MGCLO

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

Gold Member
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.

• 1 person
MGCLO
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?

MGCLO
I figured out the acceleration part but I cant figure out the Tension!

Mentor
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.

As per the rules you agreed to when you created your account, you need to show your working so far before we will help.

MGCLO
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!

Mentor
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?

• 1 person
MGCLO
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 ?

Mentor
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.

Mentor
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).

MGCLO
ok I showed my work for that part. can anyone tell me if 367.5 is correct?

Mentor
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

MGCLO
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 ???

MGCLO

Mentor
Yes, provided your solution for the acceleration is correct.

MGCLO
Yes, provided your solution for the acceleration is correct.

how about the tension on the cord between A and B
ft=152.25 ???

Mentor
how about the tension on the cord between A and B
ft=152.25 ???

That is what you calculated.

MGCLO
That is what you calculated.

yes.

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

MGCLO
That is what you calculated.

yes.

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

so? is it correct or not!

Mentor
so? is it correct or not!
Yes.