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PHYS 1401 Austin Community College General Physics I Torques Lab Report

PHYS 1401

Austin Community College

PHYS

### Question Description

Hi please I need you to watch the video to have the correct measurement

The video calls out all of the measurements as he goes through it and provides all of the values at the end.

PHYS 1401 Torque balance lab.pdf

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Name______________________ Date_____________ PHYS 1401 General Physics I Torques Equipment Fulcrum and attachment Meter Stick Digital Scale Mass Set 3 String loops Unknown Mass Theory For a body to be in static equilibrium, two conditions have to be met:  F  0 and   0 where F is force  is torque. (The torque is the force times the lever arm, r) The first condition,  F  0 , is concerned with translational equilibrium and ensures that the object is not moving linearly or is moving at a uniform linear velocity. The second condition,   0 , is concerned with rotational equilibrium and ensures that the object is not rotating or rotating at a uniform angular velocity. Procedure (For each setup, draw a scale diagram of the setup) 1. Measure the mass of the meter stick without attachment and then place the meter stick on a support stand. Adjust the meter stick through the attachment until the stick is balanced on the stand. The balancing point is called the center of mass of the meter stick. Record this point. 2. Case 1. With the meter stick on the support stand, suspend a gram mass m1  100 g at the 15-cm mark on the meter stick. Then adjust the lever arm for a gram mass m2  200 g at the other end of the meter stick. 3. Record the mass and position x as read on the meter stick and then record the lever arms. Compute the torques and find a percent difference between the clockwise (cw) torques and counterclockwise (cc) torques. 4. Case 2. Place m1  100 g at the 20-cm mark and m2  200 g is at the 60-cm mark on the meter stick. Experimentally determine the position for m 3  50 g so that the system is in equilibrium. Follow a procedure similar to steps 2 and 3. Compute the percent difference of the clockwise and counterclockwise torques. 5. Case 3. Place an unknown mass at the 10 cm mark of the meter stick. Suspend from the other side a counter mass m2  300 g and adjust its position until the system is in static equilibrium. Using,   0 , calculate the unknown mass m1. Remove the unknown mass and determine its mass on the laboratory balance. This is the accepted mass. Calculate % error. 6. Case 4. Suspend a mass m1  100 g at or near the zero end of the meter stick. Move the meter stick in the support clamp until the meter stick is in equilibrium. Record this new equilibrium position as xo’. Using the total mass of the meter stick, calculate the clockwise and counterclockwise torques, and then calculate a percent difference. In this calculation, you will include the mass of the meter stick as if it were concentrated at its center of mass, xo (around the 50 cm mark) calculating the lever arm to the new pivot point. Data Table Mass of meter stick, mmeterstick ______________________ Balancing point (center of mass) of meter stick, x0 = ______________ * Attach a sheet to the laboratory report showing diagram and calculations for each use. Values Moment (lever) arms Results *(see note above) Case 1 m1 = 100 g x1 = 15 cm r1 = ___________ cc = ___________ m2 = 200 g x2 =_________ r2 = ___________ cw = ___________ Perc.Diff. __________ Case 2 m1 = 100 g x1 = 20 cm r1 = ___________ cc = ___________ m2 = 200 g x2 = 60 cm r2 = ___________ cw = ___________ m3 = 50 g x3 = _______ r3 = ___________ Perc.Diff. __________ x1 = 10 cm r1 = ___________ m1 ____________ r2 = ___________ (measured on scale/ accepted) Case 3 m1 =? m2 = 300 g x2 =_________ m1 ____________ (calculated, experimental) Perc. error _________ Case 4 r1 = ___________ cc = ___________ mmeterstick = _______ xo = _______ r2 = ___________ cw = ___________ m1 = 100 g x1 = ________ xo’ = _______ % Error  Perc.Diff. __________ exp erimental  accepted *100 accepted % Difference   cc   cw *100  cc   cw      2   ...
Purchase answer to see full attachment Hey there! Here's the lab sheet. I uploaded a Word version and a PDF version in case the equations don't get encoded in the Word file. Let me know if you have any questions! Thank you so much! 😀

PHYS 1401 General Physics I
Torques
Data Table
Mass of meter stick, 𝑚𝑚𝑒𝑡𝑒𝑟 𝑠𝑡𝑖𝑐𝑘 = 87 g
Balancing point (center of mass) of meter stick, 𝑥0 = 50.5 cm
Case 1
m1 = 100 g
m2 = 200 g

Values
x1 = 15 cm
x2 = 68.3 cm

Moment (lever) arms
r1 = 35.5 cm
r2 = 17.8 cm

Results
τcc = 0.3479 Nm
τcw = 0.3489 Nm
Perc. Diff. = 0.28 %

Values
x1 = 20 cm
x2 = 60 cm
x3 = 73.5 cm

Moment (lever) arms
r1 = 30.5 cm
r2 = 9.5 cm
r3 = 23.0 cm

Results
τcc = 0.2989 Nm
τcw = 0.1862 Nm + 0.1127 Nm
Perc. Diff. = 0.00 %

Values

Moment (lever) arms

Results
m1 = 226 g
(measured on scale/accepted)
m1 = 225.9 g
(calculated, experimental)
Perc. Error = 0.0442 %

Case 2
m1 = 100 g
m2 = 200 g
m3 = 50 g
Case 3
m1 = ?

x1 = 10 cm

r1 = 40.5

m2 = 300 g

x2 = 81.0 cm

r2 = 30.5

Case 4
Values
m1 = 100 g
x1 = 0.3 cm
mmeterstick = 87 g
xo = 50.5 cm
xo’ = 23.7 cm
Diagrams and Calculations

Moment (lever) arms
r1 = 23.4 cm
r2 = 26.8 cm

Results
τcc = 0.2293 Nm
τcw = 0.2285 Nm
Perc. Diff. = 0.349 %

Case 1
x2 = 68.3 cm

x1 = 15 cm

r1 = 35.5 cm

r2 = 17.8
cm
m2 = 200 g

m1 = 100 g
F1 = m1 x g

F2 = m2 x g
xo = 50.5 cm

*All measurements are converted to their SI Unit counterparts. The equation for torque is given by 𝜏 =
𝑟𝐹 sin 𝜃. Since the moment (lever) arms and the forces are perpendicular, sin θ = sin 90° = 1. Therefore,
the sin θ expression is omitted in the calculations.

𝜏𝑐𝑐 = 𝑟1 𝑥 𝐹1 = 35.5 𝑐𝑚 (
𝜏𝑐𝑐

1𝑚
1 𝑘𝑔
𝑚
) 𝑥 100 𝑔 (
) (9.8 2 )
100 𝑐𝑚
1000 𝑔
𝑠
= 0.3479 𝑁𝑚
1𝑚
1 𝑘𝑔
𝑚
) 𝑥 200 𝑔 (
) (9.8 2 )
100 𝑐𝑚
1000 𝑔
𝑠
= 0.3489 𝑁𝑚

𝜏𝑐𝑤 = 𝑟2 𝑥 𝐹2 = 17.8 𝑐𝑚 ... ejt824 (701)
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