Simple Machines B/Compound Machines C

labchick
Member
Member
Posts: 24
Joined: August 23rd, 2011, 10:09 am
Division: B
State: TN

Re: Simple Machines B/Compound Machines C

Post by labchick » February 24th, 2015, 11:19 am

A 100 meter long first-class lever has a load of of 2 kilograms 75 meters away from the fulcrum. You are applying a force of 15 newtons and the lever is in equilibrium. What is the efficiency of the machine?

User avatar
Unome
Moderator
Moderator
Posts: 4197
Joined: January 26th, 2014, 12:48 pm
Division: Grad
State: GA
Location: somewhere in the sciolyverse

Re: Simple Machines B/Compound Machines C

Post by Unome » February 24th, 2015, 3:01 pm

labchick wrote:A 100 meter long first-class lever has a load of of 2 kilograms 75 meters away from the fulcrum. You are applying a force of 15 newtons and the lever is in equilibrium. What is the efficiency of the machine?
AMA=19.6/15=1.3066
IMA=25/75=0.3333
Efficiency=AMA/IMA=1000% (with significant figures)

If, however, you mean 2 N force:
AMA=2/15=0.1333
IMA=25/75=0.3333
Efficiency=AMA/IMA=40% (with significant figures)
Userpage
Chattahoochee High School Class of 2018
Georgia Tech Class of 2022

Opinions expressed on this site are not official; the only place for official rules changes and FAQs is soinc.org.

labchick
Member
Member
Posts: 24
Joined: August 23rd, 2011, 10:09 am
Division: B
State: TN

Re: Simple Machines B/Compound Machines C

Post by labchick » February 24th, 2015, 3:33 pm

Unome wrote:
labchick wrote:A 100 meter long first-class lever has a load of of 2 kilograms 75 meters away from the fulcrum. You are applying a force of 15 newtons and the lever is in equilibrium. What is the efficiency of the machine?
AMA=19.6/15=1.3066
IMA=25/75=0.3333
Efficiency=AMA/IMA=1000% (with significant figures)

If, however, you mean 2 N force:
AMA=2/15=0.1333
IMA=25/75=0.3333
Efficiency=AMA/IMA=40% (with significant figures)
Yes, sorry, I did mean 2 N force not 2 kg. Oops. Your answer is correct. Your turn to ask a question!

User avatar
Unome
Moderator
Moderator
Posts: 4197
Joined: January 26th, 2014, 12:48 pm
Division: Grad
State: GA
Location: somewhere in the sciolyverse

Re: Simple Machines B/Compound Machines C

Post by Unome » February 24th, 2015, 5:09 pm

A person with a mass of 25 kg stands in the center of the beam and walks towards the right (from your point of view). The lever bar weighs 6 kg, distributed evenly. Past what distance from the edge of the bar will the lever tip over?
Attachments
Doulbe fulcrum lever.png
Doulbe fulcrum lever.png (7.31 KiB) Viewed 2122 times
Userpage
Chattahoochee High School Class of 2018
Georgia Tech Class of 2022

Opinions expressed on this site are not official; the only place for official rules changes and FAQs is soinc.org.

mjcox2000
Member
Member
Posts: 121
Joined: May 9th, 2014, 3:34 am
Division: Grad
State: VA

Re: Simple Machines B/Compound Machines C

Post by mjcox2000 » February 24th, 2015, 6:14 pm

Unome wrote:A person with a mass of 25 kg stands in the center of the beam and walks towards the right (from your point of view). The lever bar weighs 6 kg, distributed evenly. Past what distance from the edge of the bar will the lever tip over?
The lever is 6.4m in length, so each meter weighs 0.9375, or 15/16, kg. This means, on the right side of the right fulcrum, 33/32 kg, and on the left, 159/32 kg. This is distributed evenly, so it can be thought of as a single weight of that quantity in the middle of a massless lever arm, making the clockwise torque 363/640 kg•m and counterclockwise torque 8427/640 kg•m. Subtracting, that's a total of 8064/640, or (simplifying) 63/5, kg•m counterclockwise. Therefore, since the lever's in equilibrium, 25 kg * d meters=63/5 kg•m, so d=63/125 m, or 0.504 m. Therefore, the person can walk [i][b]0.504 m[/b][/i] to the right of the rightmost fulcrum before the lever tips; with significant figures, that's 0.5 meters, or 0.50 if you meant to make the lever weight 6.0 kg.
MIT ‘23
TJHSST ‘19
Longfellow MS

See my user page for nationals medals and event supervising experience.

