A non-uniform bar of weight W = 20.0 N is suspended at rest in a horizontal position by two massless cords. The angle phi between cord 2 and the vertical is equal to 32.5o. The bar has a length L = 6.5 m and the distance of its center of gravity from the left-hand end of the bar is 1.3 m. What is the tension in cord 2?
To find the tension in cord 2, we can analyze the forces acting on the bar and use the principles of equilibrium. We will consider the forces acting vertically and horizontally.
First, let's analyze the forces acting vertically. The downward force of the weight W will be balanced by the upward tensions in cord 1 and cord 2. Since the bar is at rest, the vertical forces must cancel each other out. Therefore, the sum of the upward forces must be equal to the weight W.
Let's denote the tension in cord 1 as T1 and the tension in cord 2 as T2. The vertical component of tension T2 (T2v) can be found using trigonometry. Since angle phi is given as 32.5 degrees, we can use the sine function to relate T2v to T2:
T2v = T2 * sin(phi)
Since the bar is at rest in a horizontal position, the weight W is balanced by the sum of the vertical components of the tensions:
W = T1 + T2v
Now let's analyze the forces acting horizontally. Since the bar is in equilibrium, there is no net horizontal force. This means that the horizontal components of the tensions must cancel each other out. The horizontal component of tension T2 (T2h) can be found using trigonometry. The angle between cord 2 and the horizontal is (90 degrees - phi). Therefore,
T2h = T2 * cos(90 - phi)
Since there is no horizontal acceleration, the horizontal forces must balance each other out:
T1 = T2h
Now we can substitute the expressions for T2v and T2h into the equations:
W = T1 + T2 * sin(phi)
T1 = T2 * cos(90 - phi)
We can solve this system of equations to find the value of T2. Plugging in the given values:
W = 20.0 N
phi = 32.5 degrees
We can now solve for T2.