When the object is stationary on an surface, the static frictional force balances the force component that ia attempting to slide the object along the surface. (3) The Eq 6-1 `(f_(s. "max")=mu_(s)F_(N))`. (3) If the component of the applied force along the surface excceds this limit on the satatic friction, the block begins to slide. (4) if the object slides, the kinetic frictional force is given by Eq. 6-2 `(f_(k)=mu_(k)F_(N))`.
Calculations: To see if the block slides (and thus to calculate the magnitude of frictional force), we must compare the applied force component `F_(x)` with the maximum magnitude `f_(s,max)` that the static friction can have.
From the triangle of components and full force shown in Fig.6-3b, we see that
`F_(x)=Fcostheta`
`=(12.0N)cos30^(@)=10.39N`.
From Eq. 6-1, we know that `f_(s,max)=mu_(s)F_(N)`, but we need the magnitude `F_(N)` of the normal force to evaluate `f_(s,max)`
(a) A force is applied to an initially stationary block. (b) The componets of the applied force. (c) The vertical force components. (d) The horzontal force compnents.
Because the normal force is vertical, we need to write Newton.s second law `(F_("net.y")=ma_(y))` for the vertical force componeds acting on the block has as a downward componet `F_(y)=Fsintheta`. And the vertical acceleration `a_(y)` is just zero. Thus, we can write Newton.s secon law as
`F_(N)-mg-Fsintheta=m(0)" "(6-4)`
which gives us
`F_(N)=mg+fsintheta" "(6-5)`
Now we can evaluate `f_(s,max)=mu_(s)F_(N)`
`f_(s,max)=mu_(s)(mg+Fsintheta)`
`=(0.700)(8.00kg)(9.8m//s^(2))+(12.0N)(sin30^(@))`
=59.08 N.
Because the magnitude `F_(x)(=10.39N)` of the force-component attempting to slide the block is less than `f_(s,max)` (59.08N), the block remains stationary. That means that the magnitude `f_(s)` of the frictional force matches `F_(x)`. From Fig 6-3 d, we can write Newton.s second law for x components as
`F_(x)-f_(s)=m(0),`
and thus `f_(s)=F_(x)=10.39N~~10.4N`.
