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Moment = F x D
= 100N
x 0.3m
= 30Nm
2. This system is balanced
Clockwise moment =
(20N x 3m) + (25N x (3m + 1m))
=
60 Nm + 100 Nm
=
160 Nm
b)
State the size of the anticlockwise moment
Anti-clockwise moment =
X N
x 2m
=
2X m
c)
Calculate the size of force X
Clockwise moment =
Anti-clockwise moment
160 Nm
= 2X
m
160
: 2 = X
80 = X
3. A
pivoted uniform bar is in equilibrium under the action of the forces shown.
Anti -Clockwise moment
= Clockwise moment
10N x 4m = (F x
2m) + (6N x (2m+2m))
40
Nm =
2F m + 24 Nm
40 Nm – 24 Nm = 2F m
16 Nm = 2F m
16 Nm : 2 m = F
8
N = F
4. A uniform beam AB of mass
20kg and length 3m rests in equilibrium in a horizontal position. The beam is
supported by a single pivot at C and a particle of mass 6kg is attached at B.
Assume: AC is X meter, so CB is (3 – X) meter
Anti -Clockwise moment
= Clockwise moment
20 kg
x X = 6 kg
x (3 – X)
20X = 18 – 6X
20X + 6X = 18
26X =
18
X = 18 : 26
X = 0.69 meter
5. a
uniform metre rule, freely pivoted at a point 20 cm from end P.
The rule is kept horizontal by means of a 120g mass
suspended 5.0 cm from end P. Use the
principle of moments to help you determine the mass of the metre rule.
Anti -Clockwise moment
= Clockwise moment
120
x 15 = MASS x 30
1800 = MASS x 30
1800 : 30 = MASS
60 gram = MASS
6. Look at this example of a balanced system
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