• Support PF! Buy your school textbooks, materials and every day products Here!

NEED HELP! Rotational dynamics and static equilibrium

  • Thread starter edr2004
  • Start date
  • #1
7
0
[/B]1. Homework Statement [/b]
A cat walks along a uniform plank that is 4.00 m long and has a mass of 7.00 kg. The plank is supported by two sawhorses, one 0.44 m from the left end of the board and the other 1.58 m from its right end. When the cat nears the right end, the plank just begins to tip. If the cat has a mass of 5.4 kg, how close to the right end of the two-by-four can it walk before the board begins to tip?

Homework Equations


F1 + F2 -mg = 0;

The Attempt at a Solution


This problem is probably really easy but I have been looking at problems similar to it in my textbook and online and I think that's making me more confused. I have tried a lot of different things to solve this

For the left side of the 2nd sawhorse I need to find the length of the left side of the board. Is the length just 4-1.50 = 2.5 or is it the distance between the two sawhorses?. 4-(1.50+.44)= 1.98?
If it is the second case:
1.98m/4 = .495(7kg*9.81m/s2) = 33.99N
so if someone can tell me if I am on the right track I would really appreciate
 

Answers and Replies

  • #2
Doc Al
Mentor
44,892
1,143
Not sure I follow what you are doing, but consider this. When the plank begins to teeter, where is the pivot point? Consider torques about that pivot point.
 
  • #3
7
0
Well.....The pivot point is the second sawhorse (1.50m from the right). I know I need to find the torque but I am not sure how. To find the torque I know I need to find the mass on the left and right of the second sawhorse but that is where I am confused. For example on the left side the lenght is 2.5m. which is 62.5% of the board (2.5/4) so the mass on the left would be 7kg*.625 = 4.375kg *9.81m/s2 = 42.92N on the left. right?
 
  • #4
Doc Al
Mentor
44,892
1,143
Well.....The pivot point is the second sawhorse (1.50m from the right).
Good.
I know I need to find the torque but I am not sure how. To find the torque I know I need to find the mass on the left and right of the second sawhorse but that is where I am confused. For example on the left side the lenght is 2.5m. which is 62.5% of the board (2.5/4) so the mass on the left would be 7kg*.625 = 4.375kg *9.81m/s2 = 42.92N on the left. right?
The easy way to find the torque exerted by the weight of the plank is to treat the weight as acting at the plank's center of mass. How far is the center of the plank from the pivot point?
 
  • #5
1
0


Hey, I was working on the same problem and I have a picture that could help visualize things: http://img250.imageshack.us/img250/4881/1137aewn4.gif [Broken]
 
Last edited by a moderator:
  • #6
7
0
I thought i had figured this problem out
torque cat - torque of the board = 0
5.4X-(7*0.42) = 0
X = .544m but this is the wrong answer. I don't know what I am doing wrong!
 
  • #7
Doc Al
Mentor
44,892
1,143
I thought i had figured this problem out
torque cat - torque of the board = 0
5.4X-(7*0.42) = 0
X = .544m but this is the wrong answer. I don't know what I am doing wrong!
You did nothing wrong--you just need to express the answer exactly as specified in the problem statement. You found the distance from the pivot point, but they asked for the distance from the end of the board. :wink:

(This kind of thing happens a lot--always read the problem carefully to make sure you are answering the right question.)
 

Related Threads on NEED HELP! Rotational dynamics and static equilibrium

Replies
1
Views
4K
Replies
4
Views
8K
Replies
1
Views
3K
  • Last Post
Replies
7
Views
5K
  • Last Post
Replies
2
Views
4K
  • Last Post
2
Replies
31
Views
3K
  • Last Post
Replies
16
Views
720
Replies
3
Views
2K
Replies
1
Views
7K
Replies
1
Views
2K
Top