Complete the statement (relates to the atwood machine)

[~christina~]

Homework Statement

Completing sentences...

When the unbalanced force increases (total mass remaining constant), the acceleration of the system increases

When the total mass that is accelerating increases (unbalanced force remains constant), the acceleration of the total system decreases

Homework Equations

F= ma ?

(this relates to the Atwood machine)

The Attempt at a Solution

above

 

[Astronuc]

Yes. F = m a applies where F is the net or unbalanced force, or for constant F, a must decrease if m increases.

http://hyperphysics.phy-astr.gsu.edu/hbase/atwd.html

 

[~christina~]

Thanks Astronuc :D

 

[Gokul43201]

This is a terrible question. In an Atwood's Machine, you can not change the total mass, and have the unbalanced forces be unchanged.

Proof:

[tex]a = \frac{M-m}{M+m}g [/tex]

[tex]\implies F_1 = Ma = \frac{M-m}{M+m}Mg ~~~~~~~(1)[/tex]

[tex]and ~F_2 = ma = \frac{M-m}{M+m}mg ~~~~~~~~~(2)[/tex]

Let us change the individual masses, so that [itex]M \longrightarrow M'~,~~m \longrightarrow m'[/itex]. Then again, we can write the equations for the unbalanced forces acting on the two blocks:

[tex]\implies F'_1 = M'a = \frac{M'-m'}{M'+m'}M'g ~~~~~(3)[/tex]

[tex]and ~F'_2 = m'a = \frac{M'-m'}{M'+m'}m'g ~~~~~~~(4)[/tex]

If the unbalanced forces are to remain unchanged, then [itex]F'_1=F_1~,~~F'_2=F_2 [/itex]. So that gives us:

[tex]\frac{M-m}{M+m}M=\frac{M'-m'}{M'+m'}M'~~~~~~~~~~~~~(5) [/tex][tex]\frac{M-m}{M+m}m=\frac{M'-m'}{M'+m'}m'~~~~~~~~~~~~~~(6) [/tex]

Dividing (5) by (6) gives:

[tex]\frac{M}{m}=\frac{M'}{m'}~\implies \frac{M}{M}=\frac{m}{m'} [/tex]

Call the latter ratios [itex]\alpha [/itex], so that we have [itex]M' =M \alpha~, ~~m'=m\alpha[/itex].

Making these substitutions in (5) and (6) gives [itex]1=\alpha[/itex].

In other words, if the unbalanced forces are to remain unchanged, the individual masses must also be unchanged.