Strings, pulleys, and inclines

Consider a block of mass $m$ which is suspended from a fixed beam by means of a string, as shown in Fig. 26. The string is assumed to be light (i.e., its mass is negligible compared to that of the block) and inextensible (i.e., its length increases by a negligible amount because of the weight of the block). The string is clearly being stretched, since it is being pulled at both ends by the block and the beam. Furthermore, the string must be being pulled by oppositely directed forces of the same magnitude, otherwise it would accelerate greatly (given that it has negligible inertia). By Newton's third law, the string exerts oppositely directed forces of equal magnitude, $T$ (say), on both the block and the beam. These forces act so as to oppose the stretching of the string: i.e., the beam experiences a downward force of magnitude $T$, whereas the block experiences an upward force of magnitude $T$. Here, $T$ is termed the tension of the string. Since $T$ is a force, it is measured in newtons. Note that, unlike a coiled spring, a string can never possess a negative tension, since this would imply that the string is trying to push its supports apart, rather than pull them together.

Figure 26: Block suspended by a string

Let us apply Newton's second law to the block. The mass of the block is $m$, and its acceleration is zero, since the block is assumed to be in equilibrium. The block is subject to two forces, a downward force $m g$ due to gravity, and an upward force $T$ due to the tension of the string. It follows that

$$T - m g = 0.$$ (97)

In other words, in equilibrium, the tension $T$ of the string equals the weight $m g$ of the block.

Figure 27 shows a slightly more complicated example in which a block of mass $m$ is suspended by three strings. The question is what are the tensions, $T$, $T_1$, and $T_2$, in these strings, assuming that the block is in equilibrium? Using analogous arguments to the previous case, we can easily demonstrate that the tension $T$ in the lowermost string is $m g$. The tensions in the two uppermost strings are obtained by applying Newton's second law of motion to the knot where all three strings meet. See Fig. 28.

Figure 27: Block suspended by three strings

There are three forces acting on the knot: the downward force $T$ due to the tension in the lower string, and the forces $T_1$ and $T_2$ due to the tensions in the upper strings. The latter two forces act along their respective strings, as indicate in the diagram. Since the knot is in equilibrium, the vector sum of all the forces acting on it must be zero.

Consider the horizontal components of the forces acting on the knot. Let components acting to the right be positive, and vice versa. The horizontal component of tension $T$ is zero, since this tension acts straight down. The horizontal component of tension $T_1$ is $T_1 \cos 60^\circ = T_1/2$, since this force subtends an angle of $60^\circ$ with respect to the horizontal (see Fig. 16). Likewise, the horizontal component of tension $T_2$ is $-T_2 \cos 30^\circ = -\sqrt{3} T_2/2$. Since the knot does not accelerate in the horizontal direction, we can equate the sum of these components to zero:

$$\frac{T_1}{2} -\frac{\sqrt{3} T_2}{2} = 0.$$ (98)

Consider the vertical components of the forces acting on the knot. Let components acting upward be positive, and vice versa. The vertical component of tension $T$ is $-T=-m g$, since this tension acts straight down. The vertical component of tension $T_1$ is $T_1 \sin 60^\circ = \sqrt{3} T_1/2$, since this force subtends an angle of $60^\circ$ with respect to the horizontal (see Fig. 16). Likewise, the vertical component of tension $T_2$ is $T_2 \sin 30^\circ = T_2/2$. Since the knot does not accelerate in the vertical direction, we can equate the sum of these components to zero:

$$-m g+\frac{\sqrt{3} T_1}{2} + \frac{T_2}{2} = 0.$$ (99)

Finally, Eqs. (98) and (99) yield

$$T_1 = \frac{\sqrt{3} m g}{2},$$ (100)

$$T_2 = \frac{m g}{2}.$$ (101)

Figure 28: Detail of Fig. 27

Consider a block of mass $m$ sliding down a smooth frictionless incline which subtends an angle $\theta$ to the horizontal, as shown in Fig 29. The weight $m g$ of the block is directed vertically downwards. However, this force can be resolved into components $m g \cos\theta$, acting perpendicular (or normal) to the incline, and $m g \sin\theta$, acting parallel to the incline. Note that the reaction of the incline to the weight of the block acts normal to the incline, and only matches the normal component of the weight: i.e., it is of magnitude $m g \cos\theta$. This is a general result: the reaction of any unyielding surface is always locally normal to that surface, directed outwards (away from the surface), and matches the normal component of any inward force applied to the surface. The block is clearly in equilibrium in the direction normal to the incline, since the normal component of the block's weight is balanced by the reaction of the incline. However, the block is subject to the unbalanced force $m g \sin\theta$ in the direction parallel to the incline, and, therefore, accelerates down the slope. Applying Newton's second law to this problem (with the coordinates shown in the figure), we obtain

$$m \frac{d^2 x}{dt^2} = m g \sin\theta,$$ (102)

which can be solved to give

$$x = x_0 + v_0 t + \frac{1}{2} g \sin\theta t^2.$$ (103)

