Question 13:
13 | (a) | 4aR sinaTcos
C
aTcos2aWsin 4ax 4Wsin
4aR sin
B
R R 4WW
B C
aTcos18aWsin4Wxsin Tcos
 5W
4asin 4sin
2aTcos2aWsin4Wxsin
 2x
T W1 tan
 a  | M1*
A1
M1*
A1
B1
M1dep*
A1
[7] | 3.3
1.1
3.3
1.1
3.3
2.1
2.2a | Moments about A for AC – correct
number of terms – allow sin/cos
confusion and sign errors
Moments about A for AB – correct
number of terms – allow sin/cos
confusion and sign errors
Resolving vertically
Correct method to eliminate R and R
B C
to obtain a linear equation in T
AG | R is the normal
C
contact force at C
R is the normal
B
contact force at B
Dependent on all
previous M marks and
B mark
Sufficient working must
be shown as AG
13 | (b) | DE2 a2a22(a)(a)cos2
T  W  DE0.25a or
0.25a
T  W  0.5DE0.125a
0.125a
 2x 4W 1 
W1 tan  2a2 2a2cos2 a
 a  a  4 
 2x 4 1 
1 tan 2asin a
 a  a 4 
x1a 8coscot1 
2 | B1
M1
A1
M1
A1
[5] | 3.1a
3.3
1.1
3.1a
2.2a | Correct application of cosine rule for
triangle DAE
Correct application of Hooke’s law for
their extension
Or A2 (rather than the next M) for
 2x 4W 1 
W1 tan 2asin a
 a  a  4 
 2x 8W 1 
or W1 tan asin a
 a  a  8 
Correct use of double-angle formula to
give an equation in terms of x, a and

AG | B1 for DE2asinor
1DE asin
2
Condone x or e for
extension
AG so sufficient
working must be shown
dx a 
 8sincosec2 0
d 2
1 
sin3 
8 6
1    
x a8cos cot 1
 
2  6 6 
3 31
x a
2
d2x a    
 8cos  2cosec2  cot 
d2 2 6  6 6

6
6a 3< 0 therefore maximum | M1*
A1
M1dep*
A1
A1
[5] | 1.1
1.1
3.4
1.1
3.2a | Attempt differentiation and set equal
to zero – must be of the form
8sincosec2with no extra terms
www
Substitute their value of  and obtain
a value for x
2.10a (allow 2.1a)
Correctly showing that value of x is a
maximum (condone in terms of
degrees)
10.39...a | Allow in degrees
throughout (30)
For reference:
2.098076…a
Or equivalent
justification that this
value is a maximum
PPMMTT
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