## Engineering Mechanics: Statics & Dynamics (14th Edition)

$a=\frac{1}{2}(1-\mu_k)g$
We can determine the required acceleration as follows: $\Sigma F_y=ma_y$ $\implies T-mg=-mg$ $\implies T=m(g-a)~~$ eq(1) Now we apply Newton's second law $\Sigma F_y=ma_y$ $\implies N-mg=0$ $\implies N=mg$ and $\Sigma F_x=ma_x$ $\implies T-\mu_k N=ma$ $\implies T-\mu_k mg=ma~~$ eq(2) We plug in the value of $T$ from eq(1) into eq(2) to obtain: $m(g-a)-\mu_k mg=ma$ This simplifies to: $a=\frac{1}{2}(1-\mu_k)g$