A mechanical system consisting of three bodies of masses m1, m2, m3 rotates about a vertical axis with a constant angular velocity
ω. Ball 3 is taken as a material point. Bodies 1, 2 are homogeneous rods.l
– length of rod 1.
Using d'Alembert's principle, create dynamic equations
equilibrium of a mechanical system.
communications. The geometric parameters of the system are known. Under the influence of active loads, a mechanical system moves from a state of rest.
Given: m1, m2, m3 – masses of bodies 1, 2, 3; Jc2x2 , Jc3x3 – moments of inertia of bodies 2,
3 relative to axes passing through their centers of mass; P – active force.
Ticket number 3
Theoretical part
Exercise 1 . Formulate third law of dynamics(law of equality of action and reaction).
forced vibrations
tion of a material point?
Task 3. Write down the basic equation for the dynamics of relative motion points for the case when the transfer motion is a translational uneven curvilinear motion, and the relative motion is rectilinear
Task 4. That is measure of inertia during translational movement
of a solid body?
Task 5. Formulate the second corollary from theorems on the motion of the center of mass of a mechanical system.
Task 6. Formulate a definition of the concept “central system”
la."
Task 7. Formulate a definition of the concept “ kinetic energy».
Task 8. Formulate a definition of the concept “possible changes”
displacement of a non-free mechanical system." Task 9. What does analytical mechanics study?
Task 10. Formulate a definition of the concept “generalized system”
la."
Practical part
Body 1 rotates relative to the O1 Z1 axis with a constant angular velocity. A point M with mass m moves along a smooth channel made in body 1.
A moving mechanical system consists of four bodies. Center of mass
body 1 has a speed V.
Determine the kinetic energy of body 4 with mass m4 depending on
Using the principle of possible displacements, determine the horizontal component of the external connection reaction at point B.
A mechanical system consisting of three bodies with masses m1, m2, m3, rotating
is relative to the vertical axis with a constant angular velocity ω. Bodies 1, 2, 3 are homogeneous rods. l 1 = l 3 = l – lengths of rods 1, 3.
Using d'Alembert's principle, create equations of dynamic equations
news of the mechanical system.
On a mechanical system consisting of three bodies, ideal
communications. The geometric parameters of the system are known. Under the influence of active
loads, the mechanical system moves from a state of rest.
Given: m1, m2, m3 – masses of bodies 1, 2, 3; Jc2x2 , Jc3x3 – moments of inertia of bodies 2, 3
relative to the axes passing through their centers of mass; P – active force; M3 – active moment.
Draw up a general equation for the dynamics of a mechanical system.
Ticket number 4
Theoretical part
Exercise 1 . Formulate a definition of the concept “ inertial reference frame».
Task 2. Under the influence of what forces do forced vibrations
tion of a material point?
Task 3. Write down the differential equation of motion of a point,
occurring under the influence of a restoring force, a disturbing force changing according to a periodic law, and a force of resistance to movement,
proportional to the first power of speed.
Task 4. That is measure of inertia during translational movement
of a solid body?
Task 5. Write down the formula to determine moment of inertia te-
la relative to the vertical axis of rotation.
Task 6. Formulate a definition of the concept “ vector arm co-
the quantity of motion of a point relative to an arbitrary center.”
Task 7. Write down the formula to determine heavy force work
sti.
Task 8. Write down formulas expressing d'Alembert's principle for a non-free immutable mechanical system in coordinate form
Task 9. Formulate a definition of the concept “possible re-
displacement of the system."
Task 10. Write down a formula expressing principle of possible re-
displacements, in vector form.
Practical part
Body 1 rotates relative to the O1 Z1 axis with a constant angular velocity
e. A point M with mass m moves along a smooth channel made in body 1.
Write down the differential equation for the relative motion of point M.
