Page 164 - Schaum's Outline of Theory and Problems of Applied Physics

CHAP. 12] SIMPLE MACHINES 149

hoist are 14 and 15 cm in radius and the hoist is 75 percent efficient, how heavy a load can be raised by a
force of 150 N? (c) If the load is to be raised by 1 m, how much chain must be pulled through the pulleys?

(a) The radius of the large upper pulley is R and that of the small one is r. In a complete rotation of the upper
pulleys, the input force F in acts through a distance equal to the circumference of the large pulley, so s in = 2π R.
In this rotation, strand 1 has been shortened by 2π R while strand 2 has been lengthened by 2πr, so the total
decrease in length is 2π R − 2πr = 2π(R − r). Since the decrease in length is shared by both strands 1 and 2,
the movable pulley is raised by half this amount and s out = π(R − r). The IMA of the system is accordingly
2π R 2R
s in
IMA = = =
s out π(R − r) R − r
(b) Since R = 15 cm and r = 14 cm, the IMA of the hoist is
2R (2)(15 cm)
IMA = = = 30
R − r 15 cm − 14 cm
At 75 percent efficiency,
AMA = (Eff)(IMA) = (0.75)(30) = 22.5

An applied force of 150 N can therefore lift a load of
F out = (AMA)(F in ) = (22.5)(150 N) = 3375 N

(c) Given that s out = 1 m and IMA = 30, we have
s in = (IMA)(s out ) = (30)(1m) = 30 m

THE LEVER
The IMA of a lever is equal to the ratio between its moment arms L in and L out (Fig. 12-5):

L in
IMA =
L out
This is because, as the lever rotates around its pivot (or fulcrum), the ratio s in /s out between the arcs through which
its ends move is equal to the ratio L in /L out of the moment arms.
The wheel and axle, belt and gear drives, and pulley systems are all developments of the lever.

F out
L in L out

Fulcrum
F in

Fig. 12-5

SOLVED PROBLEM 12.7
One end of a 200-kg crate is to be lifted from the ground by using a 3-m plank as a lever. If the maximum
force that can be applied to the plank is 350 N, where should the fulcrum be placed? Assume that the
crate’s contents are uniformly distributed, so the mass to be raised is 100 kg.
The weight of a 100-kg mass is

2
w = mg = (100 kg)(9.8 m/s ) = 980 N