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SRMcDonald · Jul 31, 2007, 03:00 PM

I haven't had any questions like this so I'm not even sure if I understand it correctly let alone know how to go about solving it.

The maximum force a grocery sack can withstand and not rip is 250 N. If 20 kg of groceries are lifted from the floor to the table with an acceleration of 5 m/s^2, will the sack hold?

First, I found the net force needed to lift the bag with an acceleration of 5 m/s^2

Fnet = m*a = 20 kg * 5 m/s^2 = 100 N

Then I found the force of gravity on it

W = m*g = 20 kg * 9.80 m/s^2 = 196

So in order to lift the sack with an acceleration of 5 m/s^2 you need apply a force of 296 N? I'm having trouble visualizing this. To figure out what force will break the bag, is it simply adding the Fapp and Fgrav (492 N) together because those are all the forces acting on it, or is Fapp (296) the only valid answer because it's accelerating in that direction?

ebaines · Jul 31, 2007, 05:17 PM

Originally Posted by SRMcDonald

So in order to lift the sack with an acceleration of 5 m/s^2 you need apply a force of 296 N? I'm having trouble visualizing this. Is it simply adding the Fapp and Fgrav together because those are all the forces acting on it, or is Fapp the only valid answer because it's accelerating in that direction?

Yes - the force needed to accelerate the bag 5m/s^2 is the sum of the force to overcome gravity plus the force to accelerate the 20kg mass. If all you applied is 100-Newtons, you wouldn't have enough force to even hold the bag up against gravity, much less accelerate it.

SRMcDonald · Jul 31, 2007, 06:16 PM

Sorry I made a confusing question. I'm wondering if the total force that you use to determine if the bag breaks or not is 492 N for this question, or is it 296 N?

I really appreciate the help. :)

ebaines · Aug 1, 2007, 05:43 AM

OK, I see. You have already figured out that you have to apply 296 N to the bag. If you think about the 20 Kg item in the bag, for it to rise at 5 m/s^2 a force of 296 N needs to be applied to it - and it's the bag that applies the force to the item - you lift on the bag, and the bag lifts on the item (as a detail here we are assuming that the bag has no mass). Every force has an equal and opposite reaction - so that means the item in turn presses against the bottom of the bag with a force of 296 N. So, the bag rips.

Capuchin · Aug 1, 2007, 05:46 AM

I can't work out how I missed this question... ebains is correct, 296N will be needed, and the bag cannot provide it.