Pulleys

Finding acceleration

If these two particles in the diagram below are released from rest, what will the acceleration be?

Answer:

The particle with the highest mass will drop, and the particle with the lowest mass will rise. Let's take the particle of 5kg mass as particle a, and the 12kg mass as particle b.

To clarify the weight of each particle we have to multiply their mass with gravity. We therefore use g = 9.8 m / s².

Weight of a = 5g

Weight of b = 12g

Now you can model an equation for each particle's acceleration and tension.

\(T - 5kg \cdot g = 5kg \cdot \text{ a [Particle a] [Equation 1]}\)

\(12kg \cdot g - T = 12kg \cdot \text{ a [Particle b] [Equation 2]}\)

You now solve this simultaneously. Add both equations to eliminate the T variable.

\(7kg \cdot g = 17kg \cdot a\)

If you take g = 9.8 m / s².

\(a = 4.0 m/s^2\)

Investigating two possibilities

Two particles of mass 8 kg and m kg are connected by a tight string passing over a smooth peg. Both particles hang vertically with one particle held at rest. The particle is released. Given that the acceleration is 5 m/s², find the mass m.

Answer:

Let us draw a diagram to suit the question.

Let us take the particle with the mass of 8kg as particle a, and the particle with unknown mass as b.

For all of this to work, the mass of particle b is either greater or less than particle a. This will determine which particle will accelerate. So we may have to investigate both possibilities.

So let's have a look at the situation where m > 8.

Resolving particle a vertically:

\(T - 8kg \cdot g = 8 kg \cdot 5 m/s^2 \qquad T = 40 N + 8 kg \cdot (9.8 m/s^2) \qquad T = 118.4N\)

Resolving particle b vertically:

\(m = \frac{118.4 N}{g - 5 m/s^2} = 24.7 kg\)

That would be the mass of particle b in a case where m > 8.

Let us now take the situation where m < 8

\(8kg \cdot g - D = 8kg \cdot 5m/s^2\)

T = 38.4N

Resolving particle b vertically:

\(m = \frac{38.4N}{5 m/s^2 + g}\)

m = 2.6 kg

We now have a mass for both scenarios. If the mass of particle a> b, b will accelerate upwards while a accelerates downwards. On the other hand, a < b will mean a will accelerate upwards while b accelerates downwards.

SUVAT

The diagram shows two particles that are connected by a light inextensible string. The 5kg particle a is on a rough horizontal surface and the 3 kg particle b hangs on the other end of the string. The string passes over a smooth light pulley. The initial speed of both particles is 2 ms^-1 and a constant frictional force of 4N works against a. The particles slow down and stop before a reaches the pulley or b hits the fall. Find:

• The acceleration of the particles.

• The distance the particle travels before they come to rest.

Figure 7. Pulley example on SUVAT

Answer:

1. \(T -4g = 5a \text{ [Equation 1] [Particle a resolved horizontally]}\)

\(3g - T = 3a \text{ [Equation 2] [Particle b resolved vertically]}\)

Adding equations together

\(-g = 8a\)

\(a = \frac{-g}{8}\)

Take g to be 9.8 ms^-2

\(a = -1.225 ms^{-2}\)

s = x

u = 2 ms^-1

v = 0

a = -1.225 ms^-2

t =?

We will use the equation that doesn't have t since information on that isn't available.

\(v^2 = u^2 + 2as\)

\(0 = 4ms^{-1} - 2 \cdot 1.225 ms^{-2}\)

\(s = \frac{4ms^{-1}}{2 \cdot 1.225 ms^{-2}}\)

\(s = 1.6 m\)