$F_x = 240\,	ext{N}$$, $$F_y = 100\,	ext{N}$$, what is the direction angle $$	heta of$$ $$F$$ from the $$x-$$axis?

Accepted Answer

$$	heta = 	an$$^{-1}$$ \left( \frac{100}{240} ight) = 22.6^\circ$$

Question 15

A force $$F is $$resolved into two components $$F_1$$ and $$F_2$$ such that the angle between them is $$120^\circ. If$$ $$F = 600\,	ext{N}$$ and $$F_1 is $$along $$30^\circ$$ from $$F$$, what is the magnitude of $$F_1$$ using the law of sines?

Accepted Answer

$$F_1$$ = \frac{600}{\sin 120^\circ} \sin 30^\circ = 346\,	ext{N}$$

Question 16

Which of the following statements is true regarding the direction of the resultant force when two forces act at a point?

Accepted Answer

The direction of the resultant is always between the directions of the two forces.

Question 17

If the $$x$$ and $$y$$ components of a force are both negative, in which quadrant does the force vector lie?

Accepted Answer

Third quadrant

Question 18

A force $$F = 250\,	ext{N}$$ acts at $$45^\circ$$ above the $$x-$$axis. What is the $$y-$$component of this force?

Accepted Answer

$$F_y = 250 \sin 45^\circ = 176.8\,	ext{N}$$

Vector Addition and Resolution of Forces in Physics

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