Cart+on+Incline+-+Lekha,+Danielle,+and+Raul

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

Research:
Website I used to find the info: []
 * Gravity is divided into two components on an inclined plane.
 * One component is normal (perpendicular) to the plane surface: FN = mg·cosθ
 * The other component is parallel to the plane surface: F|| = mg·sinθ
 * To prevent the object from crashing through the surface, the surface provides a normal force that is equal and opposite to the normal component of gravity.
 * Friction acts parallel to the plane surface and opposite to the direction of motion.
 * In a non-moving object on an inclined plane: normal component of gravity = normal force;parallel component of gravity = static friction.

Hypothesis:
A force exerted on a cart that is parallel to the incline that the cart is on is directly related to the sine of the angle of inclination, meaning as the sine of the angle of inclination increases, so does the amount of force needed to keep it in equilibrium (tension). FN = mg·cosθ which means Tension= mg sin ( θ)

**Materials:**
spring scale, triple beam balance, 831.7 g four-wheeled cart, protractor with string and washer, calculator, long wood board

Procedure:
1. Calibrate the spring scale (force gauge). 2. Attach spring scale with a string near the bottom of the cart. 3 .Use the angle measuring tool to find the degree of incline. Record the finding. 4. After finding the angle, find the force reading on the spring scale. Record the finding. 5. Raise the end of the incline in small increments to increase the __ degree __ of incline. 6.Repeat steps 3-5 after every increase. 7. After acquiring a sufficient amount of data, find the mass of the cart without the string and then with the string.

Data:
__degrees__ ) || Force (Newtons) || Mass without string- 831.7 grams Mass with string- 832.7 grams
 * Degree of Incline θ (
 * 0 || 0 ||
 * 3 || 0.6 ||
 * 5 || 0.8 ||
 * 9 || 1.2 ||
 * 13 || 1.7 ||
 * 18 || 2.4 ||
 * 26 || 3.5 ||
 * 30 || 3.8 ||
 * 35 || 4.7 ||
 * 40 || 5.0 ||
 * 45 || 5.5 ||
 * 47 || 5.8 ||
 * 50 || 6.0 ||
 * 53 || 6.3 ||
 * 55 || 6.5 ||
 * 59 || 6.8 ||
 * 65 || 7.4 ||
 * 70 || 7.8 ||
 * 75 || 7.9 ||
 * 80 || 8.0 ||
 * 85 || 8.2 ||

Data Analysis:
Force vs Degree of Inclination ( θ) (with no manipulated variables)



Force vs Sin[Degree of Inclination ( θ)] Line of Best Fit
 * slope of the graph- 8.2N
 * The y-intercept in negative because we did not use the point (0,0) on our graph.
 * Because the weight (mg) of the cart is the same in every test, we said that T=mx/mg=sin( θ), so force (tension) is directly related to sin of theta.
 * Sources of error would not re-calibrating the spring scale for every new angle and maybe be an instrumental error through the readings of the protractor as well.
 * We could have improved by re-calibrating the scale every run and including (0,0) as a point on the graph through more precise readings of the instruments.

Conclusion:
To figure out how the angle of inclination affected the force parallel to the incline to keep the cart at equilibrium, we found the force(N) at various angles(θ). Like our hypothesis stated, our experiment showed a direct relationship between the sine of the angle of incline and the force parallel to the incline. The slope of the sin(theta) vs force graph is 8.2N and the mg used for our experiment was 8.17N (.8327kg * 9.81 m/s2) therefore the percent error was 0.5%. This shows that our experiment is fairly accurate in proving our hypothesis.