= Solution
Let $m_p=0.250\,\mathrm{kg}$ and let $m_w$ be the hanging <mass>. Taking up-slope and downward respectively as positive,
$$
T-m_pg\sin\theta-\mu m_pg\cos\theta=m_pa,\qquad m_wg-T=m_wa.
$$
Eliminate the common <tension> to obtain
$$
\boxed{m_w=\frac{m_p\{a+g(\sin\theta+\mu\cos\theta)\}}{g-a}=0.130\,\mathrm{kg}.}
$$
Here $\theta=10^\circ$, $\mu=0.1$ and $a=1.60\,\mathrm{m\,s^{-2}}$. The pulley changes the direction of the string <force>; in the ideal model it does not change its magnitude.
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