Statics and Dynamics Quiz: Sign Conventions And Coordinates

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Sign Conventions And CoordinatesQuestion 1 of 9

A particle moves along a curved path in the xy-plane. An engineer sets up a standard right-handed Cartesian coordinate system with x pointing right and y pointing up. The particle's velocity components are measured as $v_x = -4 \text{ m/s}$ and $v_y = 3 \text{ m/s}$ . The engineer then rotates the coordinate system 180° about the z-axis (so the new $x'$ points left and $y'$ points down) and re-expresses the same physical velocity.

After the 180° rotation of the coordinate system, what are the components of the particle's velocity in the new $x'y'$ frame?

v_{x'} = 4 \text{ m/s}

v_{y'} = -3 \text{ m/s}

v_{x'} = -4 \text{ m/s}

v_{y'} = 3 \text{ m/s}

v_{x'} = 4 \text{ m/s}

v_{y'} = 3 \text{ m/s}

v_{x'} = -4 \text{ m/s}

v_{y'} = -3 \text{ m/s}

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Statics and Dynamics Quiz

Statics and Dynamics Quiz: Sign Conventions And Coordinates

Practice Sign Conventions And Coordinates in Statics and Dynamics with focused quiz questions that help you check what you know, review explanations, and build confidence with test-style prompts.

What this quiz covers

This quiz focuses on Sign Conventions And Coordinates, giving you a quick way to practice the rules, question types, and explanations that matter most for Statics and Dynamics.

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Try each quiz question before looking at the correct answer. Use the explanations to review missed ideas, then come back to similar questions until the pattern feels familiar.

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Question 1

After the 180° rotation of the coordinate system, what are the components of the particle's velocity in the new $x'y'$ frame?

$v_{x'} = 4 \text{ m/s}$ , $v_{y'} = -3 \text{ m/s}$ (correct answer)
$v_{x'} = -4 \text{ m/s}$ , $v_{y'} = 3 \text{ m/s}$
$v_{x'} = 4 \text{ m/s}$ , $v_{y'} = 3 \text{ m/s}$
$v_{x'} = -4 \text{ m/s}$ , $v_{y'} = -3 \text{ m/s}$

Explanation: When a coordinate system rotates, the physical velocity vector doesn't change — only how its components are measured changes. This is the core concept: rotating the frame transforms the components, not the underlying motion. A 180° rotation about the z-axis is described by the rotation matrix:

\begin{bmatrix} \cos180° & \sin180° \\ -\sin180° & \cos180° \end{bmatrix} = \begin{bmatrix} -1 & 0 \\ 0 & -1 \end{bmatrix}

Applying this to the original components

(v_x, v_y) = (-4, 3)

v_{x'} = (-1)(-4) = 4 \text{ m/s}, \quad v_{y'} = (-1)(3) = -3 \text{ m/s}

This confirms A is correct. You can also reason geometrically: the new

x'

points opposite to the old

x

, so a component that was negative becomes positive. Similarly,

y'

points opposite to

y

, flipping the sign of the y-component. B is the trap of doing nothing — it keeps the original components unchanged, as if the rotation never happened. C only flips the x-component but leaves y unchanged, which would correspond to a reflection, not a 180° rotation. D flips both signs to negative, which would be correct if the original components were both positive, but here

v_x

starts negative, so negating it gives a positive result. A useful rule of thumb: a 180° rotation always negates both components. So whatever signs you start with, flip them both. If you remember that, you can solve these problems in seconds without writing out the full matrix.

Question 2

A force $\vec{F}$ is expressed in Cartesian components as $\vec{F} = 3\hat{i} - 4\hat{j} + 0\hat{k}$ N. An analyst needs to express this force in polar coordinates $(r, \theta)$ defined in the xy-plane, where $\hat{e}_r$ points radially outward and $\hat{e}_\theta$ points in the direction of increasing $\theta$ (counterclockwise). The position of the point of application is at angle $\theta = 53.13°$ from the positive x-axis (i.e., $\cos\theta = 0.6$ , $\sin\theta = 0.8$ ).

