torques

Your 'torques' report has been received. Scroll down through the document to see any comments I might have inserted, and my final comment at the end.

Your optional message or comment:

Hello Prof. Smith,

Here is my Torques lab, review.

Tanya

Positions of the three points of application, lengths of systems B, A and C (left to right), the forces in Newtons exerted by those systems, description of the reference point:

1.7, 12.2, 19.2

7.8, 8.8, 11

-3.05, 6.6, -3.9

the left end of the rob

I divided the length of each rubber band by 4.

Net force and net force as a percent of the sum of the magnitudes of all forces:

-0.35

2.58%

Moment arms for rubber band systems B and C

10.4

7.1

Lengths in cm of force vectors in 4 cm to 1 N scale drawing, distances from the fulcrum to points B and C.

12.2, 13.2, 15.6

10.4, 7.1

Torque produced by B, torque produced by C:

-31.72, 27.69

Net torque, net torque as percent of the sum of the magnitudes of the torques:

-4.03, 6.78%

I multiplied the net torque by 100 and divided the result by the obtained magnitude of the torques.

Forces, distances from equilibrium and torques exerted by A, B, C, D:

2.625, -2.5, -2.3, 2.25

1, 4.5, 12.6, 13.2

2.625, -11.25, -28.98, 29.7

The sum of the vertical forces on the rod, and your discussion of the extent to which your picture fails to accurately describe the forces:

0.075

Because I obtained this small number, I can say that the left two forces are almost in equilibrium with the other two forces, which are on the right.

Net torque for given picture; your discussion of whether this figure could be accurate for a stationary rod:

Because I obtained this number I can make a conclusion about the torques. Since the number is positive, the net torque is acting in counterclockwise position.

For first setup: Sum of torques for your setup; magnitude of resultant and sum of magnitudes of forces; magnitude of resultant as percent of sum of magnitudes of forces; magnitude of resultant torque, sum of magnitudes of torques, magnitude of resultant torque as percent of the sum of the magnitudes:

-7.98

0.075, -2.55

2.94%

For second setup: Sum of torques for your setup; magnitude of resultant and sum of magnitudes of forces; magnitude of resultant as percent of sum of magnitudes of forces; magnitude of resultant torque, sum of magnitudes of torques, magnitude of resultant torque as percent of the sum of the magnitudes:

3.05

0.075, -2.175

3.44%

2.12, 3.05, 69.5%

In the second setup, were the forces all parallel to one another?

No, in the second setup, the forces were not all parallel to one another.

About 5 degrees

Estimated angles of the four forces; short discussion of accuracy of estimates.

85 degrees

I used a protractor to estimate the angles. I think the uncertainty could be about 2-3 degrees.

x and y coordinates of both ends of each rubber band, in cm

-4.5, -14.5, -4.5, -4.2

3.6, 1.7, 12.4, 0

1.8, -3.9, 6.7, -13.5

Lengths and forces exerted systems B, A and C:.

10.4

8.96

10.776

B: I obtained the result by subtracting the upper y coordinate from the lower one.

A: I obtained the result by finding the square root of the sum of squares of x and y length, which I found by subtracting the lower y coordinate from the upper one and the lower x coordinate form the upper one.

C: I obtained the result by finding the square root of the sum of squares of x and y length of the big triangle and than I found by using the same way of finding the unknown side of the smaller triangle. The length I found by using the same way as I used in “A” triangle. Then I just subtracted the length of the side in the small triangle from the length of the side in the big triangle.

Sines and cosines of systems B, A and C:

1, 0

0.98, 0.19

0.91, 0.44

Magnitude, angle with horizontal and angle in the plane for each force:

90, 10.93, -65.22

I followed the instructions and in the last case I obtained 24.78. I subtracted this number from 90 and used this angle as a negative equivalent for 24.78 angle.

x and y components of sketch, x and y components of force from sketch components, x and y components from magnitude, sine and cosine (lines in order B, A, C):

0, -14.5, 0, 14.5, 0, 1

2.6, 12.8, 2.6, 12.8, .95, 0.203

6.7, -13.5, 6.7, 13.5, -0.9, 0.44

Sum of x components, ideal sum, how close are you to the ideal; then the same for y components.

9.3, it is the same

40.8, it is the same

Distance of the point of action from that of the leftmost force, component perpendicular to the rod, and torque for each force:

6.9, 6.6

-25.01, 25.01

Sum of torques, ideal sum, how close are you to the ideal.

-0.14, 0, 0.14