A 1.55-kg object hangs in equilibrium at the end of a rope (taken as massless) while a wind pushes the object with a 13.3-n horizontal force. find the magnitude of the tension in the rope, and the rope/'s angle from the vertical. the acceleration due to gravity is 9.81 m/s2.

To answer this problem, we use balancing of forces: x and y components to determine the tension of the rope.

First, the vertical component of tension (Tsin theta) is equal to the weight of the object.

T * sin θ = mg = 1.55 * 9.81

T * sin θ = 15.2055

Second, the horizontal component of tension (t cos theta) is equal to the force of the wind.

T * cos θ = 13.3

Tan θ = sin θ / cos θ = 15.2055/13.3 = 1.143

we can find θ that is equal to 48.82.

T then is equal to 20.20 N