A moving mechanical system consists of five bodies. Geometric
body parameters are known. R3, r3, R5 are the corresponding radii of bodies 3, 5. The center of mass of body 1 has a speed V. Jc5x5 is the moment of inertia of body 5 relative to the axis,
passing through its center of mass.
Determine the kinetic energy of body 5 with mass m5 depending on
speed V and geometric parameters of this system.
A flat mechanical system consisting of two bodies is subject to active loads P1, P2, q, M.
Using the principle of possible movements, determine the vertical
component of the external connection reaction at point A.
A mechanical system consisting of three bodies with masses m1, m2, m3 rotates about a horizontal axis with a constant angular velocity ω.
Bodies 1, 2, 3 are homogeneous rods.
Given: m1, m2, m3, m4 – body masses; Jc2x2, Jc3x3 – moments of inertia of bodies 2, 3 relative to the axes passing through their centers of mass.
Draw up a general equation for the dynamics of a mechanical system.
Ticket number 5
Theoretical part
Exercise 1 . Write down the basic equation for the dynamics of a non-free ma-
terial point in vector form.
Task 2. Formulate a definition of the concept “ cyclic hour-
the tota of free oscillations of a point."
Task 3. Formulate a definition of the concept “internal systems”
ly".
Task 4. Write down the formula to determine main vector re-
shares of external relations.
Task 5. Formulate Steiner's theorem.
Task 6. Write down momentum theorem in vector form.
Task 7. Formulate a definition of the concept of “constant work”
force on the rectilinear movement of the point of its application.”
Task 8. Write down the formula to determine inertia forces
rial point.
Task 9. Formulate a definition of the concept “possible (ele-
mental) work of force.”
Task 10. Write down the Lagrange equation of the second kind.
Practical part
Cart 1 performs translational horizontal motion according to the law y1 = 4t3 + 2t2 + t + 1, m. A ball M with mass m moves in a smooth inclined channel of the cart.
Write down the differential equation of relative motion
The moving mechanical system consists of six bodies. Geometric
The physical parameters of the bodies are known. R2, r2, R3 are the radii of bodies 2 and 3, respectively. Jc3x3 is the moment of inertia of body 3 relative to the axis passing through its center
wt. The center of mass of body 1 has a speed V.
Determine the kinetic energy of body 3 depending on the speed V and the geometric parameters of the mechanism.
A flat mechanical system consisting of two bodies is acted upon by
active loads P1, P2, q, M.
Using the principle of possible displacements, determine the vertical component of the external connection reaction at point A.
A mechanical system consisting of two homogeneous rods 1, 2 with masses m1, m2 and a weightless thread 3 rotates relative to the horizontal
axis with constant angular velocity ω.
Using d'Alembert's principle, create dynamic equilibrium equations for a mechanical system.
Ideal connections are imposed on a mechanical system consisting of four bodies. The geometric parameters of the system are known. Under the influence of active loads, a mechanical system moves from a state of rest.
Given: m1, m2, m3, m4 – body masses; Jc2x2, Jc3x3 – moments of inertia of bodies 2, 3 relative to the axes passing through their centers of mass; P – active force.
Draw up a general equation for the dynamics of a mechanical system.
Sasha, Kolya and Dima took part in distance running competitions L= 200 m. At the start, the friends were located on adjacent paths. Sasha, who started in the first lane, finished first after t= 40 s, and Dima on the third track was Δ behind the winner t= 10 s. Determine Kolya’s speed on the second track if it is known that at the moment of Sasha’s finish all three runners were located on the same straight line. The running speeds of athletes can be considered constant over the entire distance, and the treadmill is straight.
Possible Solution
Let's find Sasha's speed: V 1 = L/ t and Dima’s speed: V 3 = L/(t + Δt)
At a moment in time t Dima is behind Sasha by a distance Δ l =(V 1 – V 3)t.
