I am trying to independently solve physics problems and I am attempting to solve [Chapter 5 problem 4-4 on page 140](https://bayanbox.ir/view/6682075492227836301/Feynman-Tips-on-Physics.pdf) from the book "Feynman's Tips on Physics".

The problem is as follows:

A painter weighing 180 lbs working from a "bosun's" chair hung down the side of a tall building desires to move in a hurry. He pulls down on a fall rope with such a force that he presses against the chair with only a force of 100 lb. The chair itself weighs 30 lb.

(a) What is the acceleration of the painter and the chair?

(b) What is the total force supported by the pulley?

The diagram is as follows:

[![enter image description here][1]][1]

**MY WORK:**

(a) The question states the **weights** of the painter and the chair are 180 lbs and 30 lbs respectively. I used this idea to perform the following step:

$$\text {Total mass of the painter and the chair} = \left(\frac{(180 + 30) \rm lbs}{g}\right) $$

He exerts a **downward** force of 100 lb on the chair.
His net motion will be **upwards**.

So the equation I formed is:

$\text{Total force from the upwards motion = Weight of man + Weight of chair + Force on the chair}$

$\frac{180 + 30}{g}\times a = $180 lb + 30 lb + 100 lb

This is giving me:

a = $\frac{31 g}{21} \frac {m}{s^2}$

**This answer does not match the answer in the textbook. I believe I have accounted for all the relevant forces acting on the free-body diagram in question. Are there any additional forces I am not accounting for?**

(b)

According to my understanding, the total force supported by the pulley is zero. My reasoning for this is as follows:

The net downwards force is $\text{Weight of man + Weight of chair + Force exerted on the chair}$ which is $\text{310 lbs}$. I am compelled to think that an **equal and opposite upwards force** of 310 lbs will be exerted on the pulley.

The tension in the pulley string came to my mind but it gets cancelled out. The 100 lbs downwards force on one side will exert an equal and opposite upwards force on the other side. Consequently, the resultant force on the string will be zero.

**This answer also does not match the answer in the textbook. Are there any forces I am forgetting to take into account or am I just not adding my forces properly?**


  [1]: https://i.sstatic.net/ZfP4b.jpg