In other words, the block accelerates down the slope with acceleration $g \sin\theta$. Note that this acceleration is less than the full acceleration due to gravity, $g$. In fact, if the incline is fairly gentle (i.e., if $\theta$ is small) then the acceleration of the block can be made much less than $g$. This was the technique used by Galileo in his pioneering studies of motion under gravity--by diluting the acceleration due to gravity, using inclined planes, he was able to obtain motion sufficiently slow for him to make accurate measurements using the crude time-keeping devices available in the 17th Century.

Figure 29: Block sliding down an incline

Consider two masses, $m_1$ and $m_2$, connected by a light inextensible string. Suppose that the first mass slides over a smooth, frictionless, horizontal table, whilst the second is suspended over the edge of the table by means of a light frictionless pulley. See Fig. 30. Since the pulley is light, we can neglect its rotational inertia in our analysis. Moreover, no force is required to turn a frictionless pulley, so we can assume that the tension $T$ of the string is the same on either side of the pulley. Let us apply Newton's second law of motion to each mass in turn. The first mass is subject to a downward force $m_1 g$, due to gravity. However, this force is completely canceled out by the upward reaction force due to the table. The mass $m_1$ is also subject to a horizontal force $T$, due to the tension in the string, which causes it to move rightwards with acceleration

$$a = \frac{T}{m_1}.$$ (104)

The second mass is subject to a downward force $m_2 g$, due to gravity, plus an upward force $T$ due to the tension in the string. These forces cause the mass to move downwards with acceleration

$$a =g - \frac{T}{m_2}.$$ (105)

Now, the rightward acceleration of the first mass must match the downward acceleration of the second, since the string which connects them is inextensible. Thus, equating the previous two expressions, we obtain

$$T = \frac{m_1 m_2}{m_1+m_2} g,$$ (106)

$$a = \frac{m_2}{m_1+m_2} g.$$ (107)

Note that the acceleration of the two coupled masses is less than the full acceleration due to gravity, $g$, since the first mass contributes to the inertia of the system, but does not contribute to the downward gravitational force which sets the system in motion.

Figure 30: Block sliding over a smooth table, pulled by a second block

Consider two masses, $m_1$ and $m_2$, connected by a light inextensible string which is suspended from a light frictionless pulley, as shown in Fig. 31. Let us again apply Newton's second law to each mass in turn. Without being given the values of $m_1$ and $m_2$, we cannot determine beforehand which mass is going to move upwards. Let us assume that mass $m_1$ is going to move upwards: if we are wrong in this assumption then we will simply obtain a negative acceleration for this mass. The first mass is subject to an upward force $T$, due to the tension in the string, and a downward force $m_1 g$, due to gravity. These forces cause the mass to move upwards with acceleration

$$a = \frac{T}{m_1} - g.$$ (108)

The second mass is subject to a downward force $m_2 g$, due to gravity, and an upward force $T$, due to the tension in the string. These forces cause the mass to move downward with acceleration

$$a =g - \frac{T}{m_2}.$$ (109)

Now, the upward acceleration of the first mass must match the downward acceleration of the second, since they are connected by an inextensible string. Hence, equating the previous two expressions, we obtain

$$T = \frac{2 m_1 m_2}{m_1+m_2} g,$$ (110)

$$a = \frac{m_2-m_1}{m_1+m_2} g.$$ (111)

As expected, the first mass accelerates upward (i.e., $a>0$) if $m_2>m_1$, and vice versa. Note that the acceleration of the system is less than the full acceleration due to gravity, $g$, since both masses contribute to the inertia of the system, but their weights partially cancel one another out. In particular, if the two masses are almost equal then the acceleration of the system becomes very much less than $g$.

Figure 31: An Atwood machine

Incidentally, the device pictured in Fig. 31 is called an Atwood machine, after the eighteenth Century English scientist George Atwood, who used it to ``slow down'' free-fall sufficiently to make accurate observations of this phenomena using the primitive time-keeping devices available in his day.