Physics Problems & Solutions: Classical Mechanics

For the system consisting of the two blocks shown in the figure, the minimum horizontal force F is applied so that block B does not fall under the influence of gravity. The masses of A and B are 16.0 kilograms and 4.00 kilograms, respectively. The horizontal surface is frictionless and the coefficient of friction between the two blocks is 0.50. The magnitude of F is most nearly

A. 50 N
B. 100 N
C. 200 N
D. 400 N
E. 1,600 N

(GR0177 #73)

Solution:

$F=(m_A+m_B)a \rightarrow a=\frac{F}{m_A+m_B}$

If B does not fall under the influence of gravity:

$\mu N=m_B g \rightarrow N=\frac{m_B g}{\mu}$

And,

$\\F=F_A+N \rightarrow N=F-F_A \\N=F-m_A a \\N =F-m_A \bigg(\frac{F}{m_A+m_B}\bigg) \\N=F\bigg(1-\frac{m_A}{m_A+m_B} \bigg)=\frac{Fm_B}{m_A+m_B}$

Therefore,

$\\\frac{m_B g}{\mu}=\frac{Fm_B}{m_A+m_B} \\\\F=\frac{(m_A+m_B)g}{\mu}=\frac{(20)(10)}{0.5}=400$ N$

Answer: D

Physics Problems & Solutions

Classical Mechanics - Newton's Law

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