4 NEWTON’S LAWS OF MOTION 4.7 Strings, pulleys, and inclines(^)
T 2
30 o
T 1
60 o
T
Figure 28: Detail of Fig. 27
tension T is −T = −m g, since this t√ension acts straight down. The vertical com-
ponent of tension T 1 is T 1 sin 60 ◦ = 3 T 1 /2, since this force subtends an angle of
60 ◦ with respect to the horizontal (see Fig. 16 ). Likewise, the vertical component
of tension T 2 is T 2 sin 30 ◦ = 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 +
√
3 T T
(^1) + 2
= 0. (4.9)
2 2
Finally, Eqs. (4.8) and (4.9) yield
T 1 =
√
3 m g
m g^2
, (4.10)
T 2 =.^ (4.11)^
2
Consider a block of mass m sliding down a smooth frictionless incline which
subtends an angle θ 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 θ, acting perpendicular (or normal) to the incline, and
m g sin θ, 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 θ. 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