Please an answer - 2 crates, a rope and friction

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The discussion centers on calculating the tension in a string connecting two weights, A (21 KG) and B (5 KG), with a kinetic friction of 0.22 between B and the table. The user initially calculated the acceleration as 2.75 m/s² and sought confirmation on the tension between A and B. After several iterations and corrections, the final tension was determined to be 152.25 N using the equation 21 KG (10 m/s²) - FT = ma, where FT represents the tension.

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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 what's the tension in the string between A and B. include steps please
 
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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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turbo said:
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 what's the point of this website?
 
I figured out the acceleration part but I can't figure out the Tension!
 
MGCLO said:
I figured out the acceleration part but I can't 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.
 
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.
 
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!
 
MGCLO said:
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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turbo said:
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.

Chestermiller said:
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
MGCLO said:
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 what's the tension in the string between A and B. include steps please

You need to show your work on the problem before you may receive tutorial help here on the PF. Please check your PMs.
 
  • #11
MGCLO said:
well what's 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
ok I showed my work for that part. can anyone tell me if 367.5 is correct?
 
  • #13
MGCLO said:
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 got to be more careful with the math.

Chet
 
  • #14
Chestermiller said:
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 got to be more careful with the math.

Chet

210-ft=57.75

ft=152.25 ?
 
  • #15
can anybody check my answer?
 
  • #16
Yes, provided your solution for the acceleration is correct.
 
  • #17
Chestermiller said:
Yes, provided your solution for the acceleration is correct.

how about the tension on the cord between A and B
ft=152.25 ?
 
  • #18
MGCLO said:
how about the tension on the cord between A and B
ft=152.25 ?

That is what you calculated.
 
  • #19
Chestermiller said:
That is what you calculated.

yes.

21KG(10M/S^2) -FT= ma
210-FT=21(2.75)
FT=152.25
 
  • #20
Chestermiller said:
That is what you calculated.

MGCLO said:
yes.

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

so? is it correct or not!
 
  • #21
MGCLO said:
so? is it correct or not!
Yes.
 

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