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Tension in of string between two boxes with friction.

  • Thread starter psychochef
  • Start date
  • #1

Homework Statement



Two boxes are connected by a massless string. The first box's mass is 16kg and second box's mass is 21kg. You pull with a total force of 185N on the 16kg box. The coefficient of friction is 0.5. Calculate the tension in the string between the two boxes.

Homework Equations



Ffr=μFn
Fnet = ma


The Attempt at a Solution



Thinking of the acceleration of the whole system (including friction)

Fnet = Fa - Ffr = ma
divide by m to get the acceleration of the whole system.

a = (Fa - Ffr)/m

m = 16+21 = 37kg

Fa = 185N

Ffr = μFn = μmg =(0.5)(37)(9.8)

a = [185-(0.5)(37)(9.8)]/37 = 0.1m/s2

Using this we should be able to calculate the Fnet of the 21kg box on the end of the string.

Fnet=ma = (21)(0.1) = 2.1N

Is the Fnet of the 21kg box equal to the tension of the string? I cant seem to get my head around this part. For me it doesnt make sense that even though I'm pulling on the whole system at 185N, only 2.1N is the tension in the string pulling on the 21kg box. Should I have taken the force of friction into account in calculating the Fnet of the 21kg box? Should I be adding or subtracting 185N from something?
 

Answers and Replies

  • #2
tiny-tim
Science Advisor
Homework Helper
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welcome to pf!

hi psychochef! welcome to pf! :smile:
Should I have taken the force of friction into account in calculating the Fnet of the 21kg box?
yes
Is the Fnet of the 21kg box equal to the tension of the string? I cant seem to get my head around this part. For me it doesnt make sense that even though I'm pulling on the whole system at 185N, only 2.1N is the tension in the string pulling on the 21kg box.
if the tension in the string was the same as the force with which you're pulling the front box, then (even without friction) the front box wouldn't move, would it? :wink:
 
  • #3
Awesome! It's good to know that I'm on the right track. So assuming I need to consider the force of Friction on the 2nd box (21kg), here's the calculations I get.

Fnet = Fa-Ffr-T = ma
solve for T:

T= Fa-Ffr-ma

Now should I use 185N as my Fa? I know that on the first box the applied force is the 185N pull, so knowing that the massless string should conserve that force should I put 185N as my applied force?

If so, heres what I get:

185-μFn-21(0.1)=T

185-(0.5)(21)(9.8)-(21)(0.1)= 80N

Does that look about right?
Thanks!
 
  • #4
tiny-tim
Science Advisor
Homework Helper
25,832
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hi psychochef! :smile:

(just got up :zzz:)
Awesome! It's good to know that I'm on the right track. So assuming I need to consider the force of Friction on the 2nd box (21kg), here's the calculations I get.

Fnet = Fa-Ffr-T = ma
no, this is wrong

draw a free body diagram for the second box …

there are only two horizontal forces on it: tension and friction

(the applied force is only applied to the first box … the question says so … it has nothing to do with the second box)

try again :smile:
 
  • #5
I got it!

T-Ffr = ma

I solved for T, got 105N, that seems much better than 2.1 or 80N.
Thanks a bunch!
 

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