Strings, pulleys, and inclines

Let us apply Newton's second law to the block. The mass of the block is , and
its acceleration is zero, since the block is assumed to be in equilibrium.
The block is subject to two forces, a downward force due to gravity, and
an upward force due to the tension of the string. It follows that

(97) |

Figure 27 shows a slightly more complicated example in which a block of mass is suspended by three strings. The question is what are the tensions, , , and , in these strings, assuming that the block is in equilibrium? Using analogous arguments to the previous case, we can easily demonstrate that the tension in the lowermost string is . 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.

There are three forces acting on the knot: the downward force due to the tension in the lower string, and the forces and 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 is
zero, since this tension acts straight down. The horizontal
component of tension is
, since this
force subtends an angle of with respect to the horizontal (see Fig. 16).
Likewise, the
horizontal component of tension is
.
Since the knot does not accelerate in the horizontal direction, we can equate
the sum of these components to zero:

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 is
, since this tension acts straight down. The vertical
component of tension is
, since this
force subtends an angle of with respect to the horizontal (see Fig. 16).
Likewise, the
vertical component of tension is
.
Since the knot does not accelerate in the vertical direction, we can equate
the sum of these components to zero:

Finally, Eqs. (98) and (99) yield

(100) | |||

(101) |

Consider a block of mass sliding down a smooth frictionless incline
which subtends an angle to the horizontal, as shown in Fig 29.
The weight of the block is directed vertically downwards. However,
this force can be resolved into components
, acting
perpendicular (or normal) to the incline, and
,
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
. 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
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

(102) |

(103) |

Consider two masses, and , 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
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 , due to gravity. However, this force
is completely canceled out by the upward reaction force due to the table.
The mass is also subject to a horizontal force , due to
the tension in the string, which causes it to move *rightwards*
with acceleration

(104) |

(105) |

(106) | |||

(107) |

Note that the acceleration of the two coupled masses is

Consider two masses, and , 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 and , we cannot determine beforehand which mass is going to
move upwards. Let us *assume* that mass 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 , due to the tension in the string, and a downward
force , due to gravity. These forces cause the mass to
move *upwards* with acceleration

(108) |

(109) |

(110) | |||

(111) |

As expected, the first mass accelerates upward (

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.