A body weight 150N is supported by a thin cord attached to point in the calling another cord is a ached to the alright and pulled horizontally antic if the supporting cord make an angle of 30 with the vertical the lension in each tring
To find the tension in each string, we can analyze the forces acting on the body.
First, let's consider the vertical forces. The weight of the body is exerting a downward force of 150N. This force is balanced by the tension in the supporting cord.
Now, let's analyze the horizontal forces. There is a horizontal tension force exerted by the second cord, which is being pulled to the right. However, since the body is not moving horizontally, the horizontal components of the tension forces in both cords must be equal.
To find the tension in each string, we can break down the forces into their vertical and horizontal components using trigonometry.
Let's denote the tension in the supporting cord as T1 and the tension in the second cord as T2.
The vertical component of the tension in the supporting cord can be found using the sine function:
T1 * sin(30°) = Weight of the body
T1 = Weight of the body / sin(30°)
T1 = 150N / sin(30°) ≈ 300N
Since the weight of the body is vertically balanced by T1, the vertical component of T2 must also be equal to the weight of the body:
T2 * sin(30°) = Weight of the body
T2 = Weight of the body / sin(30°)
T2 = 150N / sin(30°) ≈ 300N
So, the tension in each string is approximately 300N.