A homogeneous rod AB=100 cm of mass m=4kg is suspended from a fixed roof by means of two very light strings OA and O'C having the same length point C is at 20cm from B , rod AB is in a horizontal position of equilibrium
Make the inventory of the forces acting on the rod AB.
define + is up
Draw a picture
x = 0 at A , Fa up
x = 80 at C , Fc up
x = 50 at CG, mg =4*9.81=39.24 down so Fcg = -39.24 N
sum of forces = 0
Fa + Fc + Fcg = 0
so
Fa+Fc = 39.24 N
now do moments about 0 for example
Fc * 80 = 39.24 * 50
Fc = 24.5 N
so
Fa = 39.24 - 24.5 = 14.7 N
To determine the inventory of the forces acting on the rod AB, we need to analyze the forces involved in the system.
1. Gravitational Force (Weight): The rod has a mass of 4 kg, so its weight (mg) acts vertically downward. The weight force can be calculated using the formula F = mg, where g is the acceleration due to gravity (approximately 9.8 m/s²). Therefore, the weight force (W) is 4 kg * 9.8 m/s² = 39.2 N.
2. Tension in Strings: The rod is suspended from the fixed roof by means of two strings, OA and O'C. Since the rod is in a state of equilibrium, the tension in the strings acts in opposite directions and maintains the rod's horizontal position. Let's calculate the tension in each string:
a) String OA: Since the rod is symmetrically suspended at point A, the tension in string OA can be determined by considering the vertical components of the forces acting on the rod (W and the tension in OC). Since the rod is in equilibrium, the vertical component of the tension in string OA must balance the weight force. Therefore, the tension in string OA is equal to the weight force, which is 39.2 N.
b) String O'C: The tension in string O'C can be determined by analyzing the horizontal components of the forces acting on the rod. The vertical component of the tension in O'C must be equal in magnitude to the weight force (39.2 N), as explained earlier. Considering the right triangle formed by the rod, we can use trigonometry to find the horizontal component of the tension in O'C.
The length of O'C is given as 20 cm. The horizontal component of the tension in O'C is equal to that of the tension in OA since the rod is in equilibrium. Let's call this horizontal component T:
T = Tension in O'C (horizontal component)
T = Tension in OA (horizontal component)
Applying the Pythagorean theorem to the right triangle O'BC, we have:
(OC)² = (O'C)² + (BC)²
(20 cm)² = (O'C)² + (100 cm)²
400 cm² = (O'C)² + 10,000 cm²
(O'C)² = 400 cm² - 10,000 cm²
(O'C)² = -9,600 cm²
Since we obtain a negative value, it means there is no real solution for (O'C)². This implies that the horizontal component of the tension in string O'C is zero.
Therefore, the inventory of forces acting on the rod AB is as follows:
- Weight: 39.2 N (acting downwards)
- Tension in string OA: 39.2 N (acting upwards)
- Tension in string O'C: 0 N (no horizontal component)