Forces acting on a box-pulley system in an elevator?

  • #26
gneill
Mentor
20,925
2,867
You have the mass of block A appearing in two places in your work, with different accelerations. That shouldn't happen as a mass can only have one acceleration in a given direction.

For part (b), on page one of your upload images, you wrote:
upload_2017-2-5_21-16-21.png

Note the indicated line. Since block A is stationary in the chosen frame of reference (the elevator), the sum of the forces acting on it must be zero.

The same applies for part (e) where you are looking for the normal force on the elevator floor: Choosing the elevator as the frame of reference, block A is stationary on the floor, so that the sum of forces must be zero in that frame of reference. You can write the same equation using your effective value of "g".
 
  • #27
62
2
For part (e), apply the same method that you did in part (b), keeping in mind the new effective value of "g" that the masses experience.

Unfortunately, my answer for part d.) is incorrect (I checked on my Homework website), which means that my part e.) is incorrect too. What could be the issue? :frown:
 
  • #28
gneill
Mentor
20,925
2,867
Unfortunately, my answer for part d.) is incorrect (I checked on my Homework website), which means that my part e.) is incorrect too. What could be the issue? :frown:
You may have rounded your answer too aggressively :smile: You should never round intermediate values in a calculation, only rounding at the end. Keep a few extra decimal places in all intermediate results to prevent rounding/truncation errors from sneaking into your significant figures.
 
  • #29
62
2
You have the mass of block A appearing in two places in your work, with different accelerations. That shouldn't happen as a mass can only have one acceleration in a given direction.

For part (b), on page one of your upload images, you wrote:
View attachment 112620
Note the indicated line. Since block A is stationary in the chosen frame of reference (the elevator), the sum of the forces acting on it must be zero.

The same applies for part (e) where you are looking for the normal force on the elevator floor: Choosing the elevator as the frame of reference, block A is stationary on the floor, so that the sum of forces must be zero in that frame of reference. You can write the same equation using your effective value of "g".

Ok, that makes sense.

and thank you again for your clear explanation, I re-did my equation, taking into mind that the box is stationary on the floor. Also, I changed my overly-rounded answer to 11.8 N instead of 12 N. Hopefully that works.

Now,

FN + FT - FG = 0

FN + 11.8 - m(g-a) = 0

FN + 11.8 - 3.4(10-1.6) = 0

FN + 11.8 - 28.56 = 0

FN = 16.8 N
 
  • #30
gneill
Mentor
20,925
2,867
Was the mass of block A 3.4 kg or 3.8 kg? I seem to recall it being 3.8 kg.
 
  • #31
62
2
Was the mass of block A 3.4 kg or 3.8 kg? I seem to recall it being 3.8 kg.

Oh my goodness, you're absolutely right. I used the wrong value for mass. Thank you so much for catching that error. With that changed, my answer is now 20.1 N, which is the correct answer on my homework website!

Thank you so much for your help with this problem gneill. Again, your explanations helped clarify a lot. I will probably be back with more questions in the future. I hope I will receive your guidance again during those difficult times :wink: and I promise to have a free body diagram and my attempts written out beforehand when I come back.
 
  • #32
gneill
Mentor
20,925
2,867
Glad to be of help! See you next time. :smile:
 

Related Threads on Forces acting on a box-pulley system in an elevator?

Replies
2
Views
3K
  • Last Post
Replies
7
Views
10K
  • Last Post
Replies
15
Views
1K
Replies
3
Views
1K
  • Last Post
Replies
3
Views
15K
Replies
5
Views
1K
Replies
1
Views
4K
V
Replies
4
Views
766
  • Last Post
Replies
3
Views
10K
  • Last Post
Replies
2
Views
883
Top