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.

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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:

3.5cm, 13.0cm, 19.0cm

9.8cm, 10.2cm, 10.4cm

1.1N, 3.0N, 1.7N

the end of the horizontal line drawn

The forces were easily found for bands B and C as they had graphs: I used the length as the x value and found the y value. For band A, I used the calibration graph for rubber band 1 (which was one of the bands) then multiplied it by two since the bands would be exerting equal force of 1.5N each (obtained from graph for length 10.2cm).

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

3.0N=? 1.1N+1.7N: 1.1N+1.7N=2.8N

2.8N/3.0N=93.3%

Moment arms for rubber band systems B and C

9.5cm, 6.0cm

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

4.4cm, 12cm, 6.8cm

9.5cm, 6cm

Torque produced by B, torque produced by C:

+.1045mN, -.102mN

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

.0025

.0025/.2065=1.2%

Since the torque in the counterclockwise(positive) direction is greater than the torque in the clockwise (negative) direction, the net torque is going to be positive. This is a percentage of the magnitude (absolute values) of both values.

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

1.6N, 10.3cm, 0

.35N, 9.1cm, 0.665

1.3N, 10.0cm, -18.46

.89N, 9.7cm, 13.795

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:

1.6N+.89N-.35N-1.3N=0.84N

WEll, if the system truly did have a net force of 0.84 it would be accelerating, not in equilibrium. The picture is inaccurate in the sense that my forces pulling the rod up are overestimated or the amounts to simulate gravity are underestimated

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

-(2.4*0) + 3.2*5 + -(1.6*15) + 2.0*23=38 cmN.

This could be an accurate torque if the bridge or rod could withstand that amount of torque without collapsing. Torgue can be occuring without the object moving.

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:

0 + 0.665 + -18.46 + 13.795=-4.001

1.6N+.89N-.35N-1.3N=0.84N, 4.14N

20.23%

4.001, 32.92, 12.2%

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:

-6.635

.05, 5.65

.89%

6.635, 50.165, 13.2%

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

If they varied at all it was by one or two degrees. The length of the rubber bands was changed more than the angle of the rubber bands.

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

95, 260, 285, 93

I believe these aren't accurate to more than 5 or 6 degrees since I did not have a protractor and the rubber band contraption was hard to estimate a straight line.

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

(2, -2), (2, 6.8)

(9,17), (8.5, 26)

(19, 0), (17.2, 7)

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

8.8cm, .14N

9.01cm, .65N

7.23cm, .05N

Sines and cosines of systems B, A and C:

1, 0

.017, .999

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

90, 86.8, 75.6

I took the arctan of the y value divided by x to find the angles in relation to the x axis.

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):

this is a vector straight down, so it is simply 0,.56cm,0,.675N

.3cm, 2.7cm, .075 N, .675N, .03N, .649N

.1cm, .2cm, .025N, .05, .012N, .048N

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

(0-.03+.012)=.018N, the sum should be zero, I am .018 off.

.649-.675-.048=-.074N, 0, .074

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

0, .56, 0

7.2cm, .649, +4.67

13.9, .048N, -.6672

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

4.0028. This sum can be whatever it wants to be since a torque is a stress placed on an object. I believe. The force on a system must be zero to be in equilibrium, but the torque does not have to be.

How long did it take you to complete this experiment?

2 and a half hours.

Optional additional comments and/or questions:

This seemed to be too long of a lab for those of us that had to do all three parts. It should be split up into two parts that can be submitted separate so it doens't all have to be done at once.

Good work. Thanks for the advice on splitting the labs. That, of course, leads to a complaint about 'too many labs', but I can get around that by splitting the form into Parts A, B and C. I'm inclined to do this on many of the longer labs.