Chapter 4 Problem 59

Problem

Determine a formula for the magnitude of the force ${\vec {F}}$ exerted on the large block $m_{C}$ so that the mass $m_{A}$ does not move relative to $m_{c}$. Ignore all friction. Assume $m_{B}$ does not make contact with $m_{C}$

Solution

The force $F$ is accelerating the total mass, since it is the only force external to the system.

If mass $m_{A}$ does not move relative to $m_{C}$ , then all the blocks have the same horizontal acceleration, and none of the blocks have vertical acceleration.

$m_{B}$:

$\sum F_{x}=F_{T}\sin \theta =m_{B}a$

$\sum F_{y}=F_{T}\cos \theta -m_{B}g=0\rightarrow F_{T}\cos \theta =m_{B}g$

Taking the squares:

$F_{T}^{2}\sin ^{2}\theta =m_{B}^{2}a^{2}$ $;F_{T}^{2}\cos ^{2}\theta =m_{B}^{2}g^{2}$

and adding them together returns the relationship

$F_{T}^{2}\left(\sin ^{2}\theta +\cos ^{2}\theta \right)=m_{B}^{2}\left(g^{2}+a^{2}\right)$

$F_{T}^{2}=m_{B}^{2}\left(g^{2}+a^{2}\right)$

turning our attention to $m_{a}$:

$\sum F_{x}=F_{T}=m_{A}a$ $\rightarrow F_{T}^{2}=m_{A}^{2}a^{2}$

$\sum F_{y}=F_{N}-m_{A}g=0$

solve for $\left|{\vec {a}}\right|=a$

$F_{T}^{2}=m_{B}^{2}\left(g^{2}+a^{2}\right)=m_{A}^{2}a^{2}$ $\rightarrow a^{2}={\frac {m_{B}^{2}g^{2}}{\left(m_{A}^{2}-m_{B}^{2}\right)}}$ $\rightarrow a={\frac {m_{B}g}{\sqrt {m_{A}^{2}-m_{B}^{2}}}}\rightarrow$

$F=\left(m_{A}+m_{B}+m_{C}\right)a={\frac {m_{B}\left(m_{A}+m_{B}+m_{C}\right)}{\sqrt {m_{A}^{2}-m_{B}^{2}}}}g$