Dynamics Boxes 2

  • Thread starter yolo123
  • Start date
  • #1
63
0
Please see picture.
Here is my method. Is it good?
T=m2g
m2g=m1a
a=m2g/m1
-n+Fpush=Ma. (n=normal force of m2 on M).
n=m2a=(m2)^2g/(m1a)

-n+Fpush=Ma.
Fpush=Ma+(m2)^2g/m
 

Attachments

  • Screen Shot 2014-04-15 at 8.47.25 PM.png
    Screen Shot 2014-04-15 at 8.47.25 PM.png
    19.7 KB · Views: 532

Answers and Replies

  • #2
SammyS
Staff Emeritus
Science Advisor
Homework Helper
Gold Member
11,377
1,038
Please see picture.
Here is my method. Is it good?
T=m2g
m2g=m1a
a=m2g/m1
-n+Fpush=Ma. (n=normal force of m2 on M).
n=m2a=(m2)^2g/(m1a)

-n+Fpush=Ma.
Fpush=Ma+(m2)^2g/m
No.

There is a horizontal force of T exerted to the left on M via the pulley.

There is also a downward force of T exerted on M, but that doesn't come into play here. It would if there was friction opposing the movement of M.

attachment.php?attachmentid=68689&d=1397609466.png
 
  • #3
63
0
Is the answer of the book wrong :O ?
 
  • #4
SammyS
Staff Emeritus
Science Advisor
Homework Helper
Gold Member
11,377
1,038
Is the answer of the book wrong :O ?

Their result looks correct.
 
  • #5
63
0
Perfect, I will come back to it!
What about this problem? I solved the force values:
I get -(5/8)M^2G/r^2 for the first attraction of the big mass and M^2G/(2r^2) for the attraction of the small mass.

Where did they get 5/16 and 1/4?
 

Attachments

  • Screen Shot 2014-04-15 at 10.43.15 PM.png
    Screen Shot 2014-04-15 at 10.43.15 PM.png
    21.5 KB · Views: 346
  • #6
SammyS
Staff Emeritus
Science Advisor
Homework Helper
Gold Member
11,377
1,038
Perfect, I will come back to it!
What about this problem? I solved the force values:
I get -(5/8)M^2G/r^2 for the first attraction of the big mass and M^2G/(2r^2) for the attraction of the small mass.

Where did they get 5/16 and 1/4?
They merely used proportions.

Gravitational force is directly proportional to mass & inversely proportional to the square of distance.

attachment.php?attachmentid=68693&d=1397616243.png
 
  • #7
63
0
Ok perfect!

Coming back to the previous question:
Would the equation for M be:

-T+Fpush-n=Ma?

Why is there a T? :O Is the pulley not frictionless?
 
  • #8
SammyS
Staff Emeritus
Science Advisor
Homework Helper
Gold Member
11,377
1,038
Ok perfect!

Coming back to the previous question:
Would the equation for M be:

-T+Fpush-n=Ma?

Why is there a T? :O Is the pulley not frictionless?

Yes, the equation is
-T+Fpush-n=Ma .​
Furthermore, T = m1a as well as m2 g .

Also, n = m2a .

Plugging those into -T+Fpush-n=Ma and solving for Fpush gives
Fpush = (M + m1 + m2)a​
in agreement with the book.

Yes, the pulley is frictionless, but there is thension T to the left on it an tension T downward on it. Both of these are transmitted to block M.
 

Related Threads on Dynamics Boxes 2

  • Last Post
Replies
3
Views
2K
  • Last Post
Replies
1
Views
1K
  • Last Post
Replies
7
Views
2K
  • Last Post
Replies
2
Views
3K
  • Last Post
Replies
2
Views
2K
  • Last Post
Replies
3
Views
1K
  • Last Post
Replies
1
Views
3K
Replies
1
Views
6K
  • Last Post
Replies
7
Views
1K
  • Last Post
Replies
3
Views
3K
Top