Forces and Newton’s Second Law Experiment Physics 131-Section L Thursdays 9:00 AM ILC S110 Abstract This experiment looked at employing Newton’s Second Law which states an object's acceleration is produced by a net force proportional to the magnitude of the object in the same direction and inversely proportional to the mass. We proved that balancing forces will cancel out each other’s vectors and lead to a balance and an acceleration will only occur as one force is at a greater magnitude than the others. Questions & Answers 1. Use your analysis from the force table. Show mathematically that the sum of the vectors is zero. (2 points) Table 1 shows our vector values obtained during the experiment Adding each of the vectors entails adding their X components since the string is pulling on the ring in a horizontal direction, the x component shows the directionality of each vector. By looking at the table the X component of A, B and C are -.04,-.01, and +.5 respectively (as seen in Table 1) if you added these three components it equals 0. 2. Calculate the tension T in the string in both cases. i.e. when m=20g and m = 40g. T=M*a Here M is the mass of the PAScar and a is the acceleration when m =20g and 40g. (2 points)Table 2 shows the values obtained at the different masses M=170g a=(m/m+M)*g Table 2 shows the value we obtained for each m=20g T=170*.712=121.04 N m=40 g T=170*1.318=224.06N 3. For both the 1st and 2nd labs, we had objects that were “falling” (a golf ball for lab 1 and the hanging mass for lab 2). Was the acceleration of both objects constant? An object in free fall has constant acceleration. Were the two objects (golf ball and hanging mass) in free fall? Explain why or why not. (2 points) In both experiments, the acceleration of both objects was constant. Gravity is the only force acting on both making them go towards the earth, bounce back up and so on until they reach a stop. In both cases, it was the only force which fits both as being in the definition of free fall. 4. For the force table, is the ring in the middle accelerating? What are the forces acting on the ring, and what is the net force acting on the ring? Is gravity acting on the ring? Why isn't the ring in free fall? If we dropped the force table, how would the horizontal and vertical components of the acceleration of the ring change? (2 points) The ring in the middle of the force table is no accelerating because F net= 0 by definition and the ring is going through no changes in direction or speed. The gravity and the three vectors, A, B and C are acting on the ring and the net force acting on the ring is 0 because the three vectors are cancelling each other out. While gravity is acting on the ring it is not in free fall because it is connected to three other pulling forces and for the ring to be considered in free fall, only the one gravity force could be acting on it which isn’t the case. If we dropped the force table, the horizontal component would go to 0 because the whole apparatus would be in free fall, meaning the would be no more pulling going on in the ring because everything is falling equally in free fall. The acceleration of the horizontal component would increase because it would be falling down quickly by the force of gravity.5. Consider the diagram below which is a model of the force table used in lab. The magnitude of vector B and C are 6 and 8, respectively. Find the magnitude and direction (relative to the horizontal) of vector A such that the ring in the middle does not move. (1 point) Magnitude of A=4.1, direction=316.97 degrees. First you must find the direction of B and C by looking at the picture with the clues given, you see that B is 30 degrees away from the horizontal, so the direction of B is 30 degress, you see that C is right on the 180 degrees line so thats it’s direction. Next you can use this information to find the x and y components of each. B. Bx=6cos(30)=5.2 By=6sin(30)=3 C. Cx=8cos(180)=-8 Cy=8sin(180)=0 We know that the sum of Ax+Bx+Cx=0 so using the numbers we obtained we can solve for Ax to equal -3. Same goes for Ay+By+Cy=0 so we can solve for Ay to equal 2.8. Using these components we can use the pythagorean theorem to solve for the magnitude of A, which equals 4.1 and then using these lengths we can take the formula of Tan=2.8/-3=-43.03 degrees or 316.97 degrees. 6. Identify all the forces acting on the PAScar and hanging mass by drawing free-body diagrams for both and clearly labeling what each force represents. (Assume the friction between the air track and PAScar is negligible) (2 points)7. What is the expression for the net force on the PAScar? Answer the same question for the hanging mass. Finally, what is the net force on the system as a whole (PAScar plus the hanging mass), and which of the three net forces that you found equals the tension in the string connecting the PAScar to the hanging mass? (2 points) For the PAScar, since gravity and the normal forces cancel each other out you are left with the tension of the rope pulling the car as being the only source of force, so Fnet= Fx For the hanging mass, only vertical components are acting on it, so Fnet=Fy=m*g in this case. As a whole the system would equal sqrt(Fx2+(m*g)2, which correlates to the tension in the string connecting the PAScar to the hanging mass. 8. List your data from the PAScar/hanging mass experiment and state your experimental value for the gravitational acceleration. Calculate the percent error for your measured value of g. Finally, state three sources of error for the PAScar/hanging mass experiment. List errors that are unique to this experiment, i.e. human, calculation, system, computer, sensor, or any other nondescript/general error will not receive credit. (2 points) Our experimental value for the gravitational acceleration is 6.79 m/s^2, since gravity is typically found to be 9.81 we can use that to find our percent error using the formula ((Experimental value-Theoretical value)/Theoretical value)*100 or ((6.79-9.81)/9.81)*100= about 30.8% error. This error could have been caused by the assumption that there would be no friction between the car and the track so we did no calculations for friction which could have slowed the free fall of the hanging mass and travels of the car. Our car was reported to be 170 g but each car could vary slightly