What are the radial and transverse components, $F_r$ and $F_\theta$ , of the force in the polar coordinate frame at $\theta = 53.13°$ ?

$F_r = 5 \text{ N}$ and $F_\theta = 0 \text{ N}$ , because the force magnitude is 5 N and it happens to be purely radial at this angle.
$F_r = -1.4 \text{ N}$ and $F_\theta = +4.8 \text{ N}$ , because the radial component is the dot product of $\vec{F}$ with $\hat{e}_r$ , and the transverse component is the dot product with $\hat{e}_\theta = +0.8\hat{i} + 0.6\hat{j}$ , taking $\hat{e}_\theta$ as having a positive x-component.
$F_r = 3 \text{ N}$ and $F_\theta = -4 \text{ N}$ , because the Cartesian components are directly transferred to the polar frame since the force lies in the xy-plane and has no z-component.
$F_r = -1.4 \text{ N}$ and $F_\theta = -4.8 \text{ N}$ , because each component is found by dotting $\vec{F}$ with the respective unit vector: $\hat{e}_r = 0.6\hat{i} + 0.8\hat{j}$ and $\hat{e}_\theta = -0.8\hat{i} + 0.6\hat{j}$ . (correct answer)

Explanation: Whenever you need to convert a vector from Cartesian to polar components, remember that polar unit vectors are angle-dependent — they rotate with

\theta

, so you must project using dot products, not simply reassign components. At angle

\theta = 53.13°

, the polar unit vectors are defined as:

\hat{e}_r = \cos\theta\,\hat{i} + \sin\theta\,\hat{j} = 0.6\hat{i} + 0.8\hat{j}

\hat{e}_\theta = -\sin\theta\,\hat{i} + \cos\theta\,\hat{j} = -0.8\hat{i} + 0.6\hat{j}

Note that

\hat{e}_\theta

points counterclockwise (increasing

\theta

), which means its x-component is negative. Now project

\vec{F} = 3\hat{i} - 4\hat{j}

F_r = \vec{F} \cdot \hat{e}_r = (3)(0.6) + (-4)(0.8) = 1.8 - 3.2 = -1.4 \text{ N}

F_\theta = \vec{F} \cdot \hat{e}_\theta = (3)(-0.8) + (-4)(0.6) = -2.4 - 2.4 = -4.8 \text{ N}

This confirms D is correct. A is wrong because even though

|\vec{F}| = 5

N, a force being purely radial requires

\vec{F}

to be parallel to

\hat{e}_r

at that angle — which it isn't. B uses the wrong sign for

\hat{e}_\theta

, flipping its x-component to positive; this is the most common trap on this type of question. C incorrectly assumes Cartesian and polar components are interchangeable, which is only true at

\theta = 0°

. As a study tip: always write out

\hat{e}_\theta = -\sin\theta\,\hat{i} + \cos\theta\,\hat{j}

explicitly before computing — the negative sign on

\sin\theta

is easy to drop and will cost you both components.

Question 3

A beam is supported at both ends, and two engineers independently set up sign conventions to compute bending moments. Engineer A defines positive bending moment as causing concave-upward (sagging) curvature, while Engineer B defines positive bending moment as causing concave-downward (hogging) curvature. They analyze the same midspan section and Engineer A calculates $M_A = +150 \text{ N·m}$ . Which statement best describes the relationship between their results and the physical state of the beam?

