[Physics] The Science of Tearing Paper-bag Handles

Question

My mother came back from a market which bags the products in paper-bags with handles, and asked me to move the bags from the trunk of the car to the house. Being the lazy human I am, I hung a few bags on each arm so I could cut the number of trips back and forth. As I was walking to the front door, the handles of a bag tore, the bag plummeting to the concrete ground. A glass jar of peppers had been smashed in to a zillion little pieces. As you might expect, my mother was furious. "You're so lazy! If you hadn't hung so many on your arm, the peppers and their jar would still be intact!"

I disagree, here's why…

Scenario Lazy

Scenario Peppers

Conclusion

Note that Bag A will have $N_A$ on it regardless of Bag B's existence. Sure, my arm had $N_A + N_B$ ($> N_A$), but it wasn't the thing that broke. So, I conclude, that the tearing of the bag was inevitable, and that the peppers' fates were written by someone other than me (e.g.: manufacturer didn't put enough glue to handle expected weight, cashier put more weight than permitted, etc).

Is my reasoning correct? Or am I missing something that proves that I'm guilty?

Answer 1

I think you are guilty. The shop assistant (or your mother) was able to load the bags into the car without causing the handles to break. They probably did not try to carry many bags at the same time.

If you hang the bags from a rod with sufficient spacing between them so that each handle hangs vertically, then the handles all bear only the weight of the bag's contents.

However, I think you probably held the bags in each hand rather than hung them from your extended arm (which would require enormous effort) or from a pole (which is unlikely to have been handy, and you are too lazy to look for one). When the bags hang from the same point the tension $T$ in the handles of the outer bags is higher than the weight $W$ of the bag, because of the large angle $\theta$ which the handle makes with the vertical. The vertical force $T\cos\theta$ provided by the handle must equal the weight $W$ of the bag's contents; the horizontal force $T\sin\theta$ is balanced by contact forces $N$ between the bags.

If the filled bags are wide, the handles of the outer bags will be at a large angle $\theta$ to the vertical, requiring a large force $T=\frac{W}{\cos\theta}$. This force tends towards infinity as the handle becomes horizontal $(\theta \to 90^{\circ})$. The outermost handles are much more likely to break than the innermost handle $(\theta = 0^{\circ})$.

Edit 1

Scenario #1 in RowanC's answer can be analysed in the same way. Assuming that the upper part of the bags have a trapezoidal shape, they spread out in an arc, with the middle bag supporting some of the weight of the outer bags.

Balancing forces on all 3 bags we get $T_2+2T_1\cos\theta=3W$. Balancing forces on the outer bags we get $2T_1(1-\cos^2\alpha)=W$ since $\theta=180^{\circ}-2\alpha$. Therefore $T_2(1-\cos^2\alpha)=(4-5\cos^2\alpha)W$.

• If $2(1-\cos^2\alpha) \lt 1$ then $T_1 \gt W$ - the outer bags bear more than their own weight. This happens for $\alpha \lt 45^{\circ}$.
• If $2(4-5\cos^2\alpha) \gt 1$ then $T_2 \gt T_1$ - the middle bag bears more weight than the outer bags. This happens when $\alpha \gt 33.2^{\circ}$.

A more thorough analysis could balance the torque on each bag.