Chapter 4 Problem 67

Problem

A block (mass $m_{A}$) lying on a fixed frictionless inclined plane is connected to a mass $m_{B}$ by a cord passing over a pulley.

(a) Determine a formula for the acceleration of the system in terms of $m_{A}$, $m_{B}$, $\theta$, and $g$.

(b) What conditions apply to masses $m_{A}$ and $m_{B}$ for the acceleration to be in one direction (say, $m_{A}$ down the plane), or in the opposite direction? Ignore the mass of the cord and pulley.

Solution

Notice there are two coordinate systems A and B! However, the connection between them is trivial due to cord, and you save time in the end.

(a)

notice that in the chosen coordinate systems, $a_{yA}=0$ and $a_{xB}=0$

$\sum F_{yA}=F_{N}-m_{A}g\cos \theta =0$ $\rightarrow F_{N}=m_{A}g\cos \theta$

$\sum F_{xA}=m_{A}g\sin \theta -F_{T}=m_{A}a_{xA}$

$\sum F_{yB}=F_{T}-m_{B}g=m_{B}a_{yB}\rightarrow F_{T}=m_{B}\left(g+a_{yB}\right)$

The connection between the blocks mean $a_{yB}=a_{xA}=a$

$m_{A}g\sin \theta -m_{B}(g+a)=m_{A}a\rightarrow m_{A}g\sin \theta -m_{B}g=m_{A}a+m_{B}a$

$a=g{\frac {\left(m_{A}\sin \theta -m_{B}\right)}{\left(m_{A}+m_{B}\right)}}$

(b)

If the acceleration is down the plane, $a>0$ :

$m_{A}\sin \theta <m_{B}$

if the acceleration is up the plane $a<0$

$m_{A}\sin \theta >m_{B}$