Calculate the theoretical acceleration of the lightest spherical object using Newton's second law. Then compare this to the averages when the distance relative to the ramp was 10, 20, and 30 cm using percent error calculations. (Include a free body diagram of the set-up in your solution) SPEHRICAL OBECT 1 MASS (kg)0.0022SPEHRICAL OBECT 2 MASS (kg)0.0031SPEHRICAL OBECT 3 MASS (kg)0.0053SET-UP 1 AND 2 (LIGHTEST MASS AT DIFFERENT LENGTHS WITH CATCHERS 1 AND 2)Time from top to end of ramp andcorresponding accelerationsTrials for 10 cm Catcher 1 Knock-Trials for 20 cmCatcher 1 Knock-Trials for 30 cmCatcher 1 Knock-ConfigurationsTime 1Time 2Time 3Ave.ramp distanceback Distanceramp distanceback Distanceramp distanceback Distance10 cm ramp distance0.350.480.560.46122Calculated Acceleration1.630.870.641.051222320 cm ramp distanceCalculated Acceleration30 cm ramp distanceCalculated Acceleration0.730.560.640.643131330.751.280.981Ave.1Ave.1.67Ave.2.670.810.560.680.680.911.911.31.4Catcher 2 Knock-Catcher 2 Knock-Catcher 2 Knock-Trials for 10 cm back DistanceTrials for 20 cmback DistanceTrials for 30 cmback Distanceramp distance(cm)ramp distance(cm)ramp distance(cm)rce calculation 10 cm ramp distar 0.00231rce calculation 20 cm ramp distar 0.0031rce calculation 30 cm ramp distar 0.00742110.713.120.623.3330.432Ave.Ave.0.57Ave.2.8SET-UP 3 & 4 (2 OTHER SPHERICAL OBJECTS WITH CATCHER 1 AT 10 CM)SHPERICALlOBJECT NO.SHPERICAL2ОBJЕCT NO.3Catcher 1 Knock-Catcher 1 Knock-Trials for 10 cm back DistanceTrials for 10 cmback Distanceramp distance(cm)ramp distance(cm)121.32.0.621.53231Ave.1.53Ave.1.27

As observed, with an increase in the distance of release relative to the ramp, there was an increase…

Calculate the theoretical acceleration of the lightest spherical object using Newton's second law. Then compare this to the averages when the distance relative to the ramp was 10, 20, and 30 cm using percent error calculations. (Include a free body diagram of the set-up in your solution)

Calculate the theoretical acceleration of the lightest spherical object using Newton's second law. Then compare this to the averages when the distance relative to the ramp was 10, 20, and 30 cm using percent error calculations. (Include a free body diagram of the set-up in your solution)

International Edition---engineering Mechanics: Statics, 4th Edition

4th Edition

ISBN:9781305501607

Author:Andrew Pytel And Jaan Kiusalaas

Publisher:Andrew Pytel And Jaan Kiusalaas

Chapter1: Introduction To Statics

Section: Chapter Questions

Problem 1.12P: A differential equation encountered in the vibration of beams is d4ydx4=2D where x = distance...

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SPEHRICAL OBECT 1 MASS (kg)
0.0022
SPEHRICAL OBECT 2 MASS (kg)
0.0031
SPEHRICAL OBECT 3 MASS (kg)
0.0053
SET-UP 1 AND 2 (LIGHTEST MASS AT DIFFERENT LENGTHS WITH CATCHERS 1 AND 2)
Time from top to end of ramp and
corresponding accelerations
Trials for 10 cm Catcher 1 Knock-
Trials for 20 cm
Catcher 1 Knock-
Trials for 30 cm
Catcher 1 Knock-
Configurations
Time 1
Time 2
Time 3
Ave.
ramp distance
back Distance
ramp distance
back Distance
ramp distance
back Distance
10 cm ramp distance
0.35
0.48
0.56
0.46
1
2
2
Calculated Acceleration
1.63
0.87
0.64
1.05
1
2
2
2
3
20 cm ramp distance
Calculated Acceleration
30 cm ramp distance
Calculated Acceleration
0.73
0.56
0.64
0.64
3
1
3
1
3
3
0.75
1.28
0.98
1
Ave.
1
Ave.
1.67
Ave.
2.67
0.81
0.56
0.68
0.68
0.91
1.91
1.3
1.4
Catcher 2 Knock-
Catcher 2 Knock-
Catcher 2 Knock-
Trials for 10 cm back Distance
Trials for 20 cm
back Distance
Trials for 30 cm
back Distance
ramp distance
(cm)
ramp distance
(cm)
ramp distance
(cm)
rce calculation 10 cm ramp distar 0.00231
rce calculation 20 cm ramp distar 0.0031
rce calculation 30 cm ramp distar 0.00742
1
1
0.7
1
3.1
2
0.6
2
3.3
3
3
0.4
3
2
Ave.
Ave.
0.57
Ave.
2.8
SET-UP 3 & 4 (2 OTHER SPHERICAL OBJECTS WITH CATCHER 1 AT 10 CM)
SHPERICAL
lOBJECT NO.
SHPERICAL
2
ОBJЕCT NO.
3
Catcher 1 Knock-
Catcher 1 Knock-
Trials for 10 cm back Distance
Trials for 10 cm
back Distance
ramp distance
(cm)
ramp distance
(cm)
1
2
1.3
2.
0.6
2
1.5
3
2
3
1
Ave.
1.53
Ave.
1.27

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A differential equation encountered in the vibration of beams is d4ydx4=2D where x = distance measured along the beam; [x]=[L] y = displacement of the beam; [y]=[L] = circular frequency of vibration; []=[T1] = mass of the beam per unit length; []=[ML1] D = bending rigidity of beam; [D]=[FL2] Show that the equation is dimensionally homogeneous. Note that [F]=[MLT2] see Eq. (1.21:).