The figure shows an arrangement in which four disks are suspended by cords. The longer, top cord loops over a frictionless pulley and pulls with a force of magnitude 82.6 N on the wall to which it is attached. The tensions in the shorter cords are T1 = 61.4 N, T2 = 47.2 N, and T3 = 9.5 N. What are the masses of (a) disk A, (b) disk B, (c) disk C, and (d) disk D?
Tjank
To find the masses of the disks, we can use Newton's second law, which states that the net force acting on an object is equal to the product of its mass and acceleration.
Let's denote the mass of disk A as m_A, the mass of disk B as m_B, the mass of disk C as m_C, and the mass of disk D as m_D.
First, let's consider the forces acting on each disk:
For disk A:
- The tension in cord T1 pulls upwards with a force of magnitude T1.
- The weight of disk A pulls downward with a force of magnitude m_A * g, where g is the acceleration due to gravity (approximately 9.8 m/s^2).
Using Newton's second law, we can write the equation for disk A:
T1 - m_A * g = 0 (since the acceleration of disk A is zero, as it is in equilibrium)
For disk B:
- The tension in cord T1 pulls upwards with a force of magnitude T1.
- The tension in cord T2 pulls upwards with a force of magnitude T2.
- The weight of disk B pulls downward with a force of magnitude m_B * g.
The equation for disk B is:
T1 + T2 - m_B * g = 0
For disk C:
- The tension in cord T2 pulls upwards with a force of magnitude T2.
- The tension in cord T3 pulls upwards with a force of magnitude T3.
- The weight of disk C pulls downward with a force of magnitude m_C * g.
The equation for disk C is:
T2 + T3 - m_C * g = 0
For disk D:
- The tension in cord T3 pulls upwards with a force of magnitude T3.
- The weight of disk D pulls downward with a force of magnitude m_D * g.
The equation for disk D is:
T3 - m_D * g = 0
Now, let's solve these equations one by one:
From the equation for disk A:
T1 - m_A * g = 0
m_A = T1 / g
From the equation for disk C:
T2 + T3 - m_C * g = 0
m_C = (T2 + T3) / g
From the equation for disk D:
T3 - m_D * g = 0
m_D = T3 / g
Now, let's substitute the given values to find the masses:
m_A = T1 / g = 61.4 N / 9.8 m/s^2 ≈ 6.27 kg
m_C = (T2 + T3) / g = (47.2 N + 9.5 N) / 9.8 m/s^2 ≈ 5.71 kg
m_D = T3 / g = 9.5 N / 9.8 m/s^2 ≈ 0.97 kg
To find the mass of disk B, we need to use the equation for disk B:
T1 + T2 - m_B * g = 0
m_B = (T1 + T2) / g = (61.4 N + 47.2 N) / 9.8 m/s^2 ≈ 11.75 kg
Therefore, the masses of the disks are:
(a) Disk A: Approximately 6.27 kg
(b) Disk B: Approximately 11.75 kg
(c) Disk C: Approximately 5.71 kg
(d) Disk D: Approximately 0.97 kg