# Why is mg sin theta in the x-sum for an inclined plane?

## Block Sliding Down an Inclined Plane

Consider the block sliding down an inclined plane (with friction) illustrated below. When we apply Newton’s second law to the box, we set the sum of the x-components of the forces equal to max and the sum of the y-components equal to may.

One of the forces is weight, which equals mg (mass times gravitational acceleration).

The x-component of mg has a sine function, whereas the y-component of mg has a cosine function:

• (mg)x = mg sin θ
• (mg)y = mg cos θ

The question is this: Why does the sine function appear in the x-sum and the cosine function appear in the y-sum?

First I’ll clarify the question, and then I’ll explain the answer.

When you first learn vector addition, you develop the habit of using cosine for the x-component and sine for the y-component.

It’s backwards for weight on an incline. Look at the equations above: (mg)x has sin θ, whereas (mg)y has cos θ.

Now I’ll explain why it’s backwards. The reason has to do with the free-body diagram (FBD).

There are three forces acting on the box:

• The weight (mg) of the box acts straight down (toward the center of the earth). Gravity points down.
• Normal force (N) is perpendicular to the surface. For my picture above, normal force is diagonally up and to the left, as shown in the FBD below.
• Friction (f) opposes the velocity. For a box sliding down the incline, friction acts up the incline. It’s conventional in physics to setup a coordinate system with +x along the acceleration. When you do this, the y-component of acceleration equals zero: ay = 0.

(If instead you want x to be horizontal and y to be vertical, neither ax nor ay will be zero, and the math will be more complicated. Physics is already hard enough: Do yourself a favor and choose x to be down the incline in order to make the math simpler.)

Since the box slides down the incline, we choose +x to be down the incline. Since x and y must be perpendicular, this means that y will be along the normal force. You can see x and y labeled in the diagram above.

What we call θ in the inclined plane problem is the angle between the incline and the horizontal. This was labeled in the original diagram (way up above).

You can also see θ in the FBD above. Why is θ between weight (mg) and the negative y-axis? That has to do with geometry. Consider the diagram below. The diagram above combines the previous two diagrams together.

• The triangle on the left involves the inclined plane, the horizontal, and the weight (mg).
• The triangle on the right shows the weight vector resolved into x- and y-components.

The angle on the left is θ, labeled just as it was in the original diagram.

The three angles of any triangle add up to 180º. The left triangle is a right triangle: One angle equals 90º. Therefore, the remaining two angles must add up to 90º. That’s why the top angle in the left triangle equals (90º – θ).

Next, the x- and y-axes make a 90º angle, so the top angle in the left triangle plus the top angle in the right triangle must add up to 90º. That’s why the top angle in the right triangle equals θ.

If you’re struggling to see these angles, try it with numbers. Try it with θ = 30º. Then two of the left angles will be 30º and 90º (since it’s a right triangle). Since 30º + 90º = 120º, the remaining angle must be 60º (since the three angles of a triangle must add up to 180º). The three angles in the left triangle are then 30º, 60º, and 90º. The top angle of the right triangle must then be 30º (since 60º + 30º = 90º), since the x- and y-axes are perpendicular.

Once you finally understand the geometry, you can see why (mg)x has sin θ and (mg)y has cos θ. Look at the above diagram again:

• (mg)x, which is parallel to the x-axis (along the incline), is opposite to θ in the right triangle. Since sine equals opposite over hypotenuse, (mg)x = mg sin θ.
• (mg)y, which is parallel to the y-axis (along the normal), is adjacent to θ in the right triangle. Since cosine equals adjacent over hypotenuse, (mg)y = mg cos θ.

If you’re still not seeing this, don’t give up: Many students don’t fully grasp this after trying just one or two times. However, once you spend enough time studying the geometry and thinking through the explanations, there will probably come a point where it finally makes sense (otherwise, you should ask for help).

It’s important to understand this. Many physics problems involve an inclined plane, so this can come up in different contexts during a physics course (even on other topics later in the semester).

