The ${\mu }_{\mathrm{k}}=0.30$wood box in FIGURE $P6.56$slides down a vertical wood wall while you push on it at a $2.0\mathrm{kg}$angle. What magnitude of force should you apply to cause the box to slide down at a constant speed?

FIGURE $P6.56$

Mass of the wood box: $2.0\mathrm{kg}$.

The angle of the force applied: ${45}^{\circ }$.

Coefficient of friction between wood and wood : localid="1647680350969" $0.20$.

A figure:

$\begin{array}{r}F=mg\\ F_f=\mu N=\mu F_h\end{array}$

The force is applied at an angle of ${45}^{\circ }$We will dissolve the applied force into horizontal and vertical components as

$\begin{array}{r}{F}_h=F\cos {45}^{\circ }\\ F_v=F\sin {45}^{\circ }\end{array}$

According to the free body diagram.

Now,

There are three forces acting at a time on the wood box. These forces are

1. Gravitation force $(F=mg)\cdots$downward

2. Vertical component of the force $(F_v=F\sin {45}^{\circ })\cdots$upward

3. Frictional force (in opposite of motion, $\left(F_f\right)$-upward

As, the box is sliding downward with constant speed. So

$\begin{array}{rl}& F_v+F_f-mg=0\\ & mg=F_v+F_f\\ & mg=F\sin {45}^{\circ }+\mu N\\ & mg=F\sin {45}^{\circ }+\mu F\cos {45}^{\circ }\\ & mg=F(\sin {45}^{\circ }+\mu \cos {45}^{\circ })\\ & 2\mathrm{kg}\times 9.8\mathrm{m}/{\mathrm{s}}^2=F(0.707+0.2\times 0.707)\\ & \frac{19.6}{0.8434}=F=23.10\mathrm{N}\end{array}$

$23.10N$Force should be applied to the box to slide down at constant speed.

$23.10N$Force should be applied to the box to slide down at constant speed.