Engineer B must calculate $M_B = +150 \text{ N·m}$ as well, because the physical bending moment at the section is an objective quantity independent of sign convention, and both engineers must report the same signed value.
Engineer B must calculate $M_B = +150 \text{ N·m}$ , because sign conventions affect only the direction of the moment vector, not its sense of rotation, and both engineers observe the same clockwise internal moment at the section face.
Engineer B must calculate $M_B = -150 \text{ N·m}$ only if the beam is statically determinate; for indeterminate beams, the moment sign depends on the chosen redundant and cannot be related by simple negation.
Engineer B must calculate $M_B = -150 \text{ N·m}$ , because the two sign conventions are opposite, so any moment that is positive under one convention is negative under the other, while the physical deformation (sagging) remains the same. (correct answer)

Explanation: Whenever you see a question involving sign conventions in structural analysis, remind yourself of a crucial distinction: the physical reality of the beam (how it actually curves, where it sags or hogs) never changes, but the signed number you assign to that reality depends entirely on your chosen convention. Here, Engineer A's convention assigns a positive sign to sagging (concave-up) curvature. Engineer B's convention is the exact opposite — positive means hogging (concave-down). Since both engineers are looking at the same midspan section with the same physical sagging deformation, and their conventions are mirror images of each other, Engineer B must report

M_B = -150 \text{ N·m}

. The magnitude stays 150 N·m because the physical intensity of bending hasn't changed; only the sign flips because the labeling system flipped. This makes D correct. A is wrong because it conflates the physical moment with the signed number. The magnitude is objective, but the sign is a bookkeeping choice — two engineers using opposite conventions will always report opposite signs for the same physical state. B is wrong because it misidentifies what sign conventions affect. They don't merely change the moment vector's direction while preserving a "sense of rotation" label — they directly govern how you assign + or − to a given curvature, meaning the reported signed values will differ. C is wrong because the relationship between sign conventions has nothing to do with static determinacy. Whether a beam is determinate or indeterminate, if two conventions are exact opposites, their signed results are always negatives of each other. Study tip: Always separate magnitude (physics) from sign (bookkeeping). When two conventions are exact opposites, their signed outputs are negatives of each other — every time, regardless of beam type or loading.

Question 4

In a planar dynamics problem, a particle's position is described using polar coordinates where $r$ is the radial distance from the origin and $\theta$ is measured counterclockwise from the positive x-axis. The particle moves such that $r = 2t$ m and $\theta = t^2$ rad, where $t$ is time in seconds. At $t = 1$ s, the acceleration is needed.

Which expression correctly gives the radial component of acceleration $a_r$ at $t = 1$ s in polar coordinates?

$a_r = \ddot{r} = 0 \text{ m/s}^2$ , because the radial acceleration depends only on the second time-derivative of $r$ , and since $r = 2t$ is linear, $\ddot{r} = 0$ .
$a_r = \ddot{r} - r\dot{\theta}^2 = 0 - (2)(1)^2 = -2 \text{ m/s}^2$ , using the polar acceleration formula but evaluating $\dot{\theta}$ as $\theta$ rather than its time derivative.
$a_r = \ddot{r} - r\dot{\theta}^2 = 0 - (2)(2)^2 = -8 \text{ m/s}^2$ , using the full polar acceleration formula with $\dot{\theta} = 2t$ evaluated at $t = 1$ s. (correct answer)
$a_r = \ddot{r} + r\dot{\theta}^2 = 0 + (2)(2)^2 = +8 \text{ m/s}^2$ , because the centripetal term in the radial direction is additive when the particle moves in the counterclockwise direction.

Explanation: Whenever you encounter acceleration in polar coordinates, resist the instinct to treat it like Cartesian coordinates. The radial acceleration is not simply

\ddot{r}

. Because the coordinate directions themselves rotate, extra terms appear. The complete formula is:

a_r = \ddot{r} - r\dot{\theta}^2

The second term,

-r\dot{\theta}^2

, is the centripetal correction — it accounts for the curvature of the path and always points inward (negative radial direction), regardless of whether motion is clockwise or counterclockwise. Applying this to the problem: since

r = 2t

, you get

\dot{r} = 2

and

\ddot{r} = 0

. Since

\theta = t^2

, the angular velocity is

\dot{\theta} = \frac{d\theta}{dt} = 2t

, which at

t = 1

s gives

\dot{\theta} = 2

rad/s. Also,

r(1) = 2

m. Substituting:

a_r = 0 - (2)(2)^2 = -8 \text{ m/s}^2

This confirms C is correct. A is the most tempting trap — it ignores the

-r\dot{\theta}^2

term entirely, treating polar acceleration like a simple second derivative. This is only valid when there's no angular motion. B uses the right formula but substitutes

