Using the Head to tad Method, calculate the sum of the vectors generated when yo

Question

Using the Head to tad Method, calculate the sum of the vectors generated when you hang 200 g at 30 degree, 100 g at 120 degree and 150 g at 300 degree on the Force Table. Give your answer forces in Newtons and angles in degrees. Results: F = Angle = Calculate the mass that you have to hang on a fourth pulley on the "Force Table" to equilibrate the resultant force. Give your answer mass in kilograms and the angle in degrees. Results: m = Angle = Friction. In the figure you find the set up used in our lab to determine the coefficients of friction (second method). Put the name in each one of the forces in the sketch and the formulas to calculate them. (Example: Weight = mg) Using the set up given in the figure, we get that the mass start to slide when the angle is 21 degree. Which is the static friction coefficient. Is it possible that the kinetic friction coefficient for those surfaces to be 0.3? Why (Explain)? In the experiment for Newton's Second Law, a team of students, using a total moving mass of 1.222 kg obtained the following data: Complete the table.

Explanation / Answer

I will answer question 1 and 4.

Question 1.

We need to calculate the x and y components from the distance given. We have the data of A and B to C., so I will calculate dx and dy from A to B and from B to C. Then, we calculate the total distance with these values, and then, the time with the velocity of the plane:

From A to B:
dx = dcos30 = 1500 cos30 = 1299 km
dy = dsin30 = 1500 sin30 = 750 km

From B to C:
dx = 2000cos120 = -1000 km
dy = 2000sin120 = 1732 km

Sum dx and dy:
dxt = 1299-1000 = 299 km
dyt = 750+1732 = 2482 km

The net distance would be:
D = (dxt2 + dy2)1/2
D = (2992 + 24822)1/2
D = 2500 km

The time would be:
t = 2500 / 800 = 3.125 h

Question 4.
F1 = 0.05 * 9.8 = 0.49 N
F2 = 0.07 * 9.8 = 0.686 N
F3 = 0.11 * 9.8 = 1.078 N
F4 = 0.13 * 9.8 = 1.274 N

average of acceleration:
a1 = 0.4+0.405 / 2 = 0.4025 m/s2
a2 = 0.54 m/s2
a3 = 0.715 m/s2
a4 = 0.8655 m/s2

calculated acceleration:
a1 = 0.49 / 1.222 = 0.401 m/s2
a2 = 0.686 / 1.222 = 0.561 m/s2
a3 = 1.078 / 1.222 = 0.882 m/s2
a4 = 1.274 / 1.222 = 1.041 m/s2

%error:
%a1 = (0.4025-0.401/0.4025) * 100 = 0.37%
%a2 = 3.74%
%a3 = 18.93%
%a4 = 16.86%

Hope this helps

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