Calculating Effort of a Lever

  • Thread starter joe465
  • Start date
  • Tags
    Lever
In summary, the effort required to lift the load can be calculated using the formula F=mg, where F is the force, m is the mass, and g is the gravitational acceleration. In this case, the load is equal to 200kg and the force required to lift it is 1962N. However, in this scenario, the formula for a type 1 lever is not applicable as the load is at a larger distance from the pivot point than the effort. Using a general equation, the effort can be calculated to be 9810N. The mechanical advantage of this lever is 0.2, which means that the load is 5 times larger than the effort. The velocity ratio is 0.2, and the
  • #1
joe465
94
0

Homework Statement



Calculate the effort required to lift the load.

Homework Equations



F=mg
large distance × small effort = small distance × large load

The Attempt at a Solution



The load of the crate=

F=mg
F=200*9.81
F=1962N

large distance × small effort = small distance × large load

Now this forumla does not represent what is happening as the load it has a large distance (5m) and the effort is small distance (1m) away.

Is there another forumla for calculating this problem or another method?

Thanks, Joe
 

Attachments

  • Calculate effort.jpg
    Calculate effort.jpg
    6.3 KB · Views: 2,523
Physics news on Phys.org
  • #2
Hi Joe! :wink:
joe465 said:
large distance × small effort = small distance × large load

Now this forumla does not represent what is happening as the load it has a large distance (5m) and the effort is small distance (1m) away.

(that formula is written for a type 1 lever, this is a type 2 lever)

it really ought to be written as a general https://www.physicsforums.com/library.php?do=view_item&itemid=64" equation:

effort distance × effort = load distance × load​

try that :smile:
 
Last edited by a moderator:
  • #3
Thankyou so:

effort distance × effort = load distance × load

1*effort=5*1962

effort=5*1962/1

effort = 9810N

I hope this is right:)
 
  • #4
Looks good! :biggrin:
 
  • #5
Thanks, the next question was calculate the mechanical advantage which is:

MA=load/effort

MA=1962/9810

MA=0.2

It just seems quite low and as far as i know there is no unit for MA
 
  • #6
joe465 said:
MA=0.2

It just seems quite low …/QUOTE]

Well, the relative distance is 1/5, so the mechanical advantage (disadvantage!) is 1/5 …

that's how it works :wink:

(and yes, there's no units, MA is a ratio, so it's a dimensionless ordinary number)
 
  • #7
Thankyou very much.

Calculate the velocity ratio for this system.If the work put in is 3000J, calculate the usefull output.

Velocity Ratio = distance moved by effort/distance moved by load

1/5=0.2 not sure if this is right, the course material is a bit shoddy with this

If it is right then:

Efficiency=MA/VR

Efficiency=0.2/0.2

1*100=100%

Doesnt seem right to me?

I noticed you said this was a second order lever, my course suggests this is a third order since the effort is greater than the load?

Cheers, Joe
 
  • #8
Hi Joe! :smile:
joe465 said:
Velocity Ratio = distance moved by effort/distance moved by load

1/5=0.2 not sure if this is right, the course material is a bit shoddy with this

Yes, that looks ok. :smile:
If it is right then:

Efficiency=MA/VR

Efficiency=0.2/0.2

1*100=100%

Doesnt seem right to me?

Can't see anything wrong with it.

The efficiency of a lever should always be 100% unless there's some friction or other obstruction.

(See http://en.wikipedia.org/wiki/Mechanical_advantage#Efficiency")
I noticed you said this was a second order lever, my course suggests this is a third order since the effort is greater than the load?

oops! :rolleyes: I should have checked! :redface:

Yes, looking at http://en.wikipedia.org/wiki/Lever#Classes", you're right, it is a Class 3 lever. :smile:
 
Last edited by a moderator:
  • #9
Thanks, i guess the 3000J was thrown in there to distract
 
  • #10
Ive been looking over the picture and i don't see how it is a lever, especially since the pivet point is at the end?

i thought a lever had to have two ends, the image looks like one with a pivet point on the end?

Wouldnt this make my maths wrong?

Shouldnt the pivet point be about the hydraulic ram?

Would you be able to explain this better to me? sorry for being such a pain

Thanks, Joe
 
  • #11
Hi Joe! :smile:
joe465 said:
Ive been looking over the picture and i don't see how it is a lever, especially since the pivet point is at the end?

i thought a lever had to have two ends, the image looks like one with a pivet point on the end?

No, a lever needs only three things, a load, an effort (or force), and a fulcrum (or pivot) …

they can be anywhere

see "Professor Beaker's" explanation, with diagrams of the three types, at http://www.professorbeaker.com/lever_fact.html" [Broken] :wink:
Shouldnt the pivet point be about the hydraulic ram?

no the pivot point (the fulcrum) is whatever is fixed (ie doesn't move)
 
Last edited by a moderator:
  • #12
Thanks,

That means that the load is 6m away from the fulcrum(pivet point) and the effort from the ram 1m from pivet point.

This would make my maths wrong then?

Since : effort distance × effort = load distance × load

I guess the distances are from the pivet point?

I have edited the original picture to include load and effort hopefully putting them in the correct place.

This means the idea of the ram is to increase the angle of the lift to allow the crate to slide down?