\theta = t^2 = 1

rad instead of its time derivative

\dot{\theta} = 2t = 2

rad/s. Always differentiate

\theta

to find

\dot{\theta}

— never plug in the angle itself. D flips the sign of the centripetal term, wrongly claiming counterclockwise motion makes it additive. The sign is always negative in the radial formula — direction of rotation doesn't change this. Your study tip: memorize

a_r = \ddot{r} - r\dot{\theta}^2

and

a_\theta = r\ddot{\theta} + 2\dot{r}\dot{\theta}

as a pair. On exam day, immediately write these down and carefully differentiate each kinematic function before substituting.

Question 5

In a 3D statics problem, three forces act on a particle: $\vec{F}_1 = 2\hat{i} + 3\hat{j} - 1\hat{k}$ N, $\vec{F}_2 = -2\hat{i} + 0\hat{j} + 4\hat{k}$ N, and $\vec{F}_3 = 0\hat{i} - 3\hat{j} + p\hat{k}$ N. A student claims that for equilibrium, $p = -3$ N. Which of the following correctly evaluates this claim?

The student is correct: summing the k-components gives $-1 + 4 + p = 0$ , so $p = -3$ , and the i- and j-components also sum to zero independently, confirming full equilibrium. (correct answer)
The student is incorrect: equilibrium requires all three component equations to be satisfied. The i-components sum to zero and j-components sum to zero, but $-1 + 4 + p = 0$ yields $p = -3$ , which does satisfy the k-equation, so the student's value is numerically right but the claim should explicitly verify all three axes.
The student is incorrect: the k-component equation is $1 + 4 + p = 0$ (using unsigned magnitudes), giving $p = -5$ , and the student forgot to account for the absolute values of the force components.
The student is incorrect: the correct equilibrium condition in 3D is that the magnitude of the resultant equals zero, which requires solving $|\vec{F}_1 + \vec{F}_2 + \vec{F}_3| = 0$ as a scalar equation; the component-wise approach the student used is only valid in 2D problems.

Explanation: When a particle is in static equilibrium in 3D, Newton's first law requires the resultant force vector to be exactly zero — meaning each component (i, j, and k) must independently sum to zero. This gives you three separate scalar equations to check, not one combined condition. Let's verify all three axes using the given forces: i-components:

2 + (-2) + 0 = 0

✓ j-components:

3 + 0 + (-3) = 0

✓ k-components:

-1 + 4 + p = 0 \Rightarrow p = -3

✓ All three equations are satisfied with

p = -3

, confirming the student's claim is fully correct. Answer A is right. Answer B is subtly wrong in its framing — it says the student's value is "numerically right" but implies the claim is somehow incomplete or incorrect. In fact, the student's value is correct, and equilibrium is confirmed. B penalizes the student for something that isn't actually an error. Answer C introduces a serious misconception: you never use unsigned magnitudes (absolute values) when summing force components. The signs carry physical meaning — they encode direction. Dropping signs and using magnitudes would only be valid when computing the scalar magnitude of a single vector, not when applying equilibrium. Answer D is flatly false. The component-wise approach is not limited to 2D — it is the standard, rigorous method in 3D statics. Requiring

|\vec{F}_{resultant}| = 0

is equivalent, but solving it as a single scalar equation is far less practical and reveals nothing new. Study tip: Always check all three component equations in 3D equilibrium problems — but remember that satisfying all three is the complete condition. Don't let distractors convince you that a correct method is somehow insufficient.

