Lab: Force Probe
- Entire class attempted an abstract.
This lab's objective is to verify that a free body diagram describes a static system consisting of a weight suspended from two strings. Force probes (spring scales) were used to measure tension. This lab generated three equations. These equations involved tensions, mass, and angles. Each equation can be verified by comparing the LHS (left-hand-side) to the RHS (right-hand-side).
The first equation is that the tensions multiplied by the cosines of the angles are equal. The angle sine (trying to describe equation in word..."degree of the sin..." Um, so we are just describing equations, or what
Guy writes equantions
- Calibrate the force probe with a 5 newton weight.
- The yellow string interferes slightly, so we used a binder clip to fix the issue. Therefore, we null the force probe by .5 newtons on the right side. And also rotated the right force probe.
off by a factor of
- 2.45/1.829=1.34 or 1.829/2.45=.7465= 25%error
in order to calibrate the force probes we rotated the right force probe approximately 2 inches so that the angle of the strings attached matched. The calibration was done using a 250g weight where each force probed equaled approximately 1 newton. For greater accuracy the strings were bound together with a binder clip. Each force probe carries the capacity for 5 newtons. Continuing the experiment we will now test the force probes using a force probe meter and a Lab Quest device to record then analyze our data.
Free body diagram: Beware the physical triangles!
Be able to derive:
- .. and ..
Practice FBD13:06, 9 October 2017 (UTC)
- Practice the 3-tension fbd until you can do it in your notes. The goal is to be able to derive the fundamental equations. The students wrote an abstract/summary:
- We made a visual (Free Body Diagram) to explain what T2 is and the components were used to solve for T2. Am more advanced version is in my notes to study for the tests.
- Class rewrote: as follows
- We made a FBD (Free Body Diagram) for a mass suspended by two strings, one of which was horizontal. The horizontal string had tension T1, while the other string with tension T2 was oriented an angle θ above the horizontal. To solve this for T2 we used the fact that the vertical (y) component was equal to mg, since the other tension exerted no force in that direction.
Our goal was to verify each side of an equation through math calculations. To do so, we had to calculate the percent error (30%) or off by a factor of 1.34. (Experimental-True Value)/(True Value). In another case we got 180=185.
- F=mg= ≈ 5N, since g≈10m/2