Two blocks over a pulley

  • Thread starter oneamp
  • Start date
  • #1
oneamp
219
0

Homework Statement



The figure shows a 100 kg block being released from rest from a height of 1.0 m. It then takes 0.90 s to reach the floor. What is the mass of the block on the left?

Illustration shows a pulley connected to the ceiling, with a block of 100 kg on the right, and a block of unknown mass on the left.


Homework Equations





The Attempt at a Solution



I used this to find acceleration of the system:
0 = 1 + (0.5)(a) [(0.90)^2]
a = (-2.5) m/s^2

I drew a diagram. F_G on the bottom, 100 * 9.8 = 980N down.

F = ma so (100)(2.5) = 250, and that must mean that the vector pointing up in my diagram will be 730N (980 - 250 = 750).

So, the 750 comes from the other block. F = ma again, solve for m I get 292 kg, which is not the right answer.

Where did I go wrong?

Thank you
 

Answers and Replies

  • #2
nasu
3,957
582

Homework Statement



The figure shows a 100 kg block being released from rest from a height of 1.0 m. It then takes 0.90 s to reach the floor. What is the mass of the block on the left?

Illustration shows a pulley connected to the ceiling, with a block of 100 kg on the right, and a block of unknown mass on the left.


Homework Equations





The Attempt at a Solution



I used this to find acceleration of the system:
0 = 1 + (0.5)(a) [(0.90)^2]
a = (-2.5) m/s^2

I drew a diagram. F_G on the bottom, 100 * 9.8 = 980N down.

F = ma so (100)(2.5) = 250, and that must mean that the vector pointing up in my diagram will be 730N (980 - 250 = 750).
Up to here is (almost) OK.
You found the tension in the rope. It is not clear if it's 730N or 750N (it depends on taking g=9.8 or 10 m/s^2).


So, the 750 comes from the other block. F = ma again, solve for m I get 292 kg, which is not the right answer.

Where did I go wrong?

Thank you
Now write Newton's second law for the other block. You know the tension in the rope and the acceleration (same as for the first block).
 
  • #3
Legaldose
74
6
The way I deal with Atwood problems is to first translate the vertical forces to horizontal ones, it makes it easier for me to deal with the forces involved, and gets rid of the confusion of having the pulley. Here's a picture of what I mean. Try applying Newton's Laws to the system on the right.
 

Attachments

  • grr.jpg
    grr.jpg
    23.6 KB · Views: 2,386
  • #4
oneamp
219
0
Thank you :)
 

Suggested for: Two blocks over a pulley

Replies
4
Views
683
Replies
1
Views
6K
  • Last Post
Replies
3
Views
2K
  • Last Post
Replies
6
Views
9K
  • Last Post
Replies
4
Views
2K
  • Last Post
Replies
2
Views
2K
  • Last Post
Replies
5
Views
3K
  • Last Post
Replies
14
Views
4K
  • Last Post
Replies
1
Views
2K
Replies
5
Views
17K
Top