Forces and equilibrium - Orthopaedic Biomaterials in Research and Practice - page 18

vector as its diameter; any two vectors that meet on the circumference
of such a circle are orthogonal components o f the v ec tor serving as the
diameter. Lay out the lines of action of − R and

R and measure R.

R = 35 N.
Numeric. From Figure 1.8:

Y h = 49 × cos 15° = 47.3 N

Y v = 9.8 + 29.4 = 39.2 N

Y = 61.4 N

tan -1 ( Y v / Y h ) = tan -1 (39.2/47.3) = 39.6°

Therefore, R = Y cos (39.6° + 15°) = 35.6 N

PROBLEM 1.8
The wheel in Figure 1.13a is mounted in an axle at P and loaded, through
two cables, at A by a 3 kg mass and at B by a 5 kg mass. What is the
resultant moment at P ? If the wheel is free to rotate on its axle, in which
direction does it turn?

ANSWER:
Graphic. Construct moment “boxes” on the lines from A to P (3 ×
9.8 N × 0.15 m) and from B to P (5 × 9.8 N × 0.1 m) (Figure 1.13b).
Partition up into units of 0.25 N ∙ m. Then the force at A contributes
about −18 squares and the one at B about +20 squares. Therefore, the net
moment is +2 × 0.25 or +0.5 N ∙ m (i.e., counterclockwise rotation).

29.4 N

(b)

0.15 m

(a)

M A at P = −4.41 Nm
( ~ 18 × 0.25 Nm)

49 N

+ P

0.1 m

B

3 kg M B at P = +4.9 Nm

( ~ 20 × 0.25 Nm)

Net M = +0.49 Nm
(CCW rotation)

1 cm = 10 N
1 cm = 10 cm
= 29.4 N

5 kg

FIGUre 1.13 Wheel and cables (Problem 1.8).