Vijesh should apply force $textbf{perpendicular}$ to the door plane and as $textbf{far away from the hinges}$ as possible.

Given:

$$
theta =55,,^{o}
$$

The lever is the same in both cases, the only difference is the projection of the force that is actively creating the torque:

$$
F_1cdot 1=F_2cdot cos (55,,^{o})
$$

This gives:

$$
F_2=frac{1}{cos (55,,^{o})}cdot F_1
$$

Finally:

$$
boxed{F_2=1.74cdot F_1}
$$

Push $74,,%$ harder.

Wheel with a mass of $m=12,,rm{kg}$ and diameter of $d=2.4,,rm{m}$ is being pulled on by two persons, each in their own direction. Person pulling in clockwise direction is pulling with force of $F_1=43,,rm{N}$ and the other in clockwise direction with force of $F_2=67,,rm{N}$. We need to calculate the net torque on the wheel.

Net torque on the wheel is external factor, meaning that it depends on the amount of net force exerted to it and its distance from the center. $textbf{It does not depend on the propertis of the wheel such as mass.}$ We can calculate the net torque as a combination of all torques acting on it:

$$
tau_{net}=tau_1+tau_2
$$

Since they are acting in opposite directions, torques will have opposite direction and therefore value. We will stick to the regular convention regarding the positive or negative torque definition where clockwise is negative, and counterclockwise is positive direction:

$$
tau_{net}=- F_1cdot (0.5cdot d) + F_2cdot (0.5cdot d)
$$

$$
tau_{net}=- 43cdot 0.5cdot 2.4 + 67cdot 0.5cdot 2.4
$$

Finally we get.

$$
boxed{tau_{net}=28.8,,rm{Nm}}
$$

$$
tau_{net}=28.8,,rm{Nm}
$$

From least to greatest we have:

– Sphere $(frac{2}{5} mr^2)$

– Solid disk $(frac{1}{2} mr^2)$

– Wheel with mass on the rim $(mr^2)$

The advantage of using the one with the least moment of inertia comes from the equation:

$$
tau=Ialpha
$$

Which shows that with equal torque we can achieve higher angular acceleration which will enable us to reach wanted angular velocity quicker or that we can use less torque.

Given:

$$
begin{align*}
F&=13,,rm{N}\
r&=0.15,,rm{m}\
omega &=14,,rm{rev/min}\
t&=4.5,,rm{s}
end{align*}
$$

First, lets calculate the torque:

$$
begin{align*}
tau &=Fr\
tau &=13cdot 0.15\
tau &=1.95,,rm{Nm}
end{align*}
$$

Now lets calculate the angular acceleration:

$$
begin{align*}
alpha &=frac{omega}{t}\
alpha &=frac{14cdot 2cdot pi}{60cdot 4.5}\
alpha &=0.3258,,rm{rad/s^2}
end{align*}
$$

Finally we can calculate moment of inertia:

$$
begin{align*}
I &=frac{tau}{alpha}\
I &=frac{1.95}{0.3258}\
end{align*}
$$

$$
boxed{I=5.96,,rm{kgm^2}}
$$

$$
I=5.96,,rm{kgm^2}
$$

In the first case, the ball will slide downhill because the force of friction is causing $textbf{negligible}$ torque which is not enough to cause significant angular acceleration, and therefore the ball will slide faster than it will rotate.

In the second case, the force of friction will cause $textbf{significant}$ torque which will then cause higher angular acceleration and therfore the ball will reach bigger angular velocitiy. This will cause the ball to rotate faster, rotating then the exact speed it is moving downhill.