Question 6

A particle's position in cylindrical coordinates is given as $(r, \theta, z) = (2 \text{ m}, 60°, 3 \text{ m})$ . Its velocity in cylindrical coordinates is $\vec{v} = \dot{r}\hat{e}_r + r\dot{\theta}\hat{e}_\theta + \dot{z}\hat{k}$ , with $\dot{r} = 1 \text{ m/s}$ , $\dot{\theta} = 2 \text{ rad/s}$ , and $\dot{z} = -1 \text{ m/s}$ .

What is the velocity vector expressed in Cartesian components $(v_x, v_y, v_z)$ ?

$v_x = 1\cos60° - 4\sin60° \approx -2.96 \text{ m/s}$ , $v_y = 1\sin60° + 4\cos60° \approx 2.87 \text{ m/s}$ , $v_z = -1 \text{ m/s}$ (correct answer)
$v_x = \dot{r}\cos60° = 0.5 \text{ m/s}$ , $v_y = \dot{r}\sin60° \approx 0.866 \text{ m/s}$ , $v_z = -1 \text{ m/s}$
$v_x = 1\cos60° + 4\cos60° = 2.5 \text{ m/s}$ , $v_y = 1\sin60° + 4\sin60° \approx 4.33 \text{ m/s}$ , $v_z = -1 \text{ m/s}$
$v_x = 1\cos60° + 4\sin60° \approx 3.96 \text{ m/s}$ , $v_y = 1\sin60° - 4\cos60° \approx -1.13 \text{ m/s}$ , $v_z = -1 \text{ m/s}$

Explanation: When converting velocity from cylindrical to Cartesian coordinates, you need to remember that both the radial and tangential unit vectors project onto the x- and y-axes. The key relationships are:

\hat{e}_r = \cos\theta\,\hat{i} + \sin\theta\,\hat{j}, \qquad \hat{e}_\theta = -\sin\theta\,\hat{i} + \cos\theta\,\hat{j}

Notice the minus sign on the x-component of

\hat{e}_\theta

— this is where most errors occur. With

\dot{r} = 1

m/s,

r\dot{\theta} = 2(2) = 4

m/s, and

\theta = 60°

, substituting gives:

v_x = \dot{r}\cos60° - r\dot{\theta}\sin60° = 1(0.5) - 4(0.866) \approx -2.96 \text{ m/s}

v_y = \dot{r}\sin60° + r\dot{\theta}\cos60° = 1(0.866) + 4(0.5) \approx 2.87 \text{ m/s}

v_z = -1 \text{ m/s}

This confirms A is correct. B fails by ignoring the tangential term

r\dot{\theta}

entirely — it treats velocity as if only radial motion exists. C uses

+\sin60°

for both components of

\hat{e}_\theta

instead of

-\sin60°

for

v_x

, missing the critical minus sign. D flips the signs, applying

+\sin60°

v_x

and

-\cos60°

v_y

— essentially reversing the correct formula. Study tip: Always write out the unit vector conversions explicitly before computing. The minus sign in

\hat{e}_\theta = -\sin\theta\,\hat{i} + \cos\theta\,\hat{j}

is easy to drop under pressure — memorize it as a pair with

\hat{e}_r

Question 7

A dynamics problem is solved using normal-tangential (n-t) coordinates for a particle moving along a circular arc of radius $R$ . The particle moves with increasing speed in the counterclockwise direction. The unit vector $\hat{e}_t$ is defined as tangent to the path in the direction of motion, and $\hat{e}_n$ is defined as pointing toward the center of curvature. Which statement about the acceleration vector in this coordinate system is correct?

