Aganju's answer is excellent, but I'd like to add one thing: The radius used is Earth's radius, which is applicable if you're at the equator. At higher latitudes, the effective radius is smaller by a factor $\cos\theta$, where $\theta$ is the latitude in radians.
For iMerchant, who seem to be located in Vancouver at
$$\theta = 49.3^{\circ}\times\pi/180 = 0.86\,\mathrm{rad},$$
the relevant radius is
$$
R_\mathrm{eff} = R_\oplus\cos\theta = 4150\,\mathrm{km},
$$
so at the $T=1.4\,\mathrm{h}$ of Aganju's answer (1.5 h is just a little too slow), the centrifugal (or -petal, if you prefer) acceleration would be only $g_\mathrm{spin} = 6.5\,\mathrm{m}\,\mathrm{s}^{-2}$.
Moreover, whereas Earth's acceleration is directed toward the center of Earth, in Vancouver the centrifugal acceleration would be directed at an angle $\theta$ toward South. Thus, only a component $g_\mathrm{spin} \cos\theta = 4.2\,\mathrm{m}\,\mathrm{s}^{-2}$ would point upward.
This is not enough to send iMerchant into space. However, a component $g_\mathrm{spin} \sin\theta = 4.9\,\mathrm{m}\,\mathrm{s}^{-2}$, or roughly half a G, would point horizontally toward South, enough to send him tumbling down the street, or at least walk awkwardly.
