Velocity Ratio and Lever Confirmation

AI Thread Summary
The discussion revolves around calculating the velocity ratio of a lever, with participants debating whether the correct ratio is 1/5 or 1/6 based on the distances from the fulcrum. One participant initially believes the ratio should be 1/5, while another insists that it should be 1/6, as the effort is 1m from the fulcrum and the load is 6m away. Despite using the formula for velocity ratio, confusion persists due to conflicting feedback on the correctness of their answers. Participants suggest verifying the diagram used, as errors in the course materials may contribute to the misunderstanding. Ultimately, the consensus leans towards 1/6 being the correct answer, but the participant continues to receive incorrect marks.
junkie_ball
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Homework Statement



Calculate the velocity ratio of the attached lever.


Homework Equations



Velocity Ratio = Distance moved by effort * Distance moved by load

The Attempt at a Solution



Ok I'm having some confusion over this question. I cannot figure out whether the 'Distance moved by the load should be 6 or 5? I am assuming it should be 1/5 as for every 1m the effort moves the load moves by 5.

Or should i be basing my answer taking the dimensions from the fulcrum in which case it will be 1/6?
 

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  • Velocity Ration of the Lever.jpg
    Velocity Ration of the Lever.jpg
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hi junkie_ball! :smile:

hint: for a given small angle θ, how far does the load actually move? :wink:
 
tiny-tim said:
hi junkie_ball! :smile:

hint: for a given small angle θ, how far does the load actually move? :wink:

Tiny-tim thanks for the hint. I have been looking at this problem again today and i believed the answer originally should have been 1/6 as the effort is 1m away from the fulcrum and the load 6m away from the fulcrum. This came back marked as incorrect though.

This is why i considered the 1/5 although to me using the equation VR= Distance moved by effort * Distance moved by load I'm still inclined to say 1/6 as in my original diagram if the effort distance moved by 1 then the load distance would move by 6? I'm so confused by this which is maddening as the maths is so simple! :cry: Anymore pointers to help me understand where my error is will be greatly appreciated as it's the last question i need to resubmit to pass this unit!
 
junkie_ball said:
Tiny-tim thanks for the hint. I have been looking at this problem again today and i believed the answer originally should have been 1/6 as the effort is 1m away from the fulcrum and the load 6m away from the fulcrum. This came back marked as incorrect though.

your answer of 1/6 looks correct to me :confused:

velocity ratio = distance moved by effort / distance moved by load,

which from the diagram (i suppose you're definitely using the correct diagram?) is obviously 1/6

(the only alternative answer i can think of is 0.166)
 
tiny-tim said:
your answer of 1/6 looks correct to me :confused:

velocity ratio = distance moved by effort / distance moved by load,

which from the diagram (i suppose you're definitely using the correct diagram?) is obviously 1/6

(the only alternative answer i can think of is 0.166)


Sorry for the late reply i was waiting for my work to come back marked. Again i used the formula above and went with 1/6 or 0.166 as my answer and again its come back as incorrect. the diagram is definitely right (well what they have printed. This is ICS and have had quite a few errors in the text up to now.) I'm lost on this. You ask the tutor advice and they refer you to the page in the course materials which just gives the equation velocity ratio = distance moved by effort / distance moved by load which to me is 1/6. Arr
 
hi junkie_ball! :smile:

your answer of 1/6 is obviously correct

if you're allowed as many tries as you like, then i'd suggest trying 1/5 or -1/5 (but it's presumably the wrong diagram, so the answer could be anything :redface:)
 
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