Applying Newton’s second law, we set the sum of the x-components of the forces equal to mass times the x-component of acceleration:

∑ Fx = max

(mg)x – f = max

Similarly, we set the sum of the y-components of the forces equal to mass times the y-component of acceleration:

∑ Fy = may

N – (mg)y = may

(Friction is in the x-sum because it’s along the incline. The minus sign before friction indicates that friction acts up the incline—opposing the velocity—while we chose +x to point down the incline. Normal force is in the y-sum because that’s the direction of our +y-axis.)

Recall that ay = 0 (that was our reason for choosing +x to be down the incline). The reason for this is that the box doesn’t accelerate towards or away from the incline (it stays on the incline). The box accelerates down the incline, along x (with no y-component). Setting ay = 0 in the above equation, we get:

N – (mg)y = 0

N = (mg)y

Recall that (mg)y = mg cos θ (that was the whole point of this article). Therefore, the normal force is:

N = mg cos θ

Friction force equals the coefficient of kinetic (since the block is sliding in this problem) friction (μ) times normal force.

f = μN

Plug in the expression for normal force:

f = μmg cos θ

Substitute this expression into the equation from the x-sum.

(mg)x – f = max

(mg)x – μmg cos θ = max

Recall that (mg)x = mg sin θ (that was the whole point of this article).

mg sin θ – μmg cos θ = max

Divide both sides of the equation by mass. The mass of the box cancels out in this problem: It has no effect on the acceleration.

g sin θ – μg cos θ = ax

Chris McMullen, Ph.D.

Author of:

• Essential Physics Study Guide Workbook (3 volume series, available at the calculus or trig level)
• 100 Physics Examples (3 volume series, available at the calculus or trig level)
• The Improve Your Math Fluency series of Math Workbooks

Click here to visit my Amazon author page.

# How to Write Sources of Error ## Sources of Error in Physics

• learn how to identify sources of error for a physics experiment
• describe common mistakes that students make in physics lab reports
• provide examples of how to describe sources of error

## What Are Sources of Error?

In everyday English, the words “error” and “mistake” may seem similar.

However, in physics, these two words have very different meanings:

• An error is something that affects results, which was not plausible to avoid (given the conditions of the experiment) or account for.
• A mistake is something that affects results, which should reasonably have been avoided.

We will see examples of each in the remainder of this article.

## Common Incorrect Answers

Part of learning how to write a good sources of error section includes learning what not to do.

Following are some common incorrect answers that students tend to include in their sources of error section.

• Human error. The problem with this phrase is that it’s way too vague. It may be okay if the nature of the error is human in origin (provided that it’s an inherent error and not a mistake), but it’s not okay to be express the error in vague terms. Advice: Don’t write the phrase “human error” anywhere on your lab report.
• Round-off error. This problem with this is that students almost never have enough precision in their answers for round-off error to be significant. Even a cheap calculator provides at least 8 figures, whereas most first-year physics experiments yield results where only 2-3 of those digits are significant figures. You can definitely find more significant sources of error to describe instead of round-off error.
• Incorrect technique. If you used equipment incorrectly or followed the procedures incorrectly, for example, these are mistakes—they are not sources of error. A source of error is something that you could not plausibly expect to avoid.
• Incorrect calculations. If there are mistakes in your calculations, these are not sources of error. Calculation mistakes are something that students are expected to avoid. Sure, some students make mistakes, but mistakes are not sources of error.
• Accidental problem. If an accident occurred during the experiment, which could plausibly be avoided by repeating the experiment, this is not a source of error. For example, if you perform the Atwood’s machine lab and the two masses collide, it’s a mistake to keep the data. Simply redo the experiment, taking care to release the masses so that they don’t collide.
• Lab partner. Don’t blame your lab partner—at least, not in your report—because it isn’t a valid source of error.

