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Please an answer - 2 crates, a rope and friction

  • Thread starter MGCLO
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  • #1
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please an answer -- 2 crates, a rope and friction

here is the problem image


http://tinypic.com/r/qxrqdu/5




Weight A is connected via a string to Weight B. Weight B is connected to a string to Weight C. Mass of A is 21 KG. Mass of B is 5 KG. Mass of C is 10 KG. the kinetic friction between the table and the 5 KG weight is 0.22

i got 2.75 M/S^2 for acceleration is that correct?

i need to know whats the tension in the string between A and B. include steps plz
 

Answers and Replies

  • #2
turbo
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Can you show and explain your work? There are lots of talented people here, but they are not allowed to do your homework for you.
 
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  • #3
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Can you show and explain your work? There are lots of talented people here, but they are not allowed to do your homework for you.
well whats the point of this website?
 
  • #4
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I figured out the acceleration part but I cant figure out the Tension!
 
  • #5
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I figured out the acceleration part but I cant figure out the Tension!
To get the tension in the string between A and B, draw a free body diagram on A, and write down the Newton's 2nd Law force balance on A.
 
  • #6
Ibix
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As per the rules you agreed to when you created your account, you need to show your working so far before we will help.

We will help you to do your homework. We will not do it for you.
 
  • #7
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well since we don't have an angle. then 210-ft= 2.75(5)
so 210-ft= 13.75

-ft = 13.75-210
ft= 196.25 ?
is that correct!
 
  • #8
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well since we don't have an angle. then 210-ft= 2.75(5)
so 210-ft= 13.75

-ft = 13.75-210
ft= 196.25 ?
is that correct!
No. If you are doing a force balance on weight A, shouldn't the 5kg on the right hand side of your equation be 21kg?
 
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  • #9
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Can you show and explain your work? There are lots of talented people here, but they are not allowed to do your homework for you.
No. If you are doing a force balance on weight A, shouldn't the 5kg on the right hand side of your equation be 21kg?
correct my bad got confused. using different weights in one problem does not make sense either.

so 210-ft= 2.75(210)

I got ft= 367.5 ?
 
  • #10
berkeman
Mentor
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here is the problem image


http://tinypic.com/r/qxrqdu/5




Weight A is connected via a string to Weight B. Weight B is connected to a string to Weight C. Mass of A is 21 KG. Mass of B is 5 KG. Mass of C is 10 KG. the kinetic friction between the table and the 5 KG weight is 0.22

i got 2.75 M/S^2 for acceleration is that correct?

i need to know whats the tension in the string between A and B. include steps plz
You need to show your work on the problem before you may receive tutorial help here on the PF. Please check your PMs.
 
  • #11
berkeman
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well whats the point of this website?
The point of this website is to help you learn how to learn. If we just spoon-feed you the answers, that will no help you to improve in your understanding of the subject, and will not help you to be able to figure out problems on your own (which is how real life works).
 
  • #12
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ok I showed my work for that part. can anyone tell me if 367.5 is correct?
 
  • #13
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correct my bad got confused. using different weights in one problem does not make sense either.

so 210-ft= 2.75(210)

I got ft= 367.5 ?
No. It's not 2.75(210), it's 2.75(21). Also, you did the algebra wrong. You have the right idea, but you gotta be more careful with the math.

Chet
 
  • #14
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No. It's not 2.75(210), it's 2.75(21). Also, you did the algebra wrong. You have the right idea, but you gotta be more careful with the math.

Chet
210-ft=57.75

ft=152.25 ???
 
  • #15
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can anybody check my answer?
 
  • #16
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Yes, provided your solution for the acceleration is correct.
 
  • #17
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Yes, provided your solution for the acceleration is correct.
how about the tension on the cord between A and B
ft=152.25 ???
 
  • #18
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how about the tension on the cord between A and B
ft=152.25 ???
That is what you calculated.
 
  • #19
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That is what you calculated.
yes.

21KG(10M/S^2) -FT= ma
210-FT=21(2.75)
FT=152.25
 
  • #20
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  • #21
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