Cart+on+Incline-+Jonathan+Hannings,+Daniel+Hayes,+Sanjana+Sreenath,+&+Nathan+Campbell

Cart on Incline
 * Title of Lab: **

Jonathan Hannings, Daniel Hayes, Sanjana Sreenath, and Nathan Campbell
 * Researchers: **


 * Research Question: **

How does the angle of inclination affect the force (parallel to the incline) that is required to keep a cart in equilibrium?

Free body diagram picture source: [] According to Newton's second law: F=ma Since the cart is not accelerating, sum of the net forces =0 Therefore, the upward forces= downward forces N=Normal force leftward forces=rightward forces Tension= where mg is constant
 * Research: **


 * Hypothesis: **

The sine of the angle of inclination and the magnitude of the tension are directly related. (Tension = mgsin θ)

Materials: cart, incline, string, spring scale and a protractor
 * Procedure: **
 * A board is set up at an angle of 10 degrees (using a protractor to measure the angle)
 * A cart is placed on the board, with minimal friction between the two surfaces
 * A string is tied to end of the cart, and a spring scale is attached to the string
 * The car is released down the incline, until it reaches a point of equilibrium (acceleration=0)
 * The force indicated on the attached spring scale is recorded (in Newtons)
 * Repeat the process at 10 degree intervals, starting at 10 degrees and ending at 80 degrees

Inclination (degrees) || Tension (N) || mass of cart= 497g
 * Data: **
 * Angle of
 * 0 || 0 ||
 * 10 || 0.8 ||
 * 20 || 1.5 ||
 * 30 || 2.4 ||
 * 40 || 3.1 ||
 * 50 || 3.6 ||
 * 60 || 4.2 ||
 * 70 || 4.8 ||
 * 80 || 4.9 ||


 * Data Analysis: **



The above graph represents Tension vs. the sine of the angle of inclination. The vertical axis is the same and the horizontal axis represents sine of theta. The line of best fit demonstrates that tension is increasing as the sine of the theta increases. Slope of the best fit line = 5.0N The slope gives the weight of the cart (mass of the cart times acceleration) = 5.0 N Our theoretical calculation predicted a weight of 4.9 N (.497 x 9.8) % error = ((5.0-4.9)/4.9) x 100 = 2.0% y intercept is 0 N indicating that when the force is 0 N, the displacement is also 0 Sources of error: lack of multiple trials Instrument uncertainty Readability

Since the tension acting parallel to the incline and the angle of inclination are directly proportional,the experiment supports our hypothesis that tension increases as sine of theta (angle of inclination) increases. The percent error can be decreased by conducting multiple trials and calibrating the spring scale before each taking each reading.
 * Conclusion: **