## Sources of Error: What to Look for

When you identify and describe a source of error, keep the following points in mind:

• It should sound like an inherent problem that you couldn’t plausibly avoid.
• It should be significant compared to other sources of error.
• It needs to actually affect the results. For example, when a car rolls down an inclined plane, its mass cancels out in the equation for acceleration (a = g sin θ), so it would be incorrect to cite an improperly calibrated scale as a source of error.
• You should describe the source of error as precisely as possible. Try not to sound vague.
• Unless otherwise stated by the lab manual or your instructor, you should describe the source of error in detail. In addition to identifying the source of the error, you can describe how it impacts the results, or you might suggest how the experiment might be improved (but only suggest improvement sparingly—not every time you describe a source of error), for example.
• The error should be consistent with your results. For example, if you measure gravitational acceleration in a free fall experiment to be larger than 9.81 m/s2, it would be inconsistent to cite air resistance as a source of error (because air resistance would cause the measured acceleration to be less than 9.81 m/s2, not larger).
• Try not to sound hypothetical. It’s better if it sounds like your source of error is based on observations that you made during lab. For example, saying that a scale might not be calibrated properly sounds hypothetical. If instead you say that you measured the mass to be 21.4 g on one scale, but 20.8 g on another scale, you’ve established that there is a problem with the scales (but note that this is a 3% error: if your percent error is much larger than 3%, there is a more significant source of error involved). On the other hand, if you get 20.43 g on one scale, but 20.45 g on another scale, this error is probably insignificant—not worth describing (since the percent error is below 0.1%).
• Sound scientific and objective. Avoid sounding dramatic, like “the experiment was a disaster” or “there were several sources of error.” (There might indeed be several sources of error, but usually only 1-2 are dominant and the others are relatively minor. But when you say “several sources of error,” it makes the experiment seem far worse than it probably was.)
• Demonstrate good analysis skills, applying logic and reasoning. This is what instructors and TA’s hope to read when they grade sources of error: an in-depth, well-reasoned analysis.
• Be sure to use the terminology properly. You can’t expect to earn as much credit if you get words like velocity and acceleration, or force and energy, confused in your writing.
• Follow instructions. You wouldn’t believe how many students lose points, for example, when a problem says to “describe 2 sources of error,” but a student lists 5 or only focuses on 1. Surely, if you’re in a physics class, you’re capable of counting. 🙂

## Examples of Sources of Error

Here are some concrete examples of how to identify and describe sources of error.

(1) A car rolls down an incline. You measure velocity and time to determine gravitational acceleration. Your result is 9.62 m/s2.

A possible source of error is air resistance. This is consistent with your results: Since the accepted value of gravitational acceleration is 9.8 m/s2 near earth’s surface, and since air resistance results in less acceleration, your result (9.62 m/s2) is consistent with this source of error. If you place the car on a horizontal surface and give it a gentle push, you will see it slow down and come to rest, which shows that there is indeed a significant resistive force acting on the car.

(2) You setup Atwood’s machine with 14 g and 6 g masses. Your experimental acceleration is 3.74 m/s2, while your theoretical acceleration is 3.92 m/s2.

A possible source of error is the rotational inertia of the pulley. The pulley has a natural tendency to rotate with constant angular velocity, which must be overcome in order to accelerate the pulley. It turns out (most textbooks do this calculation in a chapter on rotation) that if the mass of the pulley is significant compared to the sum of the two masses that its rotational inertia will have a significant impact on the acceleration. In this example, the sum of the masses is 20 g (since 14 + 6 = 20). If the pulley has a mass of a few grams, this could be significant.

One way to reduce the effect of the pulley is to use larger masses. If you use 35 g and 15 g instead, the sum of the masses is 50 g and the pulley’s rotational inertia has a smaller effect. (If you do this lab, add up the masses used. Only describe this as a possible source of error if the pulley’s estimated mass seems significant compared to the sum of the masses.)

(3) You fire a steel ball using a projectile launcher. You launch the ball five times horizontally. Then you launch the ball at a 30° angle, missing your predicted target by 6.3 cm.

A possible source of error is inconsistency in the spring mechanism. If that’s all you write, however, it will sound hypothetical. If instead, you noticed variation in your five horizontal launches, which had a standard deviation (something you can calculate) of 4.8 cm, you can establish that the variation in the spring’s launches is significant compared to the distance (of 6.3 cm) by which you missed the target.

## You Still Have to Think

I can’t list every possible source of error for every possible experiment—this article would go on forever.

You need to apply reasoning skills.

I have shown you what not to do.

I have shown you what to look for in a source of error.

And I have given you concrete examples for specific cases.

Be confident. You can do it. 🙂