Simple newtonian mechanics problem

  • Thread starter ENgez
  • Start date
  • #1
ENgez
75
0

Homework Statement


a man with mass m=66kg is standing on top of a platform with mass M=120kg. The man is pulling himself up using a pair of ropes suspended over massless pulleys. he pulls each rope with force of F=600N and is accelarating towards the ceiling at acceleration a. g=9.8 m/sec^2. find the value of a in m/sec^2.



* - pulley

Homework Equations


[tex]\sum F = ma[/tex]



The Attempt at a Solution


I tried treating the man and the platform as one body but i got negative acceleration, as if the man was accelerating away from the ceiling. I would like to see how you guys solve this problem.
 

Attachments

  • Capture.PNG
    Capture.PNG
    26.2 KB · Views: 430

Answers and Replies

  • #2
tiny-tim
Science Advisor
Homework Helper
25,838
255
Welcome to PF!

Hi ENgez! Welcome to PF! :smile:

(have a sigma: ∑ and try using the X2 tag just above the Reply box :wink:)
I tried treating the man and the platform as one body but i got negative acceleration, as if the man was accelerating away from the ceiling.

Show us your full calculations, and then we'll see what went wrong, and we'll know how top help! :smile:
 
  • #3
ENgez
75
0
i summed up the forced on the y axis treating the man and the platform as one body :
600*2 - (m+M)*g = (m+M)*a
a=-3.707 m/sec2
 
Last edited:
  • #4
tiny-tim
Science Advisor
Homework Helper
25,838
255
Hi ENgez! :smile:
i summed up the forced on the y axis treating the man and the platform as one body :
600*2 - (m+M)*g = (m+M)*a
a=-3.707 m/sec2

Why 600 times two? :wink:
 
  • #5
ENgez
75
0
The man pulls each rope with a force of 600N, as seen in the attached picture.
 
  • #6
tiny-tim
Science Advisor
Homework Helper
25,838
255
The man pulls each rope with a force of 600N, as seen in the attached picture.

So why two? :wink:
 
  • #7
ENgez
75
0
Is this supposed to be a hint?:confused:. I think it should be two as there are two ropes.
 
  • #8
tiny-tim
Science Advisor
Homework Helper
25,838
255
Is this supposed to be a hint?:confused:. I think it should be two as there are two ropes.

How do you know it isn't one rope?

If the rope continued under the platform, so that there was only one rope, would that make any difference? If not, how can the number of ropes matter?
 
  • #9
ENgez
75
0
so you are saying that the equation should be
600-(m+M)*g=(m+M)*a
a= -6.574?
 
  • #10
tiny-tim
Science Advisor
Homework Helper
25,838
255
so you are saying that the equation should be
600-(m+M)*g=(m+M)*a
a= -6.574?

ie 600*1 ?

No, I'm saying that the number of ropes is irrelevant.

Hint: what would be the tension in each rope if the platform was stationary? :wink:
 
  • #11
ENgez
75
0
I thinks i see where you are getting here... if the platform was stationary the tension each rope felt was given by: 2T-(m+M)*g=0 => T = (m+M)*g/2. So if the man is supplying 600N of force the equation becomes 2T+600+(m+M)*g/2 = (m+M)*a?
 
Last edited:
  • #12
tiny-tim
Science Advisor
Homework Helper
25,838
255
… if the man is supplying 600N of force the equation becomes 2T+600+(m+M)*g/2 = (m+M)*a?

Sorry, that equation doesn't make sense …

if it's meant to be the equation for force on the man-and-platform, how can you include force from the man?
 
  • #13
ENgez
75
0
hmmm. let me try again... the equation for the platform-and-man is:
2T-(m+M)*g= (m+M)*a
and the equation for the man:
T-600-m*g=m*a
is this correct?
 
  • #14
tiny-tim
Science Advisor
Homework Helper
25,838
255
hmmm. let me try again... the equation for the platform-and-man is:
2T-(m+M)*g= (m+M)*a

No.

Try it for a = 0, ie the man-and-platform is stationary …

draw all the external forces on the man-and-platform.
and the equation for the man:
T-600-m*g=m*a
is this correct?

i really have no idea what that's supposed to be :confused:

(and i'm off to bed :zzz:)​
 
  • #15
ENgez
75
0
thanks alot for your help :) ill keep working on it.
 

Suggested for: Simple newtonian mechanics problem

  • Last Post
Replies
2
Views
1K
  • Last Post
Replies
5
Views
4K
Replies
2
Views
1K
  • Last Post
Replies
3
Views
2K
  • Last Post
Replies
2
Views
2K
  • Last Post
Replies
4
Views
6K
Replies
7
Views
2K
Replies
5
Views
895
Replies
27
Views
2K
Top