Newton's Laws

Both wagons will travel the same distance.

kg.

The conditions for uniform motion of the aerostat during descent and ascent are expressed respectively: ; .

a) H;

b) H;

c) H.

When the elevator moves with acceleration directed upwards (this will be at the beginning of the ascent and at the end of the descent), the scale reading is greater than ( is the mass of the person). If the acceleration is directed downwards, which will be at the beginning of the descent and at the end of the ascent, the scale reading is less than . If the elevator moves uniformly, the scale reading is equal to .

The equation of motion for the person is (equation written in projections onto the direction of ). Here is the acceleration of the elevator (algebraic quantity) and is the reaction force of the scales. In accordance with Newton's third law, the force acting on the scales from the person is equal in magnitude to : .

When squatting, it will decrease; when straightening up, it will increase.

When squatting, the acceleration of the person's center of mass is directed downwards, and when straightening up - upwards.

During the free movement of the box (movement in the field of gravity in the absence of other forces), the pressure force of the balls on the bottom and walls of the box, as well as on each other, is zero.

The body will oscillate on the spring, the amplitude of which is equal to the initial extension of the spring.

The problem is conveniently solved in a non-inertial coordinate system associated with the moving elevator (see 'Elementary Physics Textbook' edited by Acad. G. S. Landsberg, vol. I, ch. 6). In this reference frame, three forces act on the mass: the force of gravity , the inertial force balancing it, and also the spring tension force, equal at the initial moment to the force of gravity acting on the mass.

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Until the moment the body detaches from the support, it moves with acceleration under the action of three forces: the force of gravity , the spring tension force , and the support reaction force . Therefore . At the moment the body detaches from the support, the support reaction force becomes zero. Considering this and substituting the expression for into the equation of motion, we find .

m/s; H; H.

The forces acting on the weights are shown in the following figure. Let's write the equations of motion for the weights: The accelerations and of the weights are numerically equal, since the string is inextensible and the displacements of the weights are always the same (the relationships between the accelerations of the bodies of the system are determined by their kinematic connection, i.e., the connection between their displacements). In the general case, the forces and applied to the ends of the string, according to Newton's third law, are numerically equal to the forces and , are not equal to each other, i.e., the difference between these forces imparts acceleration to the string and angular acceleration to the pulley. If the mass of the string and pulley can be neglected, then and and the corresponding forces and applied to the weights can be considered equal to each other. Projecting the vectors included in equations (1) and (2) onto the directions and , coinciding with the accelerations of each of the weights (conventionally assume ), we obtain the following system of equations: ; . Solving this system for the unknowns and , we get ; . (If , we get , i.e., the weights move in the direction opposite to the assumed one.) The dynamometer reading is obviously equal to the sum of the string tension forces:

s.

See the previous problem. Each weight will travel a path equal to 5 cm.

; .

In accordance with Newton's second law, let's write the equations of motion for each of the bodies (see figure): for the first body for the second body for the load where is the string tension force; is the pressure force of the load on the body, is the acceleration of the bodies; is the reaction force of the support. According to Newton's third law, and , and are numerically equal. The tensions on the right and left are equal, since friction and the mass of the pulley are neglected. Solving the obtained system of equations, we find The force acting on the axis of the pulley,

1. N.

2. N.

1. The force , acting from the axis on the rod and according to Newton's third law numerically equal to the pressure force on the axis (see figure below, a), can be replaced by two forces and , applied to each of the weights and equal to . Considering that the rod is massless, we can, using Newton's second law, write: , , whence .

2. The force must be replaced (see figure below, b) by forces and , applied to weights and respectively. The accelerations of the weights are related by , i.e., .

; .

Let's write the equations of Newton's second law for each of the weights, projecting the forces acting on the weights (see figure) onto the direction of possible motion of the corresponding weights, i.e., along the faces of the prism (assume that weight descends): Solving this system, we find: If, upon substitution of the numerical values of masses and angles, it turns out that , it means that the entire system moves with acceleration in the direction opposite to the chosen one. The latter follows from the fact that in the absence of friction, changing the chosen direction of motion of the weights will only change the signs of the right sides of the equations derived above.

m/s; H; , when .

; ; .

From the kinematic connection, it follows that the acceleration of weight is twice the acceleration of weight (when weight moves by distance , weight moves by distance ). A force equal to is applied to weight from the strings.

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Let's conditionally isolate a small part of the liquid inside. Two forces act on this element of liquid with mass : the force of gravity and the buoyant (Archimedean) force from the side of the liquid. The resultant of these forces must be directed towards the acceleration of the vessel and equal to (see figure). Therefore, the force must form an angle with the vertical. The direction of the buoyant force acting on any isolated volume inside the liquid is perpendicular to the planes of equal pressure (=const) in the liquid, is determined by them and, in particular, is perpendicular to its surface (). The requirement of perpendicularity uniquely relates to the condition of rest of the liquid relative to the vessel. Any violation of perpendicularity is 'automatically' eliminated due to the fluidity of the liquid.

The surface of the liquid is parallel to the inclined plane.

1st method. See solution to problem 16.

2nd method. Let's solve the problem in the reference frame associated with the moving vessel (in a non-inertial system). In this case, forces , , and the inertial force act on any element of the liquid, directed opposite to the motion of the vessel (see figure). The resultafignt of these forces must be zero (since the liquid is stationary relative to the vessel). This is possible when the force is perpendicular to the inclined plane. The surface of the liquid is parallel to it.

Will not change.

When the elevator moves with acceleration , Newton's second law for a floating body is written as: where is the buoyant force; is the mass of the body.

