Solving Rotational Force in Beverage Engineering

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The discussion centers on calculating the force applied to the scored section of a beverage can when a 10 N upward force is exerted on the pull tab. Initially, there is confusion about whether the downward force must simply equal the upward force, leading to a conclusion of -10 N. However, it is noted that the pivot point at the central bolt introduces additional factors that must be considered, such as torque and the can's inertia. The need to apply Newton's second law and torque relations is emphasized to arrive at a more accurate solution. Overall, the problem requires a deeper analysis beyond the straightforward assumption of equal opposing forces.
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Homework Statement



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 10 N force, approximately what is the magnitude of the force applied to the scored section?


Homework Equations





The Attempt at a Solution



Now at first I thought that the force down just had to be the opposite of the force up, which would be -10 N. But I didn't think this problem could be that easy...

So since the tab pivots on a central bolt the torque=inertia * angular acceleration (Newtons second law)
Now we don't know the mass of the can, so we cannot find inertia or acceleration. I think that since there is nowhere to go, it must be -10 N? I just can't see the problem being that simple

Thanks! Any push in the right direction would be great.
 
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I don't think it will be -10N because you're neglecting the force due to the central bolt. You're going to have to take that into account and then probably just use Newtons second law and some torque relations.
 
Thread 'Variable mass system : water sprayed into a moving container'

Thread 'Variable mass system : water sprayed into a moving container'

Starting with the mass considerations #m(t)# is mass of water #M_{c}# mass of container and #M(t)# mass of total system $$M(t) = M_{C} + m(t)$$ $$\Rightarrow \frac{dM(t)}{dt} = \frac{dm(t)}{dt}$$ $$P_i = Mv + u \, dm$$ $$P_f = (M + dm)(v + dv)$$ $$\Delta P = M \, dv + (v - u) \, dm$$ $$F = \frac{dP}{dt} = M \frac{dv}{dt} + (v - u) \frac{dm}{dt}$$ $$F = u \frac{dm}{dt} = \rho A u^2$$ from conservation of momentum , the cannon recoils with the same force which it applies. $$\quad \frac{dm}{dt}...
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