Pull tab of beverage containers

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The discussion focuses on the engineering design of beverage containers, specifically the mechanics of pull tabs. The pull tab operates as a lever, where a force applied at one end translates to a greater force at the scored section of the can. By applying a 10N force to the pull tab, calculations show that the force exerted on the scored area is approximately 23.33N. The participants confirm the accuracy of this calculation, emphasizing the lever mechanics involved. The conversation concludes with light-hearted banter about opening a beer can.
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Beverage engineering. The pull tab was a major advance in the engineering design of beverage containers. The tab pivots on a central bolt in the can's top. When you pull upward on one end of the tab, the other end presses downward on a portion of the can's top that has been scored. If you pull upward with a 10N force, approximately what is the magnitude of the force applied to the scored section? (You will need to examine a can with a pull tab.)

After I measure the pull tab, I got:
Pull tab length is 2.5cm
From pulling zone to pivot is 1.75cm, let it = R
and from pivot to pressing zone is 0.75, let it = r

Given, pull upward 10N

Can I calculate by using
F1R = F2r
10 * 1.75 = 0.75 * F2
Fpressing = 23.33N

Is this correct?
 
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Yes it's correct, given the numbers you have for R and r.
 
did i hear the sound of a beer can being opened … ?

noppawit said:
From pulling zone to pivot is 1.75cm, let it = R
and from pivot to pressing zone is 0.75, let it = r

Given, pull upward 10N

Can I calculate by using
F1R = F2r
10 * 1.75 = 0.75 * F2
Fpressing = 23.33N

Is this correct?

Hi noppawit! :smile:

Looks good!

It's a lever, so the gearing is the ratio of the lengths …

hmm … now pull one for me! :wink:

EDIT: oh … and one for Redbelly98 too! :biggrin:
 
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