A side view of a simplified form of vertical latch B is as shown. The lower member A can be pushed forward in its horizontal channel. The sides of the channels are smooth, but at the interfaces of A and B, which are at 45° with the horizontal, there exists a static coefficient of friction µ = 0.4. What is the minimum force F (in N) that must be applied horizontally to A to start motion of the latch B if it has a mass m = 0.6 kg ?

Engineering

Physics

Friction

Question

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For equilibrium of block B
$Σ$ F_x = 0   and $Σ$ F_y = 0
$Σ$ F_y = 0;   mg + $\frac{μN}{\sqrt{2}} = \frac{N}{\sqrt{2}}$
           N = $\frac{\sqrt{2} mg}{1 - μ}$
For equilibrium of Block A
$Σ$ F_x = F – $\frac{N}{\sqrt{2}} - \frac{μN}{\sqrt{2}}$ = 0
or   F = $\frac{N}{\sqrt{2}}$ [1 + µ] = mg $\frac{(1 + μ)}{(1 - μ)}$ = 0.6 × 10 $\frac{(1 + 0.4)}{(1 - 0.4)} = \frac{6 \times 1.4}{0.6}$ = 14 N

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