A 1.50 kg object is supported by two inextensible cords that are 6.0 cm and 8.0 cm long. These cords are attached to two nails A and B that are 6.0 cm apart. Find the tension in each cord.
To find the tension in each cord, we need to analyze the forces acting on the object. Since the object is in equilibrium, the net force acting on it must be zero.
Let's break down the forces:
1. Weight (mg): The force due to gravity acting vertically downward can be calculated by multiplying the mass (m) by the acceleration due to gravity (g). In this case, the weight exerted by the object is given as (1.50 kg)(9.8 m/s^2) = 14.7 N.
2. Tension in cord AC: Let's assume the tension in cord AC is T1. Since the cords are inextensible and attached to nails A and B, the tension in cord AC will be acting along the line AB. We can resolve the tension in cord AC into horizontal and vertical components. Considering the vertical components, we have T1 * sin(Θ1) acting upward for the vertical equilibrium. Since the object is in equilibrium, the sum of the vertical components of forces must zero. Thus, T1 * sin(Θ1) = mg/2 (assuming the angles formed between the cords and the horizontal are equal).
3. Tension in cord BC: Let's assume the tension in cord BC is T2. Similar to the tension in cord AC, we can resolve the tension in cord BC into horizontal and vertical components. Considering the vertical components, we have T2 * sin(Θ2) acting upward for the vertical equilibrium. Since the object is in equilibrium, the sum of the vertical components of forces must zero. Thus, T2 * sin(Θ2) = mg/2 (assuming the angles formed between the cords and the horizontal are equal).
Now, we have two equations:
T1 * sin(Θ1) = mg/2 ----(1)
T2 * sin(Θ2) = mg/2 ----(2)
To proceed, we need to find the values of sin(Θ1) and sin(Θ2).
Using trigonometry and the given lengths, we can find the values of sin(Θ1) and sin(Θ2):
sin(Θ1) = (length of AC) / (distance between A and B).
= 6.0 cm / 6.0 cm (cm cancel out)
= 1
sin(Θ2) = (length of BC) / (distance between A and B).
= 8.0 cm / 6.0 cm
= 4/3
Substitute these values into equations (1) and (2):
T1 * 1 = mg/2 ----(3)
T2 * (4/3) = mg/2 ----(4)
Divide equation (4) by equation (3) to eliminate the mg terms:
(T2 * (4/3)) / (T1 * 1) = (mg/2) / (mg/2)
3T2 / 4T1 = 1
T2 / T1 = 4/3
From this equation, we know that the ratio of T2 to T1 is 4:3.
Now, let's sum up the vertical components of the forces to solve the tension T1:
T1 * sin(Θ1) = mg/2
T1 * 1 = mg/2
T1 = (mg/2) / 1
T1 = (1.50 kg)(9.8 m/s^2) / 2
T1 = 7.35 N
Finally, to find the tension T2, we can use the ratio we found earlier:
T2 / T1 = 4/3
T2 = (4/3) * T1
T2 = (4/3) * 7.35 N
T2 = 9.80 N
Therefore, the tension in cord AC (T1) is approximately 7.35 N, and the tension in cord BC (T2) is approximately 9.80 N.