From the fact that all three friends were on the same straight line at that moment, it follows that Kolya lagged behind Sasha by a distance Δ l/2. On the other hand Δ l/ 2 = (V 1 – V 2)t, Where V 2 – Kolya’s speed. Solving the written system of equations, we get: ÷
Evaluation criteria
- Found the speeds of Sasha and Dima (1 point for each): 2 points
- The distance by which Dima was behind Sasha at the moment in time was found t: 2 points
- It was used that the friends are located on the same straight line, and a connection was obtained between the distances by which Dima and Kolya lagged behind Sasha: 2 points
- An expression has been written for the distance by which Kolya is behind Sasha at a moment in time t, through Kolya's speed: 2 points
- The expression for Kolya's speed is obtained: 1 point
- The numerical value of Kolya's speed was obtained: 1 point
Maximum per task- 10 points.
Problem 2
A system consisting of two homogeneous rods of different densities is in equilibrium. Top rod weight m 1 = 3.6 kg. Friction is negligible. Determine at what mass m 2 lower rods such an equilibrium is possible.
Possible Solution
Let us write the moment equation for the lower rod relative to its center of gravity: 5T 1 – 2T 2 = 0, where T 1 is the reaction force on the side of the left thread, T 2 is the reaction force on the side of the right thread.
Equilibrium condition of the lower rod:
T 1 + T 2 = m 2 g
From these two equations we find:
T 1 = 2/7 *m 2 g,
– T 2 = 5/7*m 2 g.
Let us write the equation of moments for the upper rod relative to the attachment point of the left (upper) thread:
Evaluation criteria
- 5T 1 – 2T 2 = 0: 2 points
- T 1 + T 2 = m 2 g: 1 point
- T 1 = 2/7*m 2 g and T 2 = 5/7m 2 g (1 point for each force): 2 points
- Moment equation: 4 points
- m 2 = 2.1 kg: 1 point
Maximum per task – 10 points.
Problem 3
A body tied by a thread to the bottom of a vessel is immersed in liquid to 2/3 of its volume. The tension force of the thread is equal to T 1 = 12 N. In order to remove this body from the liquid by 2/3 of its volume, you need to untie the body from the bottom and apply a vertically upward force to it from above T 2 = 9 N. Determine the ratio of the densities of the liquid and the body.
Possible Solution
Let us write down the equilibrium condition of the body in the first case:
where ρ Т is the density of the body, ρ Ж is the density of the liquid, V is the volume of the body.
Let's divide one equation by another:
Evaluation criteria
- Archimedes' force in the form ρ Ж gV pogr: 1 point
- The condition for body equilibrium in the first case: 4 points
- The condition for body equilibrium in the second case: 4 points
- ρ Ж /ρ T = 2.1: 1 point
Maximum per task- 10 points
Problem 4
To maintain a constant temperature in the house T= +20 ºС firewood is constantly being added to the stove. When it gets cold, the outside air temperature drops by Δ t= 15 ºС, and to maintain the same temperature in the house you have to add firewood 1.5 times more often. Determine the air temperature outside when it gets cold. What temperature would be established in the house if firewood was added with the same frequency? Consider that the power of heat transfer from room to street is proportional to the difference in their temperatures.
Possible Solution
Let the air temperature outside before the cold snap be equal, and the heat power supplied to the house due to burning wood be equal P. Then before it gets cold:
where α is some constant proportionality coefficient.
After cold weather:
1.5ϲP = α(T – (t – Δt))
Let's divide one equation by another:
If firewood was added with the same frequency, then:
Evaluation criteria
- P = α(T – t) : 3 points
- 1.5P = α(T – (t – ∆t)): 3 points
- t – ∆t = – 25°C: 1 point
- T' = 5°C: 3 points
Maximum per task- 10 points.
Problem 5
How many times will the readings of an ideal ammeter change when the switch is closed if a constant voltage is applied to the input terminals of the circuit section?
Evaluation criteria
- Total resistance before switch closure: 3 points
- I = 7U/12R: 1.5 points
- Total resistance after closing the key: 3 points
- I′=12U/17R: 1,5 points
- I′/I= 144/119 ≈ 1.2: 1 point
Maximum per task- 10 points.