If the volume displaced by the body were occupied by liquid, a force, also equal to , would act on it from the rest of the liquid. According to Newton's second law

here , where is the volume of the submerged part of the body. Substituting the expression for found from equality (1) into equation (2), we get

hence it follows that does not depend on the acceleration of the elevator, i.e., the immersion depth of the body does not change. Note. The solution to the problem becomes very simple in the reference frame associated with the elevator (see problem 17).

; .

The dog must run with such acceleration, directed downwards, so that the force , with which the dog pushes the board back, balances the resultant of the forces acting on the board: gravity and reaction force , i.e., equals (see figure). On the dog, therefore, in the direction of its acceleration, act the force and the force from the board, equal, according to Newton's third law, to (the direction of the dog's movement does not matter). The equation of motion for the dog will be , whence . The force , with which the dog pushes the board, by its nature, is a friction force, which cannot exceed the value . The problem has a solution (the dog stops the board) if , i.e., . Hence .

1st method. See solution to problem 16.

2nd method. Let's solve the problem in the reference frame associated with the moving vessel (in a non-inertial system). In this case, forces , , and the inertial force act on any element of the liquid, directed opposite to the motion of the vessel (see figure). The resultafignt of these forces must be zero (since the liquid is stationary relative to the vessel). This is possible when the force is perpendicular to the inclined plane. The surface of the liquid is parallel to it.

.

The cylinder is at rest relative to the Earth.

s.

Since the person is stationary relative to the Earth, the forces applied to him are balanced. Thus, the tension force is numerically equal to the weight of the person . Therefore, for the weight, Newton's second law will be written as: . Hence the acceleration of the weight and is directed upwards. The time for the weight to rise is determined from the formula :

1. Both boys will reach the middle of the distance between them simultaneously after time ; if the masses of the boys are unequal, the meeting place divides the distance between them inversely proportional to the masses of the boys.

2. The gymnasts will reach the block simultaneously after time .

Since the forces applied to the boys are equal, according to Newton's third law, the accelerations, and consequently, the velocities of the boys relative to the Earth are equal. The length of the rope between the boys decreases with speed and the boys will reach the middle of the distance between them after time . If the masses of the boys are different, the accelerations and paths traveled by them relative to the Earth are inversely proportional to the masses of the boys.

• Compare the solution of this problem with the solution of problem item 1.

0.5 N; 1 N.

The magnitude of the static friction force cannot exceed , where is the normal pressure. If the body is at rest, the friction force is equal in magnitude to the sum of the projections of all forces acting on the body onto the direction of possible motion (i.e., onto the direction in which the body would start moving if the friction force were zero). If the force acting on the body in the direction of possible motion exceeds , the body moves with acceleration, and the friction force is constant and equal to .

The graph is shown below.

The graph is shown below.

The motion of a body located on an inclined plane becomes possible if , i.e., at an angle . If , the body is at rest and the friction force is equal to (see problem 23). If , then .

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1. ; .

2. ; .

1. Since weight is stationary, the tension of the string, and consequently, the friction force of the ring on the string are equal to . [Note: Error in original text, should be based on context].

2. The friction force is equal to the tension force of the string, and the acceleration of the ring relative to the block is . The equations of motion for the weights are expressed as follows:

.

The normal pressure force is (see figure, a) or (see figure, b) depending on the direction of action of force .

1. H. 2. Does not change. 3. H.

for ; for . The state of motion (rest) does not change.

The accelerations of the weights relative to the table are equal in magnitude (weights are connected by an inextensible string): . In a stationary reference frame, the acceleration of weight is directed vertically, and . The acceleration of weight has vertical and horizontal components: , . Applying Newton's second law, we get the system of equations: , , , where is the string tension; is the reaction force of the table. Solving this system, we find the answer. Note. This system of equations can be represented as , , . Thus, the upward movement of the elevator with acceleration is equivalent to a reference frame in which the acceleration of free fall is . If (for ), the answer follows from the equation . In this case, the friction force is no longer equal to and the second equation is incorrect. From this note, the answer to the second question of the problem immediately follows.

H; H; does not change.

; .

; .

See problem 28.

1. ; .

2. .

1. The pressure forces of block B on block A and block A on the table are respectively and .

2. Since block B slides on the plate, two forces act on the plate from block B: the normal pressure force and the friction force The normal pressure force of the plate on the inclined plane is The plate does not slide on the inclined plane if i.e., if or
   .

1) The acceleration of the body at all times during its motion is constant and equal to ;

2. In accordance with Newton's second law . When moving upwards, the resistance force is directed, like gravity, downwards and decreases as the body ascends (since the speed of the body decreases), during descent, the resistance force is directed upwards and increases. Therefore, the acceleration of the body at the beginning of the motion is maximum (and greater than ), decreases during ascent and becomes equal to at the highest point of the trajectory, then continues to decrease during descent and may even become zero.

The speed of the large ball will be times greater than the speed of the smaller ball.

Under steady motion, the air resistance force and the gravity force acting on the ball are equal, but and where is the density of the ball; is its diameter. Equating and , we find that .

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See solutions to problems 35 and 36.

0.29 N.

a) ;

b) m/s;

c) s;

d) m/s.

.

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On the upper weight along the inclined plane act the component of gravity , the string tension force , and the friction force . According to Newton's second law, we get . For the lower weight, we have accordingly (the weight moves with the same acceleration, and the coefficient of friction between it and the inclined plane is half as much). Subtracting the second from the first equation, we find the string tension force .