91Ó°ÊÓ

Force acting on any object is the product of the mass of the object m and the acceleration of that moving object a.

$$\begin{gathered} \underset{}{∑F=ma} \\ mg=T=ma \end{gathered}$$

Here, T is tension on the object and g is gravity.

The angular acceleration of any moving body can be described as the changing rate of angular velocity of the body with respect to the time. The unit of angular acceleration is radians per square second.

The angular acceleration of the gritstone will be by the formula of motion,

$$\begin{gathered} {\mathrm{Ó\neg }}_z={\mathrm{Ó\neg }}_{0z}+{\mathrm{Î\pm }}_{z}t \\ \underset{}{∑{\mathrm{Ï„}}_z=I{\mathrm{Î\pm }}_z} \\ {f}_k={\mathrm{μ}}_{k}n \\ {\mathrm{μ}}_k=-\frac{MR{\mathrm{Î\pm }}_z}{2n} \end{gathered}$$

Where, $n$is force, ${\mathrm{Ó\neg }}_z$final rotation speed, ${\mathrm{Ó\neg }}_{0z}$starting rotation speed, ${\mathrm{Î\pm }}_z$rotation acceleration, ${\mathrm{Ï„}}_z$torque, ${\mathrm{μ}}_k$coefficient of friction, $f_k$friction force, $I$moment of inertia, and $t$time. $MandR$mass and radius of the object.

Radius R = $0.25m$

Length l = 0.12 m

$massm=50kg$

Moment of inertia = $2.9kg.m^2$

Acceleration = $1.40m/s^2$.

From the formula of the force, the tension will be,

$$\begin{gathered} \mathrm{Ï„}=m\left(g+a\right) \\ \mathrm{Ï„}=50.00kg\left(9.80m/s^2+1.40m/s^2\right) \\ \mathrm{Ï„}=560N \end{gathered}$$

When the crank is turned in the question, end of the handle rotates about the axle in a vertical circle , the cylinder turns, and the crate is raised, so the magnitude of the force F applied tangentially to the rotating crank is required to raise the crate with an acceleration of $1.40m/s^2$will be,

localid="1667811086452" $$\begin{gathered} \underset{}{∑}{\mathrm{Ï„}}_z=I{\mathrm{Î\pm }}_z,{\mathrm{Ï„}}_z=Fl-TRand{\mathrm{Î\pm }}_z=\left(\frac{a}{R}\right) \\ Fl-TR=I\left(\frac{a}{R}\right) \\ F=\frac{\left(Ia\right)}{lR}+\frac{TR}{l} \\ F=\left(\frac{2.9k.g{m}^2×1.40m/s^2}{0.12m×0.25m}\right)+\left(\frac{560N×0.25m}{0.12m}\right) \end{gathered}$$

$F=1302N$

SO, the force will be $F=1302N$.