The tangential component $a_t = \dot{v} > 0$ points in the $+\hat{e}_t$ direction, and the normal component $a_n = v^2/R > 0$ points in the $+\hat{e}_n$ direction, so both components are positive and $\hat{e}_n$ points away from the center.
The tangential component $a_t = \dot{v} > 0$ points in the $+\hat{e}_t$ direction, and the normal component $a_n = v^2/R > 0$ points in the $+\hat{e}_n$ direction (toward the center), consistent with the standard n-t sign convention where $\hat{e}_n$ always points inward. (correct answer)
The tangential component $a_t = -\dot{v} < 0$ points in the $-\hat{e}_t$ direction because counterclockwise motion inverts the sign convention, while the normal component $a_n = v^2/R > 0$ correctly points toward the center in the $+\hat{e}_n$ direction.
The tangential component $a_t = \dot{v} > 0$ points in the $+\hat{e}_t$ direction, but the normal component must be written as $a_n = -v^2/R$ because centripetal acceleration physically opposes the outward tendency of the particle and is therefore negative in the n-t frame.

Explanation: When solving dynamics problems in normal-tangential coordinates, the key is understanding how the unit vectors are defined by convention, not by the geometry of a specific problem. In n-t coordinates,

\hat{e}_t

always points in the direction of motion, and

\hat{e}_n

always points toward the center of curvature (inward). These definitions are fixed regardless of whether motion is clockwise, counterclockwise, speeding up, or slowing down. With those definitions in place, the acceleration vector is

\vec{a} = \dot{v}\,\hat{e}_t + \frac{v^2}{R}\,\hat{e}_n

. Because the particle is speeding up,

\dot{v} > 0

, so the tangential component points in

+\hat{e}_t

. Because

v^2/R

is always positive and

\hat{e}_n

is defined to point inward, the normal component correctly represents centripetal (center-directed) acceleration. This is exactly what B states, making it correct. A is almost right but contains a critical error: it claims

\hat{e}_n

points away from the center. This directly contradicts the definition —

\hat{e}_n

always points toward the center of curvature. C incorrectly asserts that counterclockwise motion flips the sign of the tangential component. It does not —

\hat{e}_t

follows the direction of motion regardless of orientation, so

a_t = \dot{v}

remains positive. D introduces a false negative sign on the normal component. Because

\hat{e}_n

is already defined as inward, writing

a_n = -v^2/R

would incorrectly point the normal acceleration outward, double-counting the direction. A reliable study tip: in n-t coordinates, both

a_t

and

a_n

are scalar magnitudes with sign determined by

\dot{v}

and

v^2/R

respectively — never manually negate

a_n

, because the inward direction is already baked into

\hat{e}_n

Question 8

Two analysts solve the same 2D equilibrium problem involving a concurrent force system. Analyst 1 uses a standard right-handed coordinate system with x pointing right and y pointing up. Analyst 2 uses x pointing left and y pointing up (a left-handed system in 2D). Both analysts correctly identify all forces and apply equilibrium. Force $\vec{P}$ points to the right with magnitude 10 N, and force $\vec{Q}$ points upward with magnitude 8 N. A third force $\vec{R}$ must be found for equilibrium.

Analyst 1 correctly determines $\vec{R}$ has components $R_{x1} = -10 \text{ N}$ and $R_{y1} = -8 \text{ N}$ in their coordinate system. What components does Analyst 2 report for $\vec{R}$ in their coordinate system, and do both analysts' solutions represent the same physical force?

Analyst 2 reports $R_{x2} = -10 \text{ N}$ and $R_{y2} = -8 \text{ N}$ , and both represent the same physical force, because equilibrium equations are invariant under any coordinate transformation including reflections.
Analyst 2 reports $R_{x2} = +10 \text{ N}$ and $R_{y2} = +8 \text{ N}$ , and both represent the same physical force, because using a left-handed system negates all components of every force to maintain self-consistency of the equilibrium equations.
Analyst 2 reports $R_{x2} = +10 \text{ N}$ and $R_{y2} = -8 \text{ N}$ , and both represent the same physical force, because the x-axis is reflected so the x-component sign flips, while the y-axis is unchanged so the y-component sign is unchanged. (correct answer)
Analyst 2 reports $R_{x2} = -10 \text{ N}$ and $R_{y2} = +8 \text{ N}$ , and both represent the same physical force, because a left-handed coordinate system reverses the y-axis direction while keeping the x-axis direction the same.

