2.6 Motion on an Inclined Plane

FIGURE 2.28a shows a problem closely related to free fall: that of motion down a straight, but frictionless, inclined plane, such as a skier going down a slope on frictionless snow. What is the object’s acceleration? Although we’re not yet prepared to give a rigorous derivation, we can deduce the acceleration with a plausibility argument.

The figure consists of two parts labeled ("a") and (b) that depict the movement of an object on an inclined plane and its motion diagram.

FIGURE 2.28 Full Alternative Text

FIGURE 2.28b shows the free-fall acceleration $\vec{a}_{free fall}$ the object would have if the incline suddenly vanished. The free-fall acceleration points straight down. This vector can be broken into two pieces: a vector ${\vec{a}}_{∥}$ that is parallel to the incline and a vector $\vec{a}_{⊥}$ that is perpendicular to the incline. The surface of the incline somehow “blocks” $\vec{a}_{⊥}$ , through a process we will examine in Chapter 6, but ${\vec{a}}_{∥}$ is unhindered. It is this piece of $\vec{a}_{free fall}$ , parallel to the incline, that accelerates the object.

By definition, the length, or magnitude, of $\vec{a}_{free fall}$ is $g$ . Vector ${\vec{a}}_{∥}$ is opposite angle $θ$ (Greek theta), so the length, or magnitude, of ${\vec{a}}_{∥}$ must be $g sin θ$ . Consequently, the one-dimensional acceleration along the incline is

a_{s} = \pm g sin θ

(2.26)

The correct sign depends on the direction in which the ramp is tilted. Examples will illustrate.

Equation 2.26 makes sense. Suppose the plane is perfectly horizontal. If you place an object on a horizontal surface, you expect it to stay at rest with no acceleration. Equation 2.26 gives $a_{s} = 0$ when $θ = 0 °$ , in agreement with our expectations. Now suppose you tilt the plane until it becomes vertical, at $θ = 90 °$ . Without friction, an object would simply fall, in free fall, parallel to the vertical surface. Equation 2.26 gives $a_{s} = - g = a_{free fall}$ when $θ = 90 °$ , again in agreement with our expectations. Equation 2.26 gives the correct result in these limiting cases.

EXAMPLE 2.14 Measuring acceleration

In the laboratory, a 2.00-m-long track has been inclined as shown in FIGURE 2.29. Your task is to measure the acceleration of a cart on the ramp and to compare your result with what you might have expected. You have available five “photogates” that measure the cart’s speed as it passes through. You place a gate every 30 cm from a line you mark near the top of the track as the starting line. One run generates the data shown in the table. The first entry isn’t a photogate, but it is a valid data point because you know the cart’s speed is zero at the point where you release it.

The figure illustrates a cart sliding down a ramp.

FIGURE 2.29 Full Alternative Text

Distance (cm)	Speed (m/s)
0	0.00
30	0.75
60	1.15
90	1.38
120	1.56
150	1.76

NOTE Physics is an experimental science. Our knowledge of the universe is grounded in observations and measurements. Consequently, some examples and homework problems throughout this book will be based on data. Data-based homework problems require the use of a spreadsheet, graphing software, or a graphing calculator in which you can “fit” data with a straight line.

MODEL Model the cart as a particle.

VISUALIZE FIGURE 2.30 shows the pictorial representation. The track and axis are tilted at angle $θ = {tan}^{- 1} (20.0 cm / 180 cm) = 6.34 °$ . This is motion on an inclined plane, so you might expect the cart’s acceleration to be $a_{x} = g sin θ = 1.08 m / s^{2}$ .

A pictorial representation of the movement of a cart on a track inclined toward the lower right at an angle theta from the horizontal.

FIGURE 2.30 Full Alternative Text

SOLVE In analyzing data, we want to use all the data. Further, we almost always want to use graphs when we have a series of measurements. We might start by graphing speed versus distance traveled. This is shown in FIGURE 2.31a, where we’ve converted distances to meters. As expected, speed increases with distance, but the graph isn’t linear and that makes it hard to analyze.

The figure consists of two parts labeled ("a") and (b) that depict a velocity graph and a velocity squared graph.

FIGURE 2.31 Full Alternative Text

Rather than proceeding by trial and error, let’s be guided by theory. If the cart has constant acceleration—which we don’t yet know and need to confirm—the third kinematic equation tells us that velocity and displacement should be related by

v_{x}^{2} = v_{0 x}^{2} + 2 a_{x} Δ x = 2 a_{x} x

The last step was based on starting from rest $(v_{0 x} = 0)$ at the origin $(Δ x = x - x_{0} = x)$ .

Rather than graphing $v_{x}$ versus $x$ , suppose we graph $v_{x}^{2}$ versus $x$ . If we let $y = v_{x}^{2}$ , the kinematic equation reads

y = 2 a_{x} x

This is in the form of a linear equation: $y = m x + b$ , where $m$ is the slope and $b$ is the y-intercept. In this case, $m = 2 a_{x}$ and $b = 0$ . So if the cart really does have constant acceleration, a graph of $v_{x}^{2}$ versus $x$ should be linear with a y-intercept of zero. This is a prediction that we can test.

Thus our analysis has three steps:

Graph $v_{x}^{2}$ versus $x$ . If the graph is a straight line with a y-intercept of zero (or very close to zero), then we can conclude that the cart has constant acceleration on the ramp. If not, the acceleration is not constant and we cannot use the kinematic equations for constant acceleration.
If the graph has the correct shape, we can determine its slope $m$ .
Because kinematics predicts $m = 2 a_{x}$ , the acceleration must be $a_{x} = m / 2$ .

FIGURE 2.31b is the graph of $v_{x}^{2}$ versus $x$ . It does turn out to be a straight line with a y-intercept of zero, and this is the evidence we need that the cart has a constant acceleration on the ramp. To proceed, we want to determine the slope by finding the straight line that is the “best fit” to the data. This is a statistical technique, justified in a statistics class, but one that is implemented in spreadsheets and graphing calculators. The solid line in Figure 2.31b is the best-fit line for this data, and its equation is shown. We see that the slope is $m = 2.06 m / s^{2}$ . Slopes have units, and the units come not from the fitting procedure but by looking at the axes of the graph. Here the vertical axis is velocity squared, with units of $m^{2} / s^{2}$ , while the horizontal axis is position, measured in m. Thus the slope, rise over run, has units of $m / s^{2}$ .

Finally, we can determine that the cart’s acceleration was

a_{x} = \frac{m}{2} = 1.03 m / s^{2}

This is about 5% less than the $1.08 m / s^{2}$ we expected. Two possibilities come to mind. Perhaps the distances used to find the tilt angle weren’t measured accurately. Or, more likely, the cart rolls with a small bit of friction. The predicted acceleration $a_{x} = g sin θ$ is for a frictionless inclined plane; any friction would decrease the acceleration.

REVIEW The acceleration is just slightly less than predicted for a frictionless incline, so the result is reasonable.

2.6 Motion on an Inclined Plane

FIGURE 2.28 Acceleration on an inclined plane.

EXAMPLE 2.14 Measuring acceleration

FIGURE 2.29 The experimental setup.

FIGURE 2.30 The pictorial representation of the cart on the track.

FIGURE 2.31 Graphs of velocity and of velocity squared. The equation of the best-fit line is given as y = because that is how it would be shown in a spreadsheet.