{[ promptMessage ]}

Bookmark it

{[ promptMessage ]}

Report for lab M3

# Report for lab M3 - Lauren Taranow LMT827 Monday 2-4 Post...

This preview shows pages 1–2. Sign up to view the full content.

LAB REPORT FOR M3 EXPERIMENT: EQUILIBRIUM OF FORCES VI-1: The main objective of this part of the lab was to establish the degree of sensitivity that was associated with the apparatus we used to measure the forces. Essentially, we were finding the uncertainty (σ). We found that in order to create a lack of equilibrium it took the addition of a minimum of 5.0 grams added to one of the forces or a minimum adjustment of 2.0 degrees to any one of the angles of the forces. These values are recorded below. VI-2: In this section of the lab we set up the force table; first, with only 3 forces, 2 forces at 120° from each other and varied the 3 rd in an attempt to recreate equilibrium using different masses to adjust the other forces. Then, with 4 forces, 3 forces were 90° from each other and then the 4 th was varied while we added masses to create equilibrium between the forces. The table of these masses, forces, and angles are recorded in the table below. To calculate force I used the equation F 1 =m 1 g where the force is equal to mass x gravity and the units for mass are kg so the units for force are Newtons or kg/m/sec 2 . Using 3 forces 1 2 3 m 1 (kg) .150 .100 .125 Ø 1 (angle) Force 1 (kg/m/sec 2 ) 1.47N 0.98N 1.225N m 2 (kg) .100 .145 .100 Ø 2 (angle) 120° 120° 120° Force 2 (kg/m/sec 2 ) 0.98N 1.421N 0.98N m 3 (kg) .140 .125 .115 Ø 3 (angle) 220° 255° 230° Force 3 (kg/m/sec 2 ) 1.372 N 1.225N 1.127N VI-3: Each of the forces calculated in VI-2 represent a vector, for example for Force 1 using only 3 forces the vector is F 1 . We can graphically represent the resultant vector of F 1 + F 1 as an arrow drawn from the end of vector F 2 to the beginning of vector F 1 . This represents the “true” magnitude and direction of vector F 3 . The graphs of Trial 1 for both 3 forces and 4 forces are attached to the end of the lab report. For both graphs 1cm=0.1N VI-4: For the graph of the 3 forces, the resultant vector appears to be about 0.9 cm which would equate to a force of 0.09N, according to my calculations in VI-5 the resultant vector should be 0.078N. Ideally the resultant

This preview has intentionally blurred sections. Sign up to view the full version.

View Full Document
This is the end of the preview. Sign up to access the rest of the document.

{[ snackBarMessage ]}

### Page1 / 5

Report for lab M3 - Lauren Taranow LMT827 Monday 2-4 Post...

This preview shows document pages 1 - 2. Sign up to view the full document.

View Full Document
Ask a homework question - tutors are online