Question

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.

Answer 1

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