The figure below 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 80 N on the wall to which it is attached. The tensions in the shorter cords are T1 = 45.0 N, T2 = 37.3 N, and T3 = 7.7 N.

What are the masses of each disk?
(a) kg (disk A)

(b) kg (disk B)

(c) kg (disk C)

(d) kg (disk D)

There is zero probability you can get help on this without a diagram.

_80N________

||
||
( )A
||
T1 ||
( )B
||
T2 ||
( )C
||
T3 ||
( )D

I believe that is the diagram you are looking for. I need help on this one too.

_80N________

||
||
( )A
||
||T1
( )B
||
||T2
( )C
||
||T3
( )D

I believe that is the diagram you are looking for. I need help on this one too. I revised the diagram.

To determine the masses of each disk, we can use Newton's second law and analyze the forces acting on each disk.

Let's denote the mass of disk A as "mA," disk B as "mB," disk C as "mC," and disk D as "mD."

For disk A:
- The tension T1 is pulling upwards, so it has the same magnitude as the weight of disk A: T1 = mAg, where "g" is the acceleration due to gravity.
- Solving for mA: mA = T1 / g = 45.0 N / 9.8 m/s^2 = 4.59 kg.

For disk B:
- We need to consider the net force acting on it. The tension T2 is pulling downwards, and the weight of disk B is pulling downwards as well.
- The net force is given by: T2 - mBg = mBa, where "a" is the acceleration of disk B.
- Since the pulley is frictionless and disk B is connected to the pulley, the acceleration of disk B is the same as the acceleration of the system.
- Solving for mB: mB = (T2 - mBa) / g = (37.3 N - mBa) / 9.8 m/s^2.

For disk C:
- Similarly, we consider the net force and acceleration of disk C.
- We have: T2 - T3 - mCg = mCa.
- Solving for mC: mC = (T2 - T3 - mCa) / g = (37.3 N - 7.7 N - mCa) / 9.8 m/s^2.

For disk D:
- Again, we consider the net force and acceleration.
- We have: 80 N - T1 - mDg = mDa.
- Solving for mD: mD = (80 N - T1 - mDa) / g = (80 N - 45.0 N - mDa) / 9.8 m/s^2.

Now, we have four equations with four unknowns (mA, mB, mC, mD). We can solve these equations simultaneously to find the mass of each disk.

Note: The values of T1, T2, T3, and g are given in the problem statement.

(a) To find the mass of disk A (mA):
Substitute the given value of T1 (45.0 N) into the equation:
mA = T1 / g = 45.0 N / 9.8 m/s^2 = 4.59 kg.

(b) To find the mass of disk B (mB):
Substitute the given value of T2 (37.3 N) into the equation:
mB = (T2 - mBa) / g = (37.3 N - mBa) / 9.8 m/s^2.

(c) To find the mass of disk C (mC):
Substitute the given values of T2 (37.3 N) and T3 (7.7 N) into the equation:
mC = (37.3 N - 7.7 N - mCa) / 9.8 m/s^2.

(d) To find the mass of disk D (mD):
Substitute the given value of T1 (45.0 N) into the equation:
mD = (80 N - 45.0 N - mDa) / 9.8 m/s^2.

Now, you can substitute the given values of T1, T2, T3 into each equation and solve them simultaneously to find the masses of disks B, C, and D.