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Question Number 30994 by ajfour last updated on 01/Mar/18

Commented by ajfour last updated on 01/Mar/18

Commented by ajfour last updated on 01/Mar/18

$P l e a s e s o l v e . F r e e b o d y d i a g r a m s$ $i h a v e a t t a c h e d a l o n g w i t h .$
Commented by ajfour last updated on 02/Mar/18

$m g \sin α + m A \cos α - μ_{1} N = {m a}_{r e l}$ $N + m A \sin α = m g \cos α$ $R = N \cos α + M g + μ_{1} N \sin α$ $N \sin α - μ_{1} N \cos α - μ_{2} R = M A$ $\Rightarrow N = m g \cos α - m A \sin α$ $& N \sin α - μ_{1} N \cos α$ $- μ_{2} (N \cos α + M g + μ_{1} N \sin α) = M A$ $S o$ $N (\sin α - μ_{1} \cos α - μ_{2} \cos α - μ_{1} μ_{2} \sin α)$ $- μ_{2} M g = M A$ $(m g \cos α - m A \sin α) (\sin α - μ_{1} \cos α - μ_{2} \cos α - μ_{1} μ_{2} \sin α)$ $= μ_{2} M g + M A$ $A = \frac{m g \cos α (\sin α - μ_{1} \cos α - μ_{2} \cos α - μ_{1} μ_{2} \sin α) - μ_{2} M g}{M + m \sin α (\sin α - μ_{1} \cos α - μ_{2} \sin α - μ_{1} μ_{2} \sin α)}$ $= \frac{20 \times \frac{4}{5} (\frac{3}{5} - 0.35 \times \frac{4}{5} - 0.025 \times \frac{3}{5}) - 0.1 \times 80}{8 + 2 \times \frac{3}{5} (\frac{3}{5} - 0.35 \times \frac{4}{5} - 0.025 \times \frac{3}{5})}$ $= \frac{16 (0.6 - 0.28 - 0.015) - 8}{8 + 1.2 (0.6 - 0.28 - 0.015)}$ $= \frac{16 \times 0.305 - 8}{8 + 1.2 \times 0.305} < 0$ $\Rightarrow A = 0 .$
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