Question

A $1.0\mathrm{kg}$wood block is pressed against a vertical wood wall by the$12\mathrm{N}$force shown in $FIGUREP6.57$. If the block is initially at rest, will it move upward, move downward, or stay at rest?

Short answer

From the calculation, we see that the force in the downward direction is equal to the force in an upward direction, so the net force on the box is zero. Hence the box will stay at rest.

Step-by-step solution

Given

Mass of the wood block: $1.0\mathrm{kg}$.

The angle of the force applied: ${30}^{\circ }$.

Coefficient of static friction between wood and wood: $0.20$.

A figure:

Formula used.

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

Calculation

The force is applied at an angle of ${30}^{\circ }$. We will dissolve the applied force into horizontal and vertical components as,

$\begin{array}{r}{F}_h=F\cos {30}^{\circ }\\ F_v=F\sin {30}^{\circ }\end{array}$

According to the free diagram,

There are three forces acting at a time on a wooden box.

Calculation.

Three forces acting at a time on a wooden box are:

1. Gravitation force $(F=mg)\cdots$downward.
2. The vertical component of the force $(F_{\nu }=F\sin {30}^{\circ })$- upward.
3. Frictional force in the opposite of motion, $\left(F_f\right)$-upward.

$\begin{array}{rl}& mg=F_v+F_f\\ & mg=F\sin {30}^{\circ }+\mu F_k\\ & mg=F\sin {30}^{\circ }+\mu F\cos {30}^{\circ }\\ & 1\mathrm{kg}\times 9.8\mathrm{m}/{\mathrm{s}}^2=12(\sin {30}^{\circ }+0.4\times \cos {30}^{\circ })\\ & 9.8\mathrm{N}=12\mathrm{N}\left(0.8464\right)\\ & 9.8\mathrm{N}=10.15\mathrm{N}\\ & 9.8\mathrm{N}\approx 10.15\mathrm{N},\end{array}$

So, from the above calculation, we see that the force in the downward direction is equal to the force in an upward direction, so the net force on the box is zero. Hence the box will stay at rest.

Conclusion.

The woodblock will stay at rest.