User avatar
Unome
Moderator
Moderator
Posts: 4197
Joined: January 26th, 2014, 12:48 pm
Division: Grad
State: GA
Location: somewhere in the sciolyverse

Re: Simple Machines B/Compound Machines C

Post by Unome » February 25th, 2015, 8:38 am

Correct (I made a typo there; I was trying to say "past what distance from the rightmost fulcrum")
Userpage
Chattahoochee High School Class of 2018
Georgia Tech Class of 2022

Opinions expressed on this site are not official; the only place for official rules changes and FAQs is soinc.org.

mjcox2000
Member
Member
Posts: 121
Joined: May 9th, 2014, 3:34 am
Division: Grad
State: VA

Re: Simple Machines B/Compound Machines C

Post by mjcox2000 » February 26th, 2015, 4:54 am

In the image of the pulley system shown below, where there are 2 concentric circles over 2 intersecting ropes, those ropes are tied together at that point; triangles in the image symbolize fixed anchor points. How much downwards effort force must be applied to the arrow (on the rightmost rope) to lift the 10kg weight, as illustrated? Assume that this pulley system is an ideal machine, and g=9.81 m/s^2.
Complex pulley system.jpg
MIT ‘23
TJHSST ‘19
Longfellow MS

See my user page for nationals medals and event supervising experience.

JonB
Coach
Coach
Posts: 303
Joined: March 11th, 2014, 12:00 pm
Division: C
State: FL

Re: Simple Machines B/Compound Machines C

Post by JonB » February 26th, 2015, 10:37 am

mjcox2000 wrote:In the image of the pulley system shown below, where there are 2 concentric circles over 2 intersecting ropes, those ropes are tied together at that point; triangles in the image symbolize fixed anchor points. How much downwards effort force must be applied to the arrow (on the rightmost rope) to lift the 10kg weight, as illustrated? Assume that this pulley system is an ideal machine, and g=9.81 m/s^2.
Complex pulley system.jpg
From team member (not myself as the coach):
9.81N

mjcox2000
Member
Member
Posts: 121
Joined: May 9th, 2014, 3:34 am
Division: Grad
State: VA

Re: Simple Machines B/Compound Machines C

Post by mjcox2000 » February 26th, 2015, 2:12 pm

JonB wrote:
mjcox2000 wrote:In the image of the pulley system shown below, where there are 2 concentric circles over 2 intersecting ropes, those ropes are tied together at that point; triangles in the image symbolize fixed anchor points. How much downwards effort force must be applied to the arrow (on the rightmost rope) to lift the 10kg weight, as illustrated? Assume that this pulley system is an ideal machine, and g=9.81 m/s^2.
Complex pulley system.jpg
From team member (not myself as the coach):
9.81N
Correct! Your turn to post a question.
MIT ‘23
TJHSST ‘19
Longfellow MS

See my user page for nationals medals and event supervising experience.

User avatar
blindmewithscience
Member
Member
Posts: 44
Joined: October 2nd, 2014, 8:57 pm
Division: C
State: NV

Re: Simple Machines B/Compound Machines C

Post by blindmewithscience » February 26th, 2015, 6:26 pm

JonB wrote:
mjcox2000 wrote:In the image of the pulley system shown below, where there are 2 concentric circles over 2 intersecting ropes, those ropes are tied together at that point; triangles in the image symbolize fixed anchor points. How much downwards effort force must be applied to the arrow (on the rightmost rope) to lift the 10kg weight, as illustrated? Assume that this pulley system is an ideal machine, and g=9.81 m/s^2.
Complex pulley system.jpg
From team member (not myself as the coach):
9.81N
Would you be able to explain this solution? Really complex pulley systems are one of the things I'm currently having trouble with.
Nevada state SO occurs on tau/2 day. Support the correct mathematical constant with all tauists.
http://www.tauday.com/tau-manifesto

Event: Regional/States
Astronomy: x/:(
Bungee: 3/3
Compound Machines: x/1
TPS: x/:(

Locked

Return to “2015 Question Marathons”

Who is online

Users browsing this forum: No registered users and 1 guest