Thanks a lot for your help, i know there has been a lot of silly questions
 

Attachments

  • Calculate effort EDIT.jpg
    Calculate effort EDIT.jpg
    8.7 KB · Views: 1,555
  • #13
oops!

joe465 said:
That means that the load is 6m away from the fulcrum(pivet point) and the effort from the ram 1m from pivet point.

This would make my maths wrong then?

Since : effort distance × effort = load distance × load

I guess the distances are from the pivet point?

oops! I should have enlarged the diagram! :redface:

Yes, it should have been 6 instead of 5 each time.
This means the idea of the ram is to increase the angle of the lift to allow the crate to slide down?

No, I'm pretty sure the aim is to raise the box, not to move it sideways. :wink:

(btw, placing the ram where it is has no mechanical advantage, in fact it has a mechanical disadvantage, as you've calculated …

so no one in their right mind would put the ram there if they didn't have to … it's usually because there's limited space, and that's the closest the ram can get without interfering with everything else in the warehouse :biggrin:)
 
  • #14
Thanks, it will raise the box but it will also increase the angle since it will move about the pivet point
 
  • #15
The new answers should be:

effort distance * effort = load distance * load

1m * effort = 6 *1962

effort=6*1962/1

effort = 11772N

MA=Load/Effort
MA=1962/11772
MA=0.17 (2dp)

Calculate the velocity ratio for this system. If the work put in is 3000J, calculate the usefull energy output.

VR=distance moved by effort/distance moved by load
VR=1/6
VR=0.17(2dp)


Should be right now, the only one I am not 100% about is the last one, why would they ask a question that gives the same answer as the mechanical advantage?

Thanks for all your help, couldn't of done this without you
 
  • #16
joe465 said:
… the only one I am not 100% about is the last one, why would they ask a question that gives the same answer as the mechanical advantage?

You mean the useful energy output, or the efficiency ratio?

I don't know … since the efficiency seems to be 100%, the energy out should be the same as the energy in, so the question relating to it seems pointless. :confused:
 
  • #17
Maybe its trying to get you to think about if the efficiency is 100% like you said then effectively the usefull energy output will be also 3000J
 
  • #18
yes :smile:
 
  • #19
The next question is:

The system has an efficiency of 65%. Calculate the velocity ratio of the system.

Im guessing its:

Since VR=0.17 when efficiency is 100%

0.17/100 = 0.0017
0.0017*65 = 0.1105
 
  • #20
joe465 said:
Im guessing …

I don't understand why you're guessing …

weren't these topics covered in your lectures or your book? :confused:
 
  • #21
This particular bit wasn't covered, teaching material has been awfull throughout, to the extent that there are many mistakes throughout and even in the answers section there are errors. I will never study through ICS again after this.
 
  • #22
(what's ICS?)

Yes this question is weird.

So far as I know, velocity ratio is purely geometrical, ie depends purely on the shape …

if the ram goes up 1 cm, the load goes up 6 cm no matter how much energy is lost.

And the efficiency will be 1 unless there's a good reason to the contrary, given in the question. :confused:
 
  • #23
ICS is international correspondence school who are a distance learning company. 'www.icslearn.co.uk' They offer a wide range of courses but the material has so many flaws and errors and has a poor way of getting the point across which is why i am struggling so much doing this.

Theres nothing in the question or previous questions to suggest different so i presume I am right
 

1. How do you calculate effort of a lever?

To calculate the effort of a lever, you need to know the length of the lever arm, the distance from the fulcrum to the point of application of the effort, and the magnitude of the effort force. The formula for calculating effort is effort = load x load arm length / effort arm length.

2. What is the lever principle?

The lever principle states that a small effort applied over a long distance can move a larger load over a shorter distance, as long as the effort arm is longer than the load arm. This principle is based on the concept of torque, which is the product of force and distance from the fulcrum.

3. How does the position of the fulcrum affect the effort of a lever?

The position of the fulcrum affects the effort of a lever by changing the ratio of the effort arm length to the load arm length. Moving the fulcrum closer to the load decreases the effort needed to lift the load, while moving it closer to the effort increases the effort needed.

4. Can a lever have multiple fulcrums?

No, a lever can only have one fulcrum. The fulcrum is the fixed point on which the lever pivots, and having multiple fulcrums would cause the lever to be unstable and unable to function properly.

5. What are the different types of levers?

There are three types of levers: first-class, second-class, and third-class. In a first-class lever, the fulcrum is located between the effort and load. In a second-class lever, the load is between the fulcrum and effort. In a third-class lever, the effort is between the fulcrum and load.

Similar threads

  • Introductory Physics Homework Help
Replies
1
Views
6K
  • Introductory Physics Homework Help
Replies
1
Views
4K
  • Introductory Physics Homework Help
Replies
1
Views
3K
  • Introductory Physics Homework Help
Replies
5
Views
3K
  • Introductory Physics Homework Help
Replies
9
Views
13K
Replies
1
Views
1K
Replies
8
Views
3K
  • Introductory Physics Homework Help
Replies
1
Views
1K
  • Introductory Physics Homework Help
Replies
6
Views
1K
  • Mechanical Engineering
Replies
10
Views
1K
Back
Top