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
Copyright © 2017.
Chris McMullen, Ph.D.
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