A charged sphere of mass 2.8 kg is suspended from a string an electric force acts horizontally on the sphere so that the string makes angle 33 degree with vertical when at rest find magnitude of electric force and tension in string

Force, F = 17.82 N

T = 32.72 N

Explanation:

The mass of the sphere, m = 2.8 kg

Angle made by the string with the vertical, θ = 33°

The force, F, is said to act horizontally on the sphere, and the string makes angle 33° with the vertical. Check the free body diagram for the representation of the situation that is described.

Weight of the sphere, W = mg

W = 2.8 * 9.8

W = 27.44 N

Sin θ = F/T

sin 33 = F/T............(1)

cos θ = W/T

cos 33 = 27.44/T

T = 27.44/cos 33

T = 32.72 N

Put the value of T into equation (1)

sin 33 = F/32.72

F = 32.72 sin 33

F = 17.82 N

The magnitude of the Tension in the string = 32.72 N

The magnitude of the electric force on the ball = 17.82 N

Explanation:

The sketch of the free body diagram for the question is presented in the attached image to this solution.

Doing a force balance on the charged ball of sphere,

- T represents the tension in the string which acts away from the ball at angle 33° to the vertical. The tension can be split into a horizontal and a vertical component (Tₓ and Tᵧ respectively).

- Electric force, F, on the charged ball of sphere which acts horizontally to balance the the horizontal component of the tension in the string.

- The weight of the ball, W, which acts vertically downwards and balances the vertical component of the Tension.

Resolving the tension into the horizontal and vertical component

Tₓ = T sin 33° = (0.5446T) N

Tᵧ = T cos 33° = (0.8387T) N

W = mg = 2.8 × 9.8 = 27.44 N

(g = acceleration due to gravity = 9.8 m/s²)

Tᵧ = W

0.8387T = 27.44

T = (27.44/0.8387) = 32.72 N

F = Tₓ

F = 0.5446T = 0.5446 × 32.72 = 17.82 N

The magnitude of the Tension in the string = 32.72 N

The magnitude of the electric force on the ball = 17.82 N

Hope this Helps!!!

    m₁v₁ + m₂v₂ = (m₁ + m₂)vmomentum is conserved.the left-hand side of the equation represents the momentum in the system before the collision.the right-hand side represents the momentum in the system after the collisionm₁ and v₁ represents the mass and the velocity of the physics blockm₂ and v₂ represents the mass and velocity of the ball of puttyv represents the final velocity of the combined mass.the question states the two masses stick together. they combined mass has its own velocity with the body mass the sum of the two objects' massestheir combined momentum is (m₁ + m₂)vto find the magitude of (m₁ + m₂)v, we need to add the vectors on the left-hand side of the equation: m₁v₁ + m₂v₂.note the diagram. i have positioned the vectors in accordance with their directions.the magnitude of the resultant can be calculated using the pythagorean theorem since this is a right triangle situation (north and west are perpendicular)|(m₁ + m₂)v| =  √[ (m₁v₁)² + (m₂v₂)² ]|(m₁ + m₂)v| =  √[ (1.5kg·8.0m/s)² + (0.40kg·20m/s)² ]|(m₁ + m₂)v| =  14.4 kg m/sfor two significant figures (assuming that 20m/s is two), this rounds to 14 kg m/s

Newtons laws of motion