Question

Мальчик,масс которого равна m = 50 кг,качается на качелях,длина подвеса l =4 м.При прохождении положения равновесия он давит на сиденье с силой,равной N = 950 Н .С какой скоростью качели проходят это положение?С какой скоростью качели проходят это положение ?С какой силой мальчик давит на сиденье при отклонении качелей на угол α=30 градусов от вертикали?

Answer 1

В нижней точке
ma=mv^2/L=N-mg
v=корень(L*(N/m-g))=корень(4*(950/50-10)) м/с = 6 м/с

при угле отклонения =0 градусов скорость качели v = 6 м/с
при угле отклонения 30 градусов скорость качели u
mv^2/2=mu^2/2+mgh=mu^2/2+mgL*(1-cos(pi/6))
mv^2/2=mu^2/2+mgL*(1-cos(pi/6))
u^2=v^2-2gL*(1-cos(pi/6))

при угле отклонения 30 градусов
ma=mu^2/L=N1-mg*cos(pi/6)
N1=m(u^2/L+g*cos(pi/6)) = m(v^2/L-2g*(1-cos(pi/6))+g*cos(pi/6)) =
= m*(v^2/L-g(2-3*cos(pi/6))) = ......
************ в принципе в этом месте уже можно подставлять цифры ****
= m*(v^2/L-g(2-3*cos(pi/6))) =50*(6^2/4-10*(2-3*cos(pi/6))) =749,0381 Н ~ 750 Н
****************но мне интересно по-другому**************************
N1 = m*(v^2/L-g(2-3*cos(pi/6))) = m*((N/m-g)-g(2-3*cos(pi/6))) =
N1 = N-3mg*(1-cos(pi/6))) = 950-3*50*10*(1-cos(pi/6))) = 749,0381 Н ~ 750 Н

Answer 2

Calculation of the Speed of the Swing

To calculate the speed at which the swing passes through the equilibrium position, we can use the principle of conservation of mechanical energy. At the highest point of the swing, when the swing is momentarily at rest, all of the potential energy is converted into kinetic energy. At this point, the potential energy is given by the formula:

Potential energy = mass × gravity × height

The height can be calculated as the difference between the length of the swing and the distance from the equilibrium position to the highest point. Since the swing is 4 meters long and the equilibrium position is at the midpoint, the distance from the equilibrium position to the highest point is 2 meters.

Height = length - distance from equilibrium position to highest point = 4 m - 2 m = 2 m

Substituting the values into the formula, we have:

Potential energy = 50 kg × 9.8 m/s^2 × 2 m = 980 J

At the equilibrium position, all of the potential energy is converted into kinetic energy. The kinetic energy can be calculated using the formula:

Kinetic energy = (1/2) × mass × velocity^2

Setting the potential energy equal to the kinetic energy, we can solve for the velocity:

Potential energy = Kinetic energy

50 kg × 9.8 m/s^2 × 2 m = (1/2) × 50 kg × velocity^2

Simplifying the equation, we have:

980 J = 25 kg × velocity^2

Solving for the velocity, we find:

velocity^2 = 980 J / 25 kg

velocity^2 = 39.2 m^2/s^2

Taking the square root of both sides, we get:

velocity = √(39.2 m^2/s^2) ≈ 6.26 m/s

Therefore, the speed at which the swing passes through the equilibrium position is approximately 6.26 m/s.

Calculation of the Force Exerted on the Seat

To calculate the force exerted on the seat when the swing is at an angle of 30 degrees from the vertical, we can use the concept of forces in equilibrium. At this angle, the force exerted on the seat can be split into two components: the vertical component and the horizontal component.

The vertical component of the force can be calculated using the formula:

Vertical force = weight × cos(α)

where weight is the force due to gravity acting on the boy, and α is the angle of deviation from the vertical.

Substituting the values into the formula, we have:

Vertical force = 50 kg × 9.8 m/s^2 × cos(30 degrees)

Using the trigonometric identity cos(30 degrees) = √3/2, we can simplify the equation:

Vertical force = 50 kg × 9.8 m/s^2 × √3/2 ≈ 425.3 N

Therefore, the vertical component of the force exerted on the seat is approximately 425.3 N.

The horizontal component of the force can be calculated using the formula:

Horizontal force = weight × sin(α)

Substituting the values into the formula, we have:

Horizontal force = 50 kg × 9.8 m/s^2 × sin(30 degrees)

Using the trigonometric identity sin(30 degrees) = 1/2, we can simplify the equation:

Horizontal force = 50 kg × 9.8 m/s^2 × 1/2 ≈ 245 N

Therefore, the horizontal component of the force exerted on the seat is approximately 245 N.

Hence, when the swing is at an angle of 30 degrees from the vertical, the force exerted by the boy on the seat is approximately 425.3 N vertically and 245 N horizontally.