Force Grocery Bag can Hold

[delfam]

1. The problem statement, all variables and given/known data
The maximum force a grocery bag can withstand and not rip is 250N, if 20kg of groceries are lifted from the floor to a table with an acceleration of 5 m/s2, will the sack hold?


2. Relevant equations
F= ma
W= mg

3. The attempt at a solution
20kg * 9.81m/s2 = 981 N

Is that all I have to do, if the bag holds 250N, it will break then cause 981N is applied to the bag, is that right?

[mgb_phys]

20kg * 9.81m/s2 = 981 N
How on earth did you get that?

[turin]

The maximum force a grocery bag can withstand and not rip is 250N, if 20kg of groceries are lifted from the floor to a table with an acceleration of 5 m/s2, will the sack hold?

lution[/b]
20kg * 9.81m/s2 = 981 N

Is that all I have to do, ...?

No. Firstly, don't go plugging in numbers at the beginning; leave in symbolic form (and please mind the arithmetic when you do plug in numbers at the end). Secondly, I highly recommend drawing a free-body-diagram. Be careful, this is not a static problem but a dynamic problem.

[delfam]

i drew a free body diagram, but I still don't get how to do it. What equations would i use to do this.

[delfam]

How on earth did you get that?
I wrote it wrong, i did 20 * 9.81 to get 196.2, then multiplied it by the acceleration(5) to get 981 N. Is that right?

[mgb_phys]

No you need to add the accelerations
Think of the units.
F = m a so N = kg m s-2
So if you have another acceleration the only way to get the same units is
F = m (g+a)

ps sorry I thought you had hit the wrong key on the calculator and then just written down whatever it printed out - an amazing number of students assume that if the computer says a number it must be correct!

[delfam]

yeah, thats what i did, i did 5(9.81)(20) to get a force of 981 N, now is that the answer, meaning if the bag holds 250 N, it will break, or do I need to do something else?

[Delphi51]

i did 5(9.81)(20) to get a force of 981 N
Nice to show that as F = ma = 20 x 9.81 = ???
so we can see your reasoning AND the calculation. Note that there are only 2 letters in the formula, so only 2 numbers should be multiplied.

This is half the problem done, the force on the bag due to the acceleration of gravity. You must now do the second half, the force due to the bag being accelerated (lifted) by the force of the hand lifting it. The two forces BOTH act on the bag. You might think of the force of gravity acting on the contents, which in turn push down on the bottom of the bag. In order to accelerate the contents upward, the bottom of the bag must push with an additional force on the contents.

[mgb_phys]

i did 5(9.81)(20) to get a force of 981 N,
You can't do that because it would be F = m a*a so it wouldn't be N anymore

The units have to balance, so F = ma1 + ma2 = m(a+g) is allowed

[delfam]

Nice to show that as F = ma = 20 x 9.81 = ???
so we can see your reasoning AND the calculation. Note that there are only 2 letters in the formula, so only 2 numbers should be multiplied.

This is half the problem done, the force on the bag due to the acceleration of gravity. You must now do the second half, the force due to the bag being accelerated (lifted) by the force of the hand lifting it. The two forces BOTH act on the bag. You might think of the force of gravity acting on the contents, which in turn push down on the bottom of the bag. In order to accelerate the contents upward, the bottom of the bag must push with an additional force on the contents.
thats what I don't get, how do I do the second half, I don't have any more accelerations given to me, so how do I find the force due to the bag being accelerated by the force of the hand lifting it.

[Carid]

delfam,

mgb_phs has already told you what to do

F = m (g+a)

Think of it this way. Instead of the bag being on the floor you are holding it with your hand.
Since it contains 20kg of stuff it's pretty heavy.
What force are you using to counteract that weight?
Now, being a strong chap, you instantaneaously accelerate that bag upwards with an acceleration of 5m/sec²
This will require an extra force provided by your arm.
This force is different in magnitude from that opposing the weight because the acceleration is different, but the mass being accelerated is the same.