91Ó°ÊÓ

At the point of contact with the wall, the wall is going to exert a friction force f directed in the downward direction and a normal force n to the right,

Thus,

There is a net force of zero network on the roll.

Here, F is the force ,x is the horizontal axis, y is the vertical exercise, n is the normal reaction, is the net downward force, and f is the force of friction

So the balancing vertical and horizontal forces would be correct, that is,

$$\begin{gathered} F_{rod}cos\mathrm{θ}=f+w+f \\ And, \end{gathered}$$

$F_{rod}sin\mathrm{θ}=n$

The expression for the force of friction is as follow:

$f={\mathrm{μ}}_{k}n$

Substitute ${\mathrm{μ}}_{k}n$for in the equation

$$\begin{gathered} F_{rod}cos\mathrm{θ}=f+w+F \\ {F}_{rod}cos\mathrm{θ}={\mathrm{μ}}_{k}n+w+F \\ Substitute{F}_{rod}fornintheequation{F}_{rod}cos\mathrm{θ}={\mathrm{μ}}_{k}n+w+faandsolvefor{F}_{rod} \\ {F}_{rod}cos\mathrm{θ}={\mathrm{μ}}_k\left(F_{rod}\sin \mathrm{θ}\right)+w+f \\ {F}_{rod}cos\mathrm{θ}-{\mathrm{μ}}_k\left(F_{rod}sin\mathrm{θ}\right)=w+f \\ {F}_{rod}=(w+f)/(cos\mathrm{θ}-{\mathrm{μ}}_{k}sin\mathrm{θ}) \end{gathered}$$

Calculate the magnitude of the force that rod exert on the paper as follow:

$$\begin{gathered} F_{rod}=\frac{w+F}{cos\mathrm{θ}-{\mathrm{μ}}_{k}sin\mathrm{θ}} \\ \mathrm{Substituting}\mathrm{mg}\mathrm{for}\mathrm{w}\mathrm{in}\mathrm{the}\mathrm{equation} \\ {F}_{rod}=\frac{w+F}{cos\mathrm{θ}-{\mathrm{μ}}_{k}sin\mathrm{θ}} \\ {F}_{rod}=\frac{mg+F}{cos\mathrm{θ}-{\mathrm{μ}}_{k}sin\mathrm{θ}} \end{gathered}$$

$$\begin{gathered} F_{rod}=(mg+F)/(cos\mathrm{θ}-{\mathrm{μ}}_{k}sin\mathrm{θ}) \\ {F}_{r}od=\left(\right(16kg\left)\right(9.8 m/s^2)+60 N)/(cos{30}^{∘}-(0.25\left)sin{30}^{∘}\right) \\ =\left(\right(16kg\left)\right(9.8 m/s^2)+60 N)/\left(\right(cos{30}^{∘}-(0.25)sin{30}^{∘}\left)\right) \\ =292.6 N \\ =293 N \end{gathered}$$

Hence, the force that the rod exert is localid="1667809565437" $\boxed{293N}$.

With respect to the center of the roll, both the rod and the normal force exert zero torque.

The magnitude of net torque is,

$$\begin{gathered} \mathrm{Ï„}=(F-f)R \\ Alsotheexpressionfor\mathrm{Ï„}isasfollow: \\ \mathrm{Ï„}=I\mathrm{Î\pm } \\ Equatingtheequation\mathrm{Ï„}=(F-f)Rand\mathrm{Ï„}=I\mathrm{Î\pm }. \\ (F-f)R=I\mathrm{Î\pm } \end{gathered}$$

$$\begin{gathered} Thefrictionforceis,f={\mathrm{μ}}_{k}n \\ ={\mathrm{μ}}_k\left(F_{rod}sin\mathrm{θ}\right) \\ \mathrm{Substitute}0.25\mathrm{for}{\mathrm{μ}}_{\mathrm{k}},293\mathrm{N}\mathrm{for}{\mathrm{F}}_{\mathrm{rod}},\mathrm{and}{30}^{∘}\mathrm{for}\mathrm{θ}\mathrm{in}\mathrm{the}\mathrm{equation} \\ \mathrm{f}={\mathrm{μ}}_{\mathrm{k}}\left({\mathrm{F}}_{rod}sin\mathrm{θ}\right) \\ f=(0.25)\left(\right(293 N\left)sin{30}^{∘}\right) \\ =36.63 N \\ \mathrm{Rearrange}\mathrm{the}\mathrm{equation}(F-f)R=I\mathrm{Î\pm }for\mathrm{Î\pm } \\ \mathrm{Î\pm }=\frac{(F-f)R}{I} \end{gathered}$$

$$\begin{gathered} \mathrm{Substituting}60NforFand36.63Nforf,18cmforR,and0.260 kg.m^{2}forIin \\ \mathrm{the}\mathrm{equation}\mathrm{Î\pm }=\frac{(F-f)R}{I} \\ \mathrm{Î\pm }=\frac{(60 \mathrm{N}-36.63 \mathrm{N})\left(18 \mathrm{cm}\left({\displaystyle \frac{{10}^2\mathrm{m}}{1 \mathrm{cm}}}\right)\right)}{0.260 \mathrm{kg}.{\mathrm{m}}^2} \\ =16.179\mathrm{rad}/{\mathrm{s}}^2 \\ =16.2 \mathrm{rad}/{\mathrm{s}}^2 \\ \mathrm{Hence},\mathrm{the}\mathrm{angular}\mathrm{acceleration}\mathrm{is}\mathrm{equal}\mathrm{to}\boxed{16.2 \mathrm{rad}/{\mathrm{s}}^2}. \end{gathered}$$