Object in equillibrium

  • Thread starter smray7
  • Start date
  • #1
6
0

Homework Statement



A light plank 2 meters long rests on two scales with a girl laying across the plank. The mass on the left(head) is 380N and at the other end(feet), the mass is 320N. How far from the girl's feet is the center of mas?

Homework Equations



net F = 0
net T = 0

because this object is in equillibrium so all torque/forces equal zero

The Attempt at a Solution



I have free body diagrams containing total forces. I just look at this as if there are 2 blocks instead of a body.

m1g + m2g=0

m1g (2-x) + m2g(2-x)=0 i know for torque i need the force to be multiplied by some distance.

would the mass of the plank matter? that would be another unknown that i'm not even sure i could solve for. do i need to choose an rotation point?
 

Answers and Replies

  • #2
tiny-tim
Science Advisor
Homework Helper
25,832
251
welcome to pf!

hi smray7! welcome to pf! :smile:
A light plank…

would the mass of the plank matter? that would be another unknown that i'm not even sure i could solve for. do i need to choose an rotation point?
"light" is one of those code-words that have the specific meaning in exam questions that you can ignore whatever they apply to …

the question is telling you to treat the mass of the plank as zero :wink:

and any rotation points will work, but some make the maths simpler than others :smile:

(since the required answer is "the distance from the girl's feet", i'd choose the girl's feet!)
I have free body diagrams containing total forces. I just look at this as if there are 2 blocks instead of a body.

m1g + m2g=0
erm :redface: … the question is simple enough already, if you simplify it any further you'll risk losing information :wink:

there are two upward forces on the body, and one downward force (acting through the centre of mass) …

how could two forces add to zero if they're in the same direction? :smile:
 
Last edited:
  • #3
6
0
ok so maybe i was over thinking it and misreading it.
newtons are forces not mass.
and you're right, what i had before doesn't equal to zero.

now i have:

Fy: N1 + N2 - mg = 0
T: N1(2) - mg(2-x) + N2(0) = 0

this seems simple but i dont know, i cant get it.
 
  • #4
tiny-tim
Science Advisor
Homework Helper
25,832
251
hi smray7! :wink:
Fy: N1 + N2 - mg = 0
T: N1(2) - mg(2-x) + N2(0) = 0

this seems simple but i dont know, i cant get it.
but you're there

just bung the numbers in! :smile:
 
  • #5
6
0
thanks t.t.! i got it.

i do have another question on a problem involving a rotating disk.
how do i relate
[tex]\tau[/tex] = I[tex]\alpha[/tex] to F = [tex]\mu\mu[/tex]N ?

i need to find [tex]\mu[/tex] as a force of 70N is applied to the disk.

all work for torque i have is completed. i just need to do the above.
 
  • #6
tiny-tim
Science Advisor
Homework Helper
25,832
251
hi smray7! :smile:

(have an alpha: α and a mu: µ and a tau: τ :wink:)

is this a force of 70 N applied radially to the disc?

then the friction force is 70µ N, and the torque is 70µr N :smile:
 

Related Threads on Object in equillibrium

  • Last Post
Replies
6
Views
5K
  • Last Post
Replies
1
Views
1K
  • Last Post
Replies
4
Views
896
Replies
2
Views
4K
  • Last Post
Replies
6
Views
2K
  • Last Post
Replies
6
Views
9K
  • Last Post
Replies
8
Views
1K
  • Last Post
Replies
6
Views
1K
  • Last Post
Replies
12
Views
1K
Top