If the solution to a problem differs from the author’s, the expert (teacher) himself draws up evaluation criteria depending on the degree and correctness of the problem solution.
If a correct solution contains an arithmetic error, the score is reduced by 1 point.
Total for work - 50 points.
When determining the equilibrium conditions for a system of interacting solid bodies, the equilibrium problem can be solved for each body separately. The reaction (interaction) forces arising at the points of contact satisfy Newton's third law. In accordance with this, we are obliged to accept the condition that the action of one body on another is equal and opposite in direction to the action of this other body on the first.
If, when solving an equilibrium problem, we choose the same center of reduction for all bodies of the system, then for each of the bodies we obtain the following equilibrium conditions:
where, respectively, is the resulting force and moment of the resulting pair of all forces acting on a given body, except for the forces of interaction between individual bodies (internal reactions). - respectively, the resulting force and moment of the resulting pair of forces of internal reactions acting on a given body. Now performing a formal summation and taking into account that the conditions are satisfied for the internal interaction forces
we obtain the following necessary conditions for the equilibrium of a system of solid bodies:
where the summation already extends to all points of interacting bodies.
Example 35. The system consists of two homogeneous rods of length P and weight P. Both rods can rotate in the same vertical plane: the rod around its center O, and the rod around the hinge O, located on the same vertical with O at a distance
A load with a weight of weight Q is suspended from the end D of the rod. The load Q, through the rod, deflects the rod from the vertical position.
Determine the angle at the equilibrium position of the system, as well as the reaction at point O (Fig. 99).
Solution. The system under consideration consists of two solid rods under the action of a plane system of forces.
Equilibrium conditions for the first rod
can be rewritten in the form
The last equation of the first group indicates that the only reaction force is located in the plane of the drawing. Consequently, the moment of the resulting pair is directed along the axis perpendicular to the plane. Considering the equilibrium conditions of the rod, we note that the reaction at point O is located in the plane of the drawing, and the equilibrium conditions of each of the rods consist of three equations. As a result, we obtain six equilibrium equations for the system to determine the angle and reaction at points. To determine the equilibrium position of a system, it is necessary to find only one quantity - the angle
When drawing up equilibrium equations, you can notice that they contain several unknown quantities (parameter and unknown reactions). Depending
Depending on the choice of the reduction center, these equations will have a more or less complex form.
Let us first consider the equilibrium of the rod, choosing point O as the center of reduction. The equilibrium condition is that the sum of the moments of pairs from the reduction of forces Q and to point O is equal to zero (here N is the reaction force acting from the rod OA on the rod CD)
Let us now proceed to the study of the equilibrium of the rod. We choose point O as the center of reduction, so that the equilibrium condition (the sum of the moments of pairs being equal to zero when reduced to point O) takes the form
OLYMPIAD PROBLEMS
8th grade
1. Get to work!
The engineer arrived at the station at the same time every day, and at the same time a car came to pick him up from the plant, in which he drove to this plant to work. One day, an engineer arrived at the station 55 minutes earlier than usual, immediately walked towards the car and arrived at the plant 10 minutes earlier than usual. What is the speed of the car if the engineer's speed is 5 km/h?
1. Since in this case the engineer arrived at the plant 10 minutes earlier (and the car left as usual), the car would travel from the meeting point to the station in 5 minutes.
2. The engineer walked the same distance in 50 minutes (he arrived at the station 55 minutes earlier than the car would have arrived).
3. Thus, the car traveled the same distance (from the station to the meeting place), spending 10 times less time than the engineer. Consequently, its speed is 10 times greater, i.e. equal to 50 km/h.
2. System in mechanical equilibrium
The system consists of two homogeneous rods, three weightless threads, one of which is thrown over a stationary block. There is no friction in the block axis, and all threads are vertical. The mass of the upper rod m 1 = 0.5 kg. Determine the mass m 2 of the lower rod.
