Mechanics Problem 1: Pulley System

I want to start a blog series which focuses on applications of mechanics. I have been giving a math-physics workshop to high school teachers in the past two months. It is a problem-driven workshop, where concepts are discussed through problems. I am turning the concept of learning around. Usually, concepts are laid out first - laboriously - and then examples are given.

I find through this workshop that by working out problems my students - the high school teachers - understand the concepts better. I want to share this idea with you. If you find this blog series interesting, give me a shout. Thank you.

The diagram above shows a sytem of two pulleys, which is used to lift block B. The person who uses this pulley system exerts a force P to lift the block B. The two pulleys are connected by the blue cable that runs from A to C to D ... E, F and all the way to G.

Through mechanics analysis, you will find that this pulley system makes lifting B easier. P is only 50% of the weight of B. Directly lifting weight B without pulleys would mean requiring exerting a 100% of the weight B.

Our analysis relies on a formula called Newton's laws of motion, in particular the so-called Newton's second law. It states that an acceleration a of a body is equal to the sum of all forces T acting on the body divided by its mass m. That is, T = m a.

1. Let us apply T = m a to our problem. The blue cable and pulleys are assumed massless and rigid, so that the force P is transferred to the block B without dissipation. The cord that hangs the block B to the lower pulley transfers a force 2P since the cable wraps around the lower pulley and each cable section carries a force P. T = m a for the block B thus gives

2P - mg = m a.

The weight of block B, i.e., m g, is directed downward, while the force 2P is directed upward. This is the reason why they have opposite signs in the previous equation. Thus,

P = m (a + g)/2.

When the block B starts to be lifted, its acceleration a is not zero. But when it moves upward with a constant speed, a = 0. At this situation, we have P = m g/2. This shows that P equals 50% of the B's weight.

2. The pulley system thus provides a mechanical advantage. To lift B, one pulls only 50% of its weight. How is that possible? Is there a catch? The answer is yes. The block B moves 200% slower than it would if it were lifted directly without pulleys. You will spend the same amount of energy to lift the block over a certain distance regardless whether you exert a force 50% of its weight, or a 100%. The energy you spend per second is lower due to the slower upward motion of B, which makes the lifting work easier.

The key to understand this aspect is that the cable portion connected to the force P travels twice as far as the portion connected to the block. That is, x_A = -2x_B, where the negative sign means A and B move in opposite directions. This fact comes straight from the fact that the length of the cable is constant

x_A + 2x_B = constant,

so that by differentiating with time we get their velocity relation,

v_A + 2v_B = 0,

and another differentiation yields their acceleration relation,

a_A + 2a_B = 0.

These kinematic relations are constraints to the motion and arise independently of the dynamics dictated by the Newton's second law.

3. The work done is equal to force times distance travelled. Thus, when the block B moves at a constant speed

Work = P x_A = (m g/2)(2 x_B) = m g x_B.

This shows that the work needed to lift the block B remains the same: m g x_B. The conservation of energy is still obeyed. The beauty of the pulley system, however, is that it makes lifting it easier since per second the force needed is only 50%. The trade off is the distance travelled at A becomes twice as long.