****physics Help - Simple Machines ***

[Spud2388 10+ year member]

were doing a lab on simple machines and were trying to find their efficiency. Im confused on how to do this. Heres what we did.

-Pulley lab

lifted a guy that weighed 149lbs (1461.69N) in the air 1.41m using 5 pullies (3 oattached tot he ceiling and 2 attatched to his harness). We had to pull the rope 14.37m to raise him the 1.41m. How do you find the efficiency of the pullies using this data? note: we did not measure the force needed to pull on the rope.

I know you use the F(input)*d(pulled rope)=F(output)* d(he moved up)

F(input)=143.42N

d(pulled rope)=14.37m

F(output)=1461.69N

d(hemoved)=1.41m

How do you get the efficiency of this?

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[surftb15 10+ year member]

SIMPLE

efficiency = work output / work input

 

[faulkton 5,000+ posts]

2. Ferrer-Wreder et al. conducted a study to evaluate school-based intervention programs geared towards urban at-risk minority youth (2002: 168-188). As part of the design, a pretest and posttest were administered. What is one of the sources of internal validity that may be compromised through the participants’ exposure to a pretest if there was no comparison group?

 

[faulkton 5,000+ posts]

6. In a 1996 study of the effectiveness of behavioral modification programs for adolescents, the researchers evaluated the behavior of the subjects using a pre/posttest design without any control group (Al Ansari, Gouthro, Ahmad, and Steele: 469-477). The researchers were anticipating cyclical behavior patterns, so they decided to use scores of points averaging over a period of four weeks. Specifically, what potential threat to internal validity were the researchers attempting to avoid?

 

[Spud2388 10+ year member]

SIMPLE
efficiency = work output / work input

exactly. W=Fd right? W(output) =1.41m*1461.69N and W(input)= 14.37m*143.42N correct?

so (1.41*1461.96)/(14.37*143.42)= 1.00 *100%= 100% i didnt think they were 100% efficient (excluding the loss of friction)

 

[surftb15 10+ year member]

well you prob ****ed up ur data.

let me take a crack at it

 

[Spud2388 10+ year member]

see i think the reason its comin out to be 100% efficient is because i'm using an equation to find the input force, which is going to make the work output and work input equal each other.

 

[surftb15 10+ year member]

organize your data better so i dont have to read the problem again.

thanks/

mass, raised, etc,

 

[surftb15 10+ year member]

Equations you need, Plug into each.

work output = load x distance moved by load

work input = effort x distance moved by effort

efficiency = work output / work input

 

[Spud2388 10+ year member]

mass=67.72kg

distance mass moved upward=1.41m

distance we pulled rope=14.37m

Thats the information we were able to measure.

 

[Spud2388 10+ year member]

67.72kg x 9.81=664.4N

F(i)=?

d(i)=14.37m

F(o)=664.4N

d(o)=1.41m

F(i)d(i)=F(o)d(o)

F(i) x 14.37m=664.4N x 1.41m

F(i)=65.19N

W(output)/W(input) x 100=efficiency

W(output)=1.41m x 664.4N

W(input)=14.37m x 65.19N

(1.41 x 664.4)/(14.37 x 65.19) x 100=100% I know Im doing the math right, but because I used an equation to find the input force, its going to make the works equal each other. I guess I'll just compare the distances of the ropes or something and bullshit the lab. Thanks for the help though.

 

[surftb15 10+ year member]

yea now you go from there

 

[Spud2388 10+ year member]

thanks for the help, im doing it right, but because we didnt measure the force needed to pull the rope, and had to use and equation to find it, its alwasy going to make the works equal each other, thus giving me 100% efficiency.