Problem No. 5 (15 points)

The system shown in the figure is in equilibrium. It is known that the uniform rod $AB$ and the load lying on it have the same mass $m=10$ kg, and the load is located at a distance of one quarter of the length of the rod from its left end. Determine the mass $m$ of the second load suspended from the center of one of the pulleys. The threads and pulleys are weightless, and there is no friction in the pulley axes.

![](https://cdn.mathpix.com/cropped/2024_05_06_3c4b08bfdadd8858901bg-1.jpg?height=728&width=585&top_left_y=1115&top_left_x=770)
Answer: 100 kg

# Solution and grading criteria: 

Diagram with forces correctly placed.

Tension force in the thread: $T=\frac{M g}{2}$

The lever rule, written relative to point v:

$m g \cdot \frac{1}{2} A B+m g \cdot \frac{3}{4} A B+T \cdot \frac{3}{4} A B=T \cdot A B$

As a result, we get: $M=10 m=100$ kg.

A person walked slowly in a straight line at a speed of $v_{1}=1$ m/s. At time $t_{1}$, he turned strictly to the right and walked at a speed twice as fast. After another time interval $t_{1}$, he turned strictly to the right again, and his speed became three times the initial speed. After the last right turn, he ran at a speed of $4 v_{1}$ and returned to the starting point of his trajectory. Determine his average speed over the entire path.

## Answer: $2.1 \mathrm{~m} / \mathrm{s}$

## Solution and grading criteria:

From the condition, we can conclude that the person's trajectory is a rectangle.

(2 points)

![](https://cdn.mathpix.com/cropped/2024_05_06_3c4b08bfdadd8858901bg-2.jpg?height=377&width=663&top_left_y=982&top_left_x=728)

Moreover, $S_{2}=2 S_{1}$

Average speed:

$v_{\text{avg}}=\frac{S_{1}+S_{2}+S_{3}+S_{4}}{t_{1}+t_{2}+t_{3}+t_{4}}=\frac{6 S_{1}}{t_{1}+t_{1}+\frac{S_{1}}{3 v_{1}}+\frac{2 S_{1}}{4 v_{1}}}=\frac{6 S_{1}}{t_{1}+t_{1}+\frac{t_{1}}{3}+\frac{t_{1}}{2}}=\frac{6 S_{1}}{\frac{17}{6} t_{1}}=\frac{36}{17} v_{1} \approx 2.1 \mathrm{~m} / \mathrm{s}$

## Problem 7 (10 points)

A person holds one end of a light board, and the other end of the board rests on a cylinder lying on the ground. The length of the board is $L=10$ m. How far will the person walk before meeting the cylinder? The radius of the wheel is much smaller than the length of the board. There is no slipping.

![](https://cdn.mathpix.com/cropped/2024_05_06_3c4b08bfdadd8858901bg-2.jpg?height=188&width=1026&top_left_y=2213&top_left_x=535)

Answer: $20 \mathrm{~m}$

Solution and grading criteria:

The speed of the person is equal to the speed of the top point of the cylinder, which is twice the speed of the center of the cylinder.

Thus, the distance the person will walk is twice the distance the cylinder has rolled.

Obviously, $S_{\text {person }}-S_{\text {cylinder }}=L=10 \mathrm{~m}$ (3 points)

We get: $S_{\text {person }}=20$ meters