Newton's Third Law and FreeBody Diagram. Solutions provided
A force of 14.0N is applied to a block as shown in the diagram below
!
If the coefficient of friction is between the surface and the blocks is 0.35
a) What is the acceleration of the two blocks?
b) What is the force that the 3.0kg exerts on the 1.0kg block?
Solutions Given a)
$\sum F = 14.0N - 13.7N = 0.28N$
$\frac{0.28N}{4.0kg} = 0.07m/s^2 = a$
b)
$\sum F = (1.0kg)(0.07m/s^2) = 0.070N$
$F_{fric} = (0.35)(1.0kg)(9.8m/s^2) = 3.43N$
$F = 0.07N + 3.43 = 3.5N$
Confusion
Here is the thing, but my never show (probably due to print cost?) the free-body diagrams. I want to understand a systematic way of solving this, that is, draw FBD and then apply <PERSON>'s Second law. 512
I already know that they first considered the whole system as "one mass" and they applied the 2nd Law and then the third law on each. I want to do this with a Free-Body Diagram.
Here is my attempt. I am going to denote the masses as 1 and 3 in the subscripts.
Here are my freebody diagrams.
For the block with mass 3kg
!
For the block with mass 1kg
!
My equations should be
$\sum F = F - (\mu m_3 g + F_{13}) = m_3 a$
$\sum F = F_{31} - \mu m_1 g = m_1 a$
Here is the problem, if I attempt to solve this I get (add the two equations together)
$F - \mu (m_1 + m_3)g - F_{13} + F_{31} = (m_1 + m_3)a $
If I were to go ahead and plug in the values I get
$14N - 13.72N - F_{13} + F_{31} = 4a $
$0.28N - F_{13} + F_{31} = 4a $
How come I can't get the same answer this way? Am I missing another equation? Perhaps a constraint? It would seem to make sense if it happens that $|F_{13}| - |F_{31}| = 0$
But when I broke it into components like this, I am dealing with the vectorial components, not the magnitiude. So really vectorically, I need $\vec{F}{13} = -\vec{F}{31}$
So I actually get if I plug in the vector force
Thank you 512