Rigid Object in Equilibrium Problem please

[gcombina]

A 3.0-kg ball and a 1.0-kg ball are placed at opposite ends of a massless beam so that the system is in equilibrium as shown. Note: The drawing is not drawn to scale. What is the ratio of the lengths, b/a?
(a) 2.0 (c) 3.0 (e) 5.0
(b) 2.5 (d) 4.0

 

[BvU]

Well, you formulated the problem, so we know what is asked in the exercise. What is your question, what are the relevant equations and what have you attempted to solve the problem?

 

[BvU]

I see you did well in an earlier exercise which is very, similar to this one. Now there are only three forces. Pick a good point to calculate the sum of torques comfortably, and that's all!

 

[gcombina]

lb = left ball
rb = right ball
SO I GOT 2 TORQUES
-(Tlb) (Llb) + - (Trb) (Lrb) = 0
-3kga - 1kgb = 0
-3kga = 1kgb
a = 1kgb/3kg
a= .33 b

 

[BvU]

Excellent! So b/a is in the list.
If you really want to delight physicists, you explicitly make clear that you did not forget g:
$\sum$ torques = 0 w.r.t. supported point $\Leftrightarrow$
m_a * g * (-L_a) + m_b * g * (L_b) = 0 $\Leftrightarrow$ etc.

 

[gcombina]

hahaha delight physicists? I am a language major trying to get into a physician assistance program and this course is just WAY TOO HARD

 

[gcombina]

OHHHH OK OK Let me try with the m.g

 

[BvU]

The more respect you earn by passing !
Remember: physicists are lazy nerds. But not insensitive to this subtle way of flattering.

 

[BvU]

As I said, lazy. the etc means that from there it's "only" math.
Math is even more important for physicians assistants: if they can prevent the doctor from making a factor 10 mistake in their recipes even only once in their career, all the effort was well worth it!

 

[gcombina]

prescriptions, not recipes...language major )