1. Let us arrange the forces acting on each of the rods. Let us take into account that the forces applied at one point are the same. And a stationary block does not provide a gain in strength, therefore the forces acting on the thread thrown over the block are also the same on both sides. 
2. Both rods are in equilibrium without rotating. And both rods do not move, remaining at rest. Therefore, we first apply the rule of moments for each rod. Because The rods are at rest, then the resultant of the applied forces is 0. 
Water is poured into a U-shaped tube so that the distance from the water level to the top of the tube is 40 cm. Oil is added to one elbow of the tube to the top. How much will the water level rise in the second leg of the tube? The density of oil is 800 kg/m3, the density of water is 1000 kg/m3.
1. The distance between levels 1 and 2 is 40 cm. When oil is added to the left knee, the water level in it drops by a distance x (the distance between levels 2 and 3). In the right knee the water rises by the same amount, because the liquids are incompressible and the volume of water released from the left elbow is equal to the volume of water transferred to the right elbow (the cross-sections of the tubes are the same).
2. According to Pascal's law, the pressure at the same level should be the same. Let's find out the pressure in each knee at level 3:
Water is poured into a glass at room temperature 20°C to half the volume. Then the same amount of water is added to this glass at a temperature of 30°C. After thermal equilibrium was established, the temperature in the glass turned out to be 23°C. In another similar glass, pour water at a temperature of 20°C to 1/3 of the volume and add hot water at a temperature of 30°C to the top. What temperature will be established in this glass? Neglect heat losses during the establishment of equilibrium.
1. Let us denote: C - the heat capacity of the glass, c - the heat capacity of water, t 0 = 20 o C, t = 30 o C, t 1 = 23 o C, t 2 - the desired value.
2. Let us write down the heat balance equations for each case: 
5. Fuel consumption
The fuel consumption of a bus (a) depends on its speed (v) as shown in the first graph. From city A to city B the bus moves in accordance with the schedule (second schedule). Find out if the driver can do itget to your destination without refueling if the car has 25 liters of fuel in the tank?
Using the first graph, we determine fuel consumption at speeds of 20 km/h and 80 km/h. Since the dependence of fuel consumption on speed is linear, the following proportions are valid: 
Let's take into account that at a speed of 80 km/h the bus traveled 80 km, which consumed the volume of gasoline
V 1 = a 1 s 1 = (11/60) 80 = 44/3 l. At a speed of 20 km/h the bus traveled 40 km, on which it spent
V 2 = a 2 s 2 = (13/60) 40 = 26/3 l. In total, the bus consumed 70/3 liters, which is less than 25 liters. Therefore, there will be enough fuel to travel to your destination without refueling.
An aeronaut, traveling in a hot air balloon, suddenly saw that he was moving evenly downward. Then he dropped 60 kg of ballast, stored just for this occasion. The balloon, after being freed from ballast, began to rise upward at half the speed. Considering the force of air resistance to be directly proportional to the speed of the ball, determine this force during descent.
Let us arrange the forces acting on the balloon as it flies up and down: 
Since in both cases the motion is uniform, the resultant of all applied forces is zero. Then for the downward movement we have F resist + F arch = m 1 g, and for the upward movement F arch = m 2 g + F resist /2. Here we took into account that the Archimedean force does not change (the air density and the volume of the ball are the same), and the resistance force when moving upward will become 2 times less, because according to the condition, it is proportional to the speed of movement and the speed when moving up is 2 times less than when moving down. The dropped load has a mass m 1 - m 2, then we find that 3/2 F resist = (m 1 - m 2)g. Hence F resist = 400 N.
A homogeneous, flat steel rod 1 m long was bent in half at an angle of 90°. At what distance from the vertex of a right angle should the rod be hung so that the sides of the resulting angle are oriented vertically and horizontally?