Explanation: When switching between coordinate systems, the key question to ask is: which axes changed direction? Here, Analyst 2 flips only the x-axis (left instead of right), leaving the y-axis unchanged (still pointing up). This is purely a geometric transformation — the physical force

\vec{R}

doesn't change at all, only how its components are described changes. Since

\vec{R}

physically points left and downward, let's check each component independently. In Analyst 1's system,

R_{x1} = -10\text{ N}

means the force points in the negative x-direction (left). In Analyst 2's system, left is the positive x-direction, so the same leftward force now has

R_{x2} = +10\text{ N}

. The y-axis is identical in both systems, so

R_{y2} = -8\text{ N}

remains unchanged. That confirms C as correct — both analysts describe the exact same physical force. A is wrong because not all components are invariant; only components along unchanged axes stay the same. Claiming full invariance ignores the reflection. B is wrong because only the reflected axis changes sign — negating all components would require both axes to flip, which isn't what happened here. D is wrong because it reverses the logic entirely: it claims the y-axis flipped when in fact the y-axis is identical between the two systems. It's the x-axis that was reflected. A reliable strategy: when comparing coordinate systems, audit each axis independently — ask "did this axis reverse direction?" If yes, flip that component's sign. If no, keep it. Apply this axis-by-axis check and you'll never mix up which signs change.

Question 9

An engineer analyzes a block sliding on an inclined plane that makes angle $\theta = 30°$ with the horizontal. She establishes a tilted coordinate system where $x'$ points up the incline and $y'$ points perpendicular to the incline (away from the surface). The block's weight is $W = 100 \text{ N}$ acting vertically downward.

In the tilted coordinate system, which expression correctly gives the component of the weight along the $-x'$ direction (i.e., the component that drives the block down the incline)?

$W_{x'} = -W\cos\theta = -86.6 \text{ N}$ , because the weight vector projects onto the incline axis using the cosine of the inclination angle measured from the horizontal.
$W_{x'} = -W\sin\theta = -50 \text{ N}$ , because the incline axis makes angle $\theta$ with the vertical weight vector, so the projection uses sine of the inclination angle. (correct answer)
$W_{x'} = +W\sin\theta = +50 \text{ N}$ , because the component along $x'$ (up-incline positive) is positive when gravity has a component in the up-incline direction.
$W_{x'} = -W\tan\theta = -57.7 \text{ N}$ , because the ratio of the parallel to perpendicular components of weight on an incline is governed by the tangent of the inclination angle.

Explanation: When decomposing gravity in a tilted coordinate system, the key is tracking the angle between the weight vector and each axis — not defaulting to "cosine for components." Draw the geometry carefully. The weight vector

W

points straight down. The

x'

axis points up the incline, which is tilted

\theta = 30°

from horizontal. The angle between the vertical weight vector and the

x'

axis is

(90° - \theta)

, not

\theta

itself. However, the angle between the weight vector and the

y'

axis (perpendicular to incline) equals

\theta

. This means the projection onto

x'

uses

\sin\theta

W_{x'} = -W\sin\theta = -(100)(0.5) = -50 \text{ N}

. The negative sign reflects that gravity pulls in the

-x'

direction (down the incline). This confirms B. A is the classic trap — using

\cos\theta

for the along-incline component. Cosine gives the projection onto

y'

(perpendicular to the surface), not

x'

. Students reach for cosine by habit because it's used for "adjacent" components, but here the perpendicular axis is the one adjacent to the weight vector. C is the same magnitude as B but with the wrong sign. If

x'

is defined as pointing up the incline, gravity clearly drives the block down, making the component negative. D introduces tangent, which has no role in resolving a single force vector into components. Tangent relates the ratio of two components, not either component individually. Your memory anchor: on an incline, sine goes along, cosine goes perpendicular — the incline angle

\theta

is measured from horizontal, and the along-incline component always pairs with

\sin\theta