Question 12:
12 | (a) | R sin30=R cos30
A B
R cos30+R sin30=W
A B
( )
R =R 3⇒ 3 R 3 +R =2W
A B B B
R = 1W
B 2 | M1*
A1
M1*
A1
M1dep*
A1
[6] | 3.1b
1.1
3.3
1.1
3.4
1.1 | Resolve horizontally – correct number of terms, allow
sin/cos confusion (and sign errors) but no θ'spresent
where R and R are the normal contact forces at A
A B
and B respectively
Resolve vertically – correct number of terms, allow
sin/cos confusion (and sign errors) but no θ'spresent
Eliminate R to form an equation in R (or equivalent
A B
to find an expression for R )
B
(b) | 1aW + 1a(Wcosθ)= 1a(Wsinθ)+a (1Wcos(60−θ))
8 2 2 2
( )
1+ 1cosθ= 1sinθ+ 1 1cosθ+ 3sinθ
8 2 2 2 2 2
1+4cosθ=4sinθ+2cosθ+2 3sinθ
( )
⇒2 3+2 sinθ−2cosθ=1 | M1*
A2
M1dep*
A1
[5] | 3.3
1.1
1.1
3.1a
2.2a | Taking moments about A – one couple term, at least
one weight term and at least one term for the contact
force at B – dimensionally correct – condone sin/cos
confusion and allow R for 1W
B 2
A1 for two terms correct
1aW + 1a(Wcosθ)(1−tanθ)=a (1Wcos(60−θ))
8 2 2
( 2cos(θ+45))
1aW + 1a W
8 2
= 1a(Wsin30cosθ+cos30sinθ)
2
Obtain an equation in sinθand cosθonly – weight
component must be equivalent to a term in sinθand a
term in cosθ and the angle of the contact force at B
must be of the formsin(±α±θ)or cos(±α±θ)
where α=30or 60
AG – sufficient working must be shown
(c) | Considers both f(22.35)and f(22.45) where
f(θ)=± 2 ( 3+2 ) sinθ−2cosθ−1 
 
f(22.35)=−0.0114...<0and f(22.45)=0.00194...>0
change of sign indicates that θ=22.4correct to 1 decimal place | M1
A1
[2] | 1.1
2.4 | Working or correct answer for one value is sufficient
evidence of correct method but both 22.35 and 22.45
must be seen
Correct values (to at least 1 sf rot) together with
explanation (change of sign) and conclusion (as a
minimum ‘root’) oe (e.g., comparing 1.00194… and
0.98856… to 1)