Шарик радиусом ги массой и удерживается на не- подвижном шаре радиусом R невесомой нерастяжимой ни- тью длиной 4, закрепленной в верхней точке с шара (рис. 67). Других точек соприкосновения между шаром и нитью нет. Найдите силу натяжения нити. Трением пре- небречь.
To find the tension force in the string holding the ball, we can use the concept of equilibrium. In this situation, the ball is not moving and therefore the forces acting on it must be balanced. This means that the tension force in the string must be equal in magnitude to the weight of the ball. We can calculate the weight of the ball using the formula F=mg, where m is the mass of the ball and g is the acceleration due to gravity (approximately 9.8 m/s^2). However, in this case we also need to take into account the radius of the ball, as it is attached to the string and is not sitting on the sphere's surface. This means that the string is being pulled at an angle, which creates a vertical and a horizontal component of force. Using trigonometry, we can calculate that the vertical component of the tension force is equal to mgcosθ, where θ is the angle between the string and the vertical axis. To find this angle, we can use the fact that the string is 4 units long and the radius of the sphere is R. This creates a right triangle where the hypotenuse is R and one of the sides is 4 units (the length of the string). Using the Pythagorean theorem, we can solve for the angle and then substitute it back into our formula to find the vertical component of the tension force. Finally, to find the tension force itself, we need to use the Pythagorean theorem again to calculate the horizontal component and then add it to the vertical component using the formula Ftension=√(Fvertical^2+Fhorizontal^2). Keep in mind that we can neglect the effect of friction in this problem, as stated in the prompt. Therefore, the tension force in the string holding the ball of mass m and radius r on the stationary sphere of radius R is equal to Ftension=√(mgcosθ)^2+(mgsinθ)^2. This can be simplified to Ftension=√(m^2g^2cos^2θ+m^2g^2sin^2θ), which further simplifies to Ftension=√m^2g^2. Therefore, the tension force in the string is equal to Ftension= mg.
