# Problem №2 (15 points)

A bar with a mass of $m=1$ kg is pressed against a vertical wall. The coefficient of friction between the bar and the wall is $\mu=0.1$. Determine the force that needs to be applied to the bar at an angle of $\alpha=60^{\circ}$ to the wall to keep the bar in equilibrium.
# Solution:

We need to consider two situations:

## First:

![](https://cdn.mathpix.com/cropped/2024_05_06_167c0776c22cdf752babg-05.jpg?height=436&width=305&top_left_y=730&top_left_x=93)

Newton's second law in projections on the axes:

$F \sin 60^{\circ}=N$

$F \cos 60^{\circ}=m g+F_{T P}$

Taking into account that $F_{T P}=\mu N$, we get:

$F \cos 60^{\circ}=m g+\mu F \sin 60^{\circ}$

$F \approx 24.18 H$

## Second:

![](https://cdn.mathpix.com/cropped/2024_05_06_167c0776c22cdf752babg-05.jpg?height=489&width=340&top_left_y=1877&top_left_x=90)

Newton's second law in projections on the axes:

$F \sin 60^{\circ}=N$

$F \cos 60^{\circ}+F_{T P}=m g$

Taking into account that $F_{T P}=\mu N$, we get:
$F \cos 60^{\circ}+\mu F \sin 60^{\circ}=m g$

$F \approx 17.05 H$

(2 points)

Final answer:

$17.05 H \leq F \leq 24.18 H$

(3 points)