Question 4:
4 | (a) | Horizontal component of F is 0 | B1 | 1.1 | This mark can be implied if direction is stated as
vertical (oe)
e.g. 2 c o s 4 c o s 3 c o s 5 c o s ( 2 c o s )      + − − = −
e.g. 12 +12 −2(1)(1)cos(2(90−)) | M1 | 3.3 | Resolving vertically (oe) (at least four terms required)
where is the angle between AB and the upward
vertical – condone if trig. terms are not consistent (but
all four must have been resolved)
3
c o s  =  vertical component of F is 1.2
5 | A1 | 1.1 | Sight of  1 .2
Magnitude of F is 1.2
Direction of F is in the same direction as AC | A1 | 2.2a | Magnitude must be positive – allow ‘vertically
upwards’ or ‘upwards’ only (oe) for direction
(‘vertically’ only though is A0, but may be clarified
with a diagram)
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(b) | e.g. moments about A: G = 6 ( 3 s i n ) 6 ( 4 s i n )   +
or 3 ( 4 s i n ) 4 ( 4 c o s ) 3 ( 6 s i n )    + +
or 3 ( 3 s i n ) 4 ( 3 c o s ) 3 ( 4 s i n ) 4 ( 4 c o s )     + + +
or e.g. moments about O:
3 ( 3 s i n ) 3 ( 2 s i n ) 4 ( 4 c o s ) 4 ( 5 c o s )     + + +
or 4 ( 4 2 ) c o s 4 ( 3 5 ) c o s   + + + | B1 | 3.1b | Taking moments (resolved force distance (oe)) with
correct terms for their chosen point (note that if their
answer to (a) is correct then taking moments about D
or B, their F must be included too).
A correct answer implies this mark
4
G=42sin=42 =33.6
5 | B1 | 1.1 | No direction required but must be positive (seen
anywhere in their solution so ISW if 33.6 is then
stated as –33.6)
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