A 2.1 kg ball tied to a string fixed to the ceiling is pulled to one side by a force to an angle of è = 26.6

how large is F?
tension of string?
before it was released

Tsinα=F

Tcosα=mg

F=mg•tanα
T=mg/cosα

To determine the magnitude of the force, F, and the tension in the string, we can use the equilibrium conditions for the ball.

Let's break down the problem step by step:

Step 1: Draw a Free Body Diagram
Start by drawing a diagram of the ball and label all the forces acting on it. In this case, we have the force F pulling the ball to one side and the tension, T, in the string. Since the ball is at rest, the weight of the ball is balanced by the vertical component of the tension.

Step 2: Resolve Forces
Next, resolve the force F into its horizontal and vertical components. The horizontal component, Fx, will be given by Fx = F * cos(è), where è is the angle given in the problem. The vertical component, Fy, will be Fy = F * sin(è).

Step 3: Apply Equilibrium Conditions
In equilibrium, the sum of all the forces acting on the ball must be zero. Therefore, we can write the equilibrium equations for the horizontal and vertical components of the forces:

Horizontal equilibrium: Fx = 0
Vertical equilibrium: T - Fy - mg = 0, where m is the mass of the ball and g is the acceleration due to gravity.

Step 4: Solve Equations
From the horizontal equilibrium equation, we can deduce that Fx = F * cos(è) = 0. Since cos(è) cannot be zero (as it would cause Fx to be zero), we can conclude that the force F must be zero.

From the vertical equilibrium equation, we have T - F * sin(è) - mg = 0. We can rearrange the equation to find T:

T = F * sin(è) + mg.

Given the mass of the ball (m = 2.1 kg), the angle (è = 26.6°), and the acceleration due to gravity (g = 9.8 m/s^2), we can substitute these values into the equation to find the tension T.

Final Answer:
Since the force F is zero, T = mg + F * sin(è) = m * g = 2.1 kg * 9.8 m/s^2 = 20.58 N. Therefore, the tension in the string before it was released is approximately 20.58 N.