91影视

Weight of box is,$W=890\text{鈥�}N$ .

Using the first equilibrium condition given below, we can find the relation between F(external force) and$\mathbf{}{\mathit{F}}_{\mathbf{f}}$(friction force).Then, using the second equilibrium condition, we can find the ${\mathit{F}}_{\mathbf{N}}$(contact force) in terms of W (Weight of the object). Using the third equilibrium condition and the above two relations, we can find $\mathit{F}$and$\mathbf{}\mathit{渭}$ (coefficient of friction).The smallest possible force that would have to be applied directly to the box to roll it can be calculated by using equilibrium conditions for torque when$\mathit{F}$is acting in the upward right direction.

The free Body Diagram can be drawn as,

Using the condition of equilibrium in the x direction, we can write,

$鈭�F_x=0$

Substitute the values in the above expression, and we get,

$\begin{array}{c}F鈭�F_f=0\\ F=F_f\end{array}$ (1)

Using the condition of equilibrium in the x direction, we can write,

$鈭�F_y=0$

Substitute the values in the above expression, and we get,

$\begin{array}{c}{F}_N鈭�W=0\\ F_N=W\end{array}$ (2)

Now the net torque is zero, in this case, then we can write is as,

$鈭�蟿=0$

Substitute the values in the above expression, and we get,

$\begin{array}{c}F\left(L\right)鈭�W\left(\frac{L}{2}\right)=0\\ F=\frac{W}{2}\end{array}$

Substitute the values in the above expression, and we get,

$\begin{array}{c}F=890/2\\ =445\text{鈥�}N\end{array}$

Thus, the magnitude of the minimum force required is$445\text{鈥�}N$ .

Substitute the values in equations 1 and 2, and we get,

$F_N=890\text{鈥�}N$

And

$F_f=445\text{鈥�}N$

The coefficient of friction can be calculated from the formula as,

$渭=\frac{F_f}{F}$

Substitute the values in the above expression, and we get,

$\begin{array}{c}渭=\frac{445}{890}\\ =0.5\end{array}$

Thus, the minimum coefficient of friction force between box and floor is$0.5$.

Free Body Diagram

The box can be rolled with a smaller applied force if the force points upwards and to the right.Let$胃$be the angle the force makes with horizontal.The torque equation then becomes:

$\begin{array}{c}FL\cos 胃+FL\sin 胃鈭�\frac{WL}{2}=0\\ F=\frac{W}{2(\cos 胃+\sin 胃)}\end{array}$

We are choosing $胃=45掳$so that$\cos 胃+\sin 胃$will be the maximum, and we will be able to calculate the smallest possible force.

Substitute the values in the above expression, and we get,

$\begin{array}{c}F=\frac{890}{2(\cos 45掳+\sin 45掳)}\\ =315\text{鈥塏}\end{array}$

Thus, the smallest possible force is $315\text{鈥�}N$.