1. Calculate the moment of a 45 N force turning about an axis 3 m away.
Solution M = Fd ⇒ M = (45)(3) = 135 N m
2. Calculate the sum of moments acting on the metre stick in the diagram below.
50 cm 40 cm 28 N FIG. 6.9 Solution
Clockwise moments: (10)(0.25) + (6)(0.5) = 5.5 N m
Anticlockwise moments: (28)(0.1) = 2.8 N m H
Resultant moment: 5.5 – 2.8 = 2.7 N m (clockwise direction)
3. If the metre stick below is in equilibrium, calculate the missing weight.
45 cm 0 cm 7 N FIG. 6.10 Solution Let the missing weight be x.
Clockwise moments = Anticlockwise moments (in equilibrium) (8)(0.1) + (x)(0.55) = (5)(0.2) + (7)(0.45) 0.8 + 0.55x = 1 + 3.15 0.55x = 3.35 x = 6.09 N
You may now complete Exercise 6A (page 61). 60 FUSION 25 cm 5 N 55 cm 8 N 100 cm ?
man Q
FIG. 6.12
Length of pole = 2 m, weight of pole = 100 N Let weight supported by man = Q Let weight supported by woman = 3Q Let distance from man to 800 N weight = x Take moments about the position of the man: (100)(1) + (800)(x) = (3Q)(2) ⇒ 100 + 800x = 6Q
If upward forces = downward forces, 4Q = 900 N and Q = 225 N 100 + 800x = 6(225) 800x = 1 250 ⇒ x = 1.5625 m
Therefore the 800 N weight is hanging 1.5625 m from the man.
75 cm 100 cm 10 N 6 N
FIG. 6.11 12 N
Solution
Clockwise moments = Anticlockwise moments (in equilibrium)
5. Where should you hang an 800 N weight on a 2 m uniform cylindrical pole of weight 100 N, in order that a man at one end of the pole supports one-third as much weight as the woman?
Solution 2 m x 100 N 800 N (2 – x) 3Q woman 8 N 4 N
4. In the diagram below the metre stick is in equilibrium